SOLUTIONS TO
~ PROBABILITY AND STATISTICS
WITH APPLICATIONS:
A PROBLEM SOLVING TEXT
Leonard A. Asimow, Ph.D., ASA
Mark M. Maxwell, Ph.D., ASA
ACTEX PUBLICATIONS, INC.
WINSTED, CONNECTICUT
SOLUTIONS TO
PROBABILITY AND STATISTICS
WITH APPLICATIONS:
A PROBLEM
SOLVING
TEXT
Leonard A. Asimow, Ph.D., ASA
Mark M. Maxwell, Ph.D., ASA
ACTEX PUBLICATIONS, INC.
WINSTED, CONNECTICUT
Copyright © 2010, by ACTEX Publications, Inc.
All rights reserved. No portion of this book
May be reproduced in any form or by and means
Without the prior written permission of the
Copyright owner.
Requests for permission should be addressed to
ACTEX Publications
PO Box 974
Winsted, CT 06098
Manufactured in the United States of America
109539) 65437221
ISBN:
978-1-56698-722-6
TABLE OF CONTENTS
Chapter 1: Combinatorial Probability 1
amet
ie EL OD aD ymVIOGCle neha cck ystems Ser
ook cceccs ac scrretnsch mites estes tcsscase 1
1.2
Finite Discrete Models with Equally Likely Outcomes ..........c.sccscccssssssssessssessesesessssesesees 2
eM
ATU ara Ec UO Ut ON card stn sMeree cme
era. oes. doses Sage ac acsTPe Letetseoes coset cus es eee 8
TES IN RGTOS TANbY)FCGE218(6)1S 1s a SR OH yah rR
oe
10
meee DaDich IPS alliplcieXAIMINAtlOU, ten wet cemtese 5, tenth tctas ts ete roa sitec es eass aeece ee
ee 13
Chapter 2: General Rules of Probability 15
PM
CUBE THCOLV ASUL VIVAL Koll srtce sogeen toner ei ta coc ao Ateabevioscanavh cues (Eee vues: we Resa cea eee 15
Bem
COUCH LON AL ELODADIILY 6-5 es ie ctseseen irra oo is Cie chetce ogi cree AC Eon
23
MR
OCC CEC Ctra
tn cet Nese ocuscaode Guat Cap yc taet a SNe cea oa CMR aN oe
29
SMES
ee 35
Ay CS eelNCOTSIYI Reet eia 0 ees, Ws each Ty Re nD ee
AS NCO ako e a
SN Re Ue eed sea ee ee eee 38
in occy0. cen Lia
CoECOMDUINCY tee Pewee Ree
PME
Pee
IU cles aliple EX aNMDatlOl sc sc1 tae tg a ers AR Aca, rye hioacts anes eserth ioe ene: 40
Chapter 3: Discrete Random Variables 51
EMIS LeosANCOMI.Y
ALA DICS omen
eet es aa aiced Sithcee artiste ors ete ats 51
Pee
ULALIVe EroUaDIlLy LiSitiDULIONe er a
etc
eee a ee ca eres ne a2
Boe Mc cotirC ayn (ci ralel CLICCIICY aerrets eee nicer: ars ahans
ys
OC eae mer ciaeei od tos Eee a
Re
HC ASUITOSM aLISS CE SILOM erat eotner te tommeted cseeraten ty tnReocia aaaraat eek ante een cre ee 56
Pen
ONKOL ex DeCLAUON ANG. V ALANCO. sara sees eoceaceed.coasece cassdndsaphpeueetepbes nian eset ao
SomemrOiny listed
Random. Variables (ROUMnG 1) i ..jcc-+..tss avn cavesoecasncssouevasttocweansndeones 62
Soe EXO
UA IMI yA Oe LAtIN ee:PUNCLION: sxc.nsatectscsseet
sides secs(oasis siieutcnseiariensnvacnnsuaancas@nesnsomes 65
Fea
11) teeny MUNA KATIE Oar cee ocean stent ovicaog ict oie wn sieancano dnees dun ee vacua soutemen neue haeak oe 66
Chapter 4: Some Discrete Distribution 73
SP
Uist emTOUR LUE UNIOU wetge sscles rncger eeniveneeie ate ar sha ce dshy vara Snnca>yeruacieseee 73
aoe
Deraoill triale and the DYimormial DISPUTOD cose tosscxencsetievanceraesesseusumagesutens-cosacanas--nadees 76
SPM
rere OIAC
ITTGR!ISELIN eee ecg regi can 5 Coxe ascent e datspsoesdupiaa seep asuacgdth da aalia soto ene danon 78
em
ee ASV ESOT
JISCUIU LOU oponcsscacarnc gapeaanissmny smcpig-te ths wwaantaasypGrtaicduoasneduda
scien sve’ 82
Gee
CAELY et eCICOMICITIC. IST DULIOM so,ucct:csocoecrs sndeuse sui oorensaftrsantpach coatnens saataacaruruspomcanan' 83
Sm
LICH ASSET SSL PUL LULU ety orecar rie aetraks avacoedvecout creck y ck cata tanta ri dhaain sadlsea is nstuas haa hinn 84
vinesinestnsuserensny areananennrey nan are89
GIALS CoOUDArisoll OL LISCLETS LISI DULONS divyorenccenss.
BES
c Napict 4 matinic EX AIMAtION eerste css enshcdtns.svoegy re costal atactraecanaban< argon: 93
imme
Chapter 5: Calculus, Probability and Continuous Distributions 97
LLOM ae tiasciiaa rant cater cb es? IV Assaraly as cof wats dunes soon nt vusch,casiornas iN adbssacnunanaesa ise. oF
me ML eS IEE UMC
MLC erPCC ON Goat ot crepe cimn ate cicwat se by sillnish oo dieovsands inges nyunsoaatenges capsinnia svsis>oxslenes quantum cannes: ao
SOCEM COLL VESEY ASUONLstpee Cee te usta RY ol cae sav capaha unsaaautin ics vbrannntoactyts genk oc sSaencns sens104
Applications to Insurance: Deductibles and Caps.....cccicesecseseetnessssessnessessseeseesenens 105
5.5
fetapencsanacsien euapnqects og nen 107
esisatsreats
enncaacegiqayunts
Oe tCC ETAL PUG CULO Mes ecaeesagtevac
UV
sesenntsasnsoneens 111
sssentosestesseeaeesseneoeno
scisscisesccscsstsesacossces
Chapter S Sample Examination...
5-7
il ® TABLE OF CONTENTS
Chapter 6: Some Continuous Distributions 115
C7 ee UDLLORTR
ANG OMIeV ALLA IES acccas seater, ete nee eae ees Py
eee
tae pee giant tec 1
Ore)
Hee XPOUCA LDIS(T DULION tc... toes xt nee aes em
aioe oe yc ah lk och eneaete abe
116
6:3
[he Normal Distribution...
at-0 ccm hese epee
wood ocd ceo encstou ocpinsonnjaaenetee= ce 120
64 Pe ihe law or Averages and the Central Lrmit VR6 Gren ies. scstecesss-cee-ccdocesesoacvaraneudeevatene122
Orme Ne GraTniina-L)ISCITOUILL
OIL cp tere crete tne ee eerede- fag ete gine te dnc ven ats atee saa nie altuna eevee |
Wee
G Geel Ne-Beta tainty Of L1Sttt DUNOMS se eecna cess oavevees teeter er tor et ws eaunee aa ore reese wader 127
Ow
Ole © ONUTUOUS LIIStPIOULIOIS sree eastern tate neti gtentees nt tesa ea asian tacoe cies 130
GrSaar CMaptcy O-Salaple EX ALMITALIOL sa corse cr sncee ieee raed aster eer eee meine adieree tgs ecteoacernee ened 132
Chapter 7: Multivariate Distributions 135
7A,
J Olt Disiibutionstor Discrete RaiCOM VAMADIES jcessesee pevercses-cercccet
eecnecn-auneekeeteneeene 135
7.2. Conditional Distributions — The Discreté Case... iesc-c-secenscnen--ee ian
aaitaces dome ee 137
Jedme. Independence. DISCTELe, CASE ares teccasiynnt neta. ee URC ee aati alse e ee epee a
139
eee
COVarIance atid COTTCLAUOU:.«.a5
thet eoeaccts sascha cee erence Se nace tat clas eerie a ease 139
7 yee Joint DIstioutions for Continuous Random Wallabies .rec..sstece-e
ee asset eee 143
HeOme SOUCIHONAL WIStUIDUM
ONS =. He G OMINOUS ese peer ens see css nase aaa
147
Ueyeeincependence and, Covariance m thie: CONUMUCUS, CASE. 5, ..cc.cncesersseten
coestnaesesendesnca tanec 149
Pp mee Gy armale NOrmal: LISI
DU TOL scum see coe tccve: erreueeaeroe tor ae ee deseasodedia: eked tenn ne eenenenmetetee 154
fal
Cevioment Getlerating Function fOr a J OUIt DISH DULLON ys. -c oop aceeeveonanryoe
ee-cneacento 155
POMS
AO Le igieS ATL EX AIIIITALION sernorce a hleOocorenetacus tien temsuna ant ase take eased aman a ee 155
Chapter 8: A Probability Potpourri 163
Sree
be istlibution Obdslranstorimed. andor: V alialee gece. ates ee orem eden een 163
Onl meetHe WIGMENE Gener atlily PAIR
CU Oi ser gecsuce au. oc steeee te Meron cou eee ete ees oe nee 178
S cig, (COV anIaNCe! b OLAlaser sam, meee paces ec sctreenscrcee ce an easuanuecs teeneaeeee ee ee DP sa ohn 179
SAge he Coma iionin eek Oni las eerste. soe beeen cae nec ae orare oe eee
ee
179
Se
CHAplEr GS Saliple BXaMie AOD. asc cceececs ce tete crvaeiecesase rene roeee races mace ein eae 183
Chapter 9: Statistical Distributions and Estimation 187
Dales
bhessample: Mean asran PStimiatot rye ccc ctor te aecree ee eee eee
De eeu
Demerol
Mating Lee OPUlAatiOn. V alLALICOs,
ic StNGent 1 Listhi DULL
cxccac ecco es er Glatt
eee
a eee
rercsri tarts ca ieennes © oanrsts Uneaten teem
eo
certo tao
187
ae 188
ee eens bone AS 190
De Mme eetLEEDS
LAD ULL Lyn teste’ cee tvs sua cel cares astieaeoutenen aueeie rene ter osseuen vceanvee ea tee eames aeneremete eRe 191
SP am SUITIATILLS PTODOLLIOLS 5 .acace-cesoncacovsaty oxseveaeu roan Uatetens teeta aataunreacapsr ae sar tee eae eee tees 193
DEO MeL stiiiaung thea iierence, DOtWweenl IVCALS seed sectre teeet ens enisstecsteeeat waenateemetee ea treme 194
Fee
SLULENAGIIVG CLS eSALII()LS cOUZ eattgmn nrg ei eyes uate seeks oneniert ceemesits cactiacercere caeta tact eter ee anne 195
DS
COADLEr DO Sar
pLe Haima COM. cx csncngsnseit eines teas tsaneenceees auevenie aan santesesgesccsnte nena enema 196
Chapter 10: Hypothesis Testing 201
LOST Hypothesis: Vesting: Prarie work: nasa shen canchs sce seuraaite nine arth cache ttee tek eat ee en 201
10.2 Hypothesis ‘Testing for POpUlation VICeIS cencconreuseanreenastcmreeectcs¢eteeentnesee
eee nena 201
10.3 Hypothesis T estings Tor: Pop UlariOne Vartan a acccecetenigue terrence unten otters eee ene
10.4 Hypothesis Testing Tor PropGmOnSwan. anexs-c<ccvoteereanatueeteeetereet
ctaeedsere neta cereeam
10.5 Hypothesis festingetor Ditterences 1 Population ivican acececawe
sce eereeeeeene 204
10.6 CHI-SQuUare T EStS sicnsdaresstneneescsasecneea
soulcaveisin tte eee MER een nee tee etn
10.7 Chapter 10 Sample EX art
Ati00)ceiver sccasscauurs chester earn eaeanne nner chee eee 206
TABLE OF CONTENTS
® iil
Chapter 11: Theory of Estimation and Hypothesis Testing 209
SWC HESTSpO i Vie SttAtOL. .ostac tuts csr rg. mamaate MONE. <teee «slo ensecoevasccsaveneeavecveetoatascastlec 209
SU, TSUN PORE LEGS WUNRETy A ea eR Rr
ae
ge I al ol SE er
ee
ee ae
209
IS ee LOPGnL CStOltrSUNTIALOES tre oer Cores. het RUM ere sae cactsgarvactuaah abt uretaes malar mens teaten 21S
Aaa ELV
OLNGSISRPSUIT pk COL Yeates ae tect eee e eee Mack oo dit ca
ee tae dea whet oresdaw si aly,
} Deum ViOre CrcTievallla
ke lINOOG: RATION LEStG, epee stat ts peates farteoetec tec cactoxsoder seause dasa eeder atone: Pee,
LVDSGS LSS SUIE GH SIRS VW WETET 0 2c
A eo pF Ne Oe a a Pal SR
RP
er Rn
Zo)
eit icecineals WepressiOn:, asi F OTIMUIAS 2a ssucdieeseteacosnesdvosdnec-seacue-censbanettusousesvmrtes226
Bie cm ile Linea coresstOn:, HStimation, Of PAaraincteys cnctnctodetasnsracsocstecotdcsernsoceseeanrae 231
IMCS Gramm ANIC ox AITO
ate eae rate tac ene surederectectate couse eansoesaruaube caciesmeeereease aye
the Internet Archive
a
eae
Seat
ion
stin Foundat
cme
ne
HaBlea aes ns
ne
7?
https://archive.org/details/solutionstoproba0000na
SOLUTIONS TO
PROBABILITY AND STATISTICS WITH APPLICATIONS
A PROBLEM SOLVING TEXT
Ee
See
a eS
SS A hat SAE
CHAPTER 1: COMBINATORIAL
Section 1.1
Ee
OORT OES SSN LSE IS OE LOI See AK PENTOSE AY ee
PROBABILITY
The Probability Model
oe (a) We systematically list the sample space.
chicken sandwich
chicken sandwich
|coleslaw
(b)
{hamburger & potato chips & soda, hamburger & coleslaw & soda, chicken
sandwich & potato chips & soda, and chicken sandwich & coleslaw & soda}.
(Cc)
Pr(soda with meal) =
Reducing
=—
l
number of meals with a soda
4
=
=—,
total number of possible meals
12
a‘
;
has the interpretation that one of the three beverages is a
The event that doubles are rolled is the set of outcomes
1!1,22,33,44,55,66}.
There are 6 ways to roll doubles.
(b)
Pr(sum is prime) = Pr(sum equals 2, 3, 5, 7, or
11) = 36"
36
2
SOLUTIONS TO EXERCISES — CHAPTER |
1-3
(a) Legend
Third plain M&M package
First peanut M&M package
Second peanut M&M package
/B | Peanut butter M&M package
/D | Dark chocolate M&M package
Almond M&M package
There are two ways to list the sample space. If you grab two packages at the
same time, then the outcome of DA represents one package of dark M&Ms and
one package of almond M&M’s. In this case DA is the same outcome as AD
(order does not matter), and therefore only needs to be listed once in the sample
space. Then there are 28 outcomes:
VA Eo,
PONG ea NZ. 2
BaD i A, ei tos io Ng, 128, 20,
P,A, Ps Ni, Ps No, PB, PD, P,A, NM Nz, NiB, MD, MA, N2B, N2D,
NA, BD, BA, DA}.
If you envision the experiment as grabbing one package and then grabbing a
second package (in other words, order matters), there will be 56 elements in the
sample space.
(b If order doesn’t matter the answer is =. If order matters then the answer is
—
=. Of course 2 = = , so both perspectives lead to the same answer so long
as you are consistent in your treatment of the sample space.
(c)
18 036
a
56
Section 1.2 Finite Discrete Models with Equally Likely Outcomes
1-4
There are 3-2-6
= 36 current outfits
eee
Shoes
Pants
Shirts
Adding shoes will allow 4-2-6=48 different outfits, adding a shirt will allow
3-2-7=42. You increase the total number of outfits most by adding to the item that
you have the least. That is, if your mommy offered to buy pants, you could form
3-3-6=54 different outfits.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 3
1-5
**
CIEE
Red
White
Blue
(oye
ae
There are six choices for the blue die.
(c)
Pr(first die is 3 and second die is 4) = : - = +
(d) No, tree diagrams are not so useful if tracking more than 50 or so outcomes.
10°
LILILI-LILI-LJLILIL
10-23-26-23-10-10-10
One could try to use a tree diagram to systematically list all possible way to rob four
banks. The thief could rob any bank first. He has only five choices of bank to rob
second (he does not rob the bank he robbed first). Similarly, the thief has five
choices of bank to rob third (it just cannot be the bank he robbed second, but it could
be the bank he robbed first). The total is 6-5-5-5=750.
8-10-10-9-10-10-10-10-10-10
er
Tap
ov
eed
A ns
(c)
7! 1
Sere
USI
aa
(dy) 5 220
(e)
4!=24
(f)
Undefined
(g)
23°22-21
eat; aie
4 @ SOLUTIONS TO EXERCISES — CHAPTER |
(hy
3,1
(1)
ee
0)
Gat
kh)
)
7!
==
31.4!
7-6-5
32 2e1
3s
Ciel
‘EE
Wik
Cay
Ken!
2
ey
Kendra gets her own exam, the other 16 exams can be redistributed to
any of the remaining 16 students.
Kee ae
aL”
sP3
6P; = 6-5-4
12 Py
r=23.
365 Fy. 2 .5 by trial and error.
Solve 1—-==—
365!
r=4.
Solve l—- sae
yt
7
Pr[people get offon unique floors] = 11ehP;
352
3512
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 5
oe
n=5.-solve Veaamee .5_ by trial and error for the minimumn.
oo
hes
(a) 7C6=ier
lous
(b) 3012
= W2'q
T =13
13!
=
(c) 3CC| =——~
Tho
5!
(d) 5C, = 371 =10
OnCe
ol Remembertar
olet.
!
(g) Cs = 525 =10
a
ee
(1) 5Cs
aol
G) 5367C1 =
a0
= 5867
(k) HE = T=
50,=35
(1) 0Co = aie
Both expressions equal 220.
n C, =
'
n!
(n-r)!
=
n!
(n-r)!r!
= Kner
Placing r toppings on a pizza is equivalent to keeping n—r off the pizza.
6
SOLUTIONS TO EXERCISES — CHAPTER | _
1-23
6C4
1-24
The multiplication principle says that the total number of possible pizzas is
Number of ways to select toppings x ways to select size x ways to select crust
(a)
13C2-3C1-2C)
erCare Cis
(DO
1-25
= 468
aC py
(Cl)
Oy sCrscr
(a)
iss
(ob)
4Ci-5Ca-6C2
1-26
123015Ci
1-27
14C3-3C4-7C2-5C,
1-28
(a)
e252
= 210
39C2-290C 2=8550
(b) To match exactly one white ball, your lottery ticket must match one of the two
“good” white balls. There are 7C ;=2 ways to do this. The second white ball
could be any of the remaining eight “bad” balls.
16
oC ® 4Cjyes
match | match!
good
bad white
white
1-29
(10C1)
oOo oe 8550 = ee
match both
blue balls
6
(a) ——_——_
= .01997
60 Co
choose |
piece from 4
choose one ofthe chosen
choose 5 variety w/2
varieties
varieties
pieces
easy ae
(b)
6Cs >
5Cy
(15Cy)
4
choose2
Pieces for
variety w/2
pieces
= ge
= 2697
60 C6
4Cy~ §Cy + 6Cq = 410-15 = 4G = 50y
——
—
centers
forwards
Gs
—
|
ae
—~—
guards
centers
forwards
guards
starters
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
ee
15
15!
Eee = Sh3h7t =
15s
°
start 5 of the 15
1-32.
10 C3
eee
ON
7Cq
3 of remaining
7 ofthe final
10 are on the bench
7 do not dress
4
!
The number of words is ,i ;
= saa =12. The sample space of possible words is
thoop, hopo, hpoo, pooh, poho, phoo, ohop, ohpo, opho, opoh, ooph, oohp}
Pr(chosing HOOP) = a
1-33
5 ees ee : 3 : : = 30)
rT) WeLahIbS Pol ohh
!
1-34
og
la
1-35
@ 7
(a)
Deal5
rer
a
nrers
ae SSS Ses gtr
Tanase
cards to the first poker player. There are 52Cs ways to do this. Next,
deal 5 of the remaining 47 cards to the second player and continue in this
fashion. The total number of ways to distribute six 5-card poker hands is
52!
s2C5 + 47C5 + 42C5 + 3705+ 32C5-27Cs. = SE 5151515151991
(b)
The 52 playing cards need to partitioned into 6 piles of 5 and | pile of 22.
There are
2
je eA
=
Ee
EL|ETAL
a,
s to do this
this equals |
Cs + ——
2Cs
Nee eens
frat party
soup kitchen
:
Cis
° Se
stay home
:
=
25!
PAD).
AY
25!
=
Sem
ZOESI Aolor le
oNor lS:
In terms of a multinomial coefficient,
;
L
5
llth
WN
Nm
8 @ SOLUTIONS TO EXERCISES — CHAPTER |
(b) This is similar to part (a), but we need to take into account that the 5 person
teams are essentially the same.
versus team green.
That is, suppose the hoop game was team blue
If the teams are {Amy, Bree, Candie, Diane, Edna} versus
{Zoe, Yolanda, Xena, Willaminia, Venus}, then it does NOT
matter which is
team green and which group is team blue.
25 C5 + 20 Cs
Z
1-37
30 C7 - 23C7 +2
1-38
9
!
There are L4 {= sche = 9C)-7C4-3C;
Prowomen
e
total ways to assign the jobs.
assign 2 women
assign 4 of the 7
assign the remaining
both bad jobs
menthe average jobs
3 men the good jobs
————
a
—|——_
:
cet both bad 106s)
Jom)
1 Cie
—_——<———————
Note: This can also be solved as
7C4
3C3
9 Cz > 704-33
=
l
o>)ON
since the problem doesn’t involve
the distinction between average and good jobs.
Section 1.3 Sampling and Distribution
1-39
There are 16 different side dishes.
specific order:
1-40
age
Order does not matter.
The customer must select different sides in a
816
seelow:
14
5
;
first side
:
second side
a
ae
third side
+
So, for example, the following 3-side choices are equivalent:
{(1)
(2)
(3)
(4)
corn & dressing & mac-N-cheese,
corn & mac-N-cheese & dressing,
mac-N-cheese & corn & dressing,
mac-N-cheese & dressing & corn,
(5)
dressing & mac-N-cheese & corn,
(6)
2.53560
dressing & corn & mac-N-cheese}
3
nen
There are 1¢Cs = me = 560 such orders of 3 side dishes
1-4]
16° = 4096
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
1-42
# 9
This is equivalent to distributing 3 indistinguishable balls (selections) into 16 fixed
urns (side-dishes) without exclusion.
16,3-)C3 = 1g C3 =816
1-43
Ordered samples without replacement.
See Exercise 1-39 and solution.
1-44
Unordered samples with replacement.
See Exercise 1-42 and solution.
1-45
Pr(order 3 servings of the same side) = As
816
1-46
(a) 345-1Cs =7Cs
(b) The parent gives each child one dollar bill. She is left with two dollar bills
which she can distribute to her 3 children:
1-47
4C.
(a) The fundamental theorem of counting implies the solution equals:
Number of ways to distribute the Snickers x Number of ways to distribute the
gum 13Ci9-10C7
(b) After giving each trick-or-treater a Snickers and a pack of gum, she has 6
Snickers bars and 3 packs of gum to distribute in any way she likes.
6x4-1C6 - 344-103 — 9 Ce. 6 C3
1-48
Each child could receive multiple pickled beets. There are j9,s_;Cjo = 14C\o ways to
give beets.
1-49
A bridge hand consists of 13 cards dealt from a standard deck of 52 different cards.
5213
52 Cs
It depends upon the number of chicken sandwiches.
If you do not order a chicken
sandwich, then you have 5 dollars to distribute to 5 different $1 items. There are
9Cs ways to do this. If you order exactly one chicken sandwich, then you are lett
with 3 dollars with which to buy $1 items. You either get zero, one, or two chicken
sandwiches.
9 Cs 1 7C3 1 5 Cj
36 C21: 43.C21
10 # SOLUTIONS TO EXERCISES — CHAPTER |
Section 1.4 More Applications
1-53
(a)
(xt3)* = Ix4 44-23-3146.
x2 -32 £4-¥-33 41.34
Niele
454%
Ose
(b)
(2x+y)
oe|
4
1-(2x)? +5-(2x)*-
y +10-(2x)> - y?
+10-(2x)*-y?+5-(2x):
y+ +1-y?
= 32x° + 80x*ty + 80x73 y? +40x7y? +10xy* + y°
(Cy
1-54
3)
(4x) + (-Sy))P°
= 1-(4x)? +3-(4x)? -(-Sy)! +3-(4x)! -(-Sy)? +1-(-Sy)?
64x? — 240x7y + 300xy? -125y3
II
n
The key here is to write out |
n—-|
and | }and then combine using a common
r-
r
denominator.
(A oe
r—-1
—
r
(n-1)!
oe
er)!
(n-l)!r
—
(n-1)!
iF
Ter
P
(n-rybr-(r-)!
(n—-1)!\(n—-r)
(n—-r)-(n—r—-DeEr!
(n—-l)!r
Fi(n-1)!(n—-r)
(n—r)tr!
(a=r)brl
_ (a-Ibr+@-Di@=r)_
-
leand y=l.
(1+1)”
A
OF +. a
_a!
_{”)
~ (n—r)br! GY. )
1-55
betix
1-56
Selecting r items from n possible toppings to put on the pie is equivalent to
selecting n—r
to keep off
of the pie.
1-57
=
(n—r)!r!
+ nO? see
Gas
;
n
n
n+l
ae
:
Since |}+| = | Joincoefficient of
4
5
5
:
n+]
io yoes = x yitt)-s s|
= 3,876+11,628
=
15,504.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 11
Glee
n
Alternatively, -<
n
=
O88
3,876
faa
5
= 3. Thus n=19
and the coefficient of
4
ayes
= x°y!> in the expansion of ep
= (x+y) is
20
J1s.s04,
5)
(ay @=2ye5z)) = 1.x? 43-2. (-2y)43-x2+(52)
+3-x-(-2y)? +3-x-(Sy)*
+.6-x-(-2y)-(Sz) +1-(-2y)} +3-(-2y)? (52)
23 Capen
2)
= x3 —6x*y+12xy? —8y3 +15x2z
—60xyz
+ 60y2z 4+ 75xz* —150 yz? +12523
(b) (w+x—y+2z)*
= 1-w? 4+1-x? +1-(-y)? +1-(2z)?
+2:w-x+2-w-(—y)+2-w-(2z)
\\
ake (=) ex
ogee Fs Pe hee ci
Jot
oO) Net
+U\Z
-
Uy
22) 2 (—y) (2
a pa + | WZ
Zz.
Detailed calculations follow the exércise in the text. 0
Each of the five dice has 6 possible outcomes.
So the total number of ways to roll
five dice is 6°.
(a) Pr(Five dice-of-a-kind) = sts. =.00077.
ways to select 4 of
the remaining die
4 of what kind?
the Sdieto match
must be different
ee,
—__——_—,
een
6C\
5C4
(b) Pr(Four dice-of-a-kind) =
4&;
6>
5?
=
64
= 01929,
(c) There are 2 possible straights, 1-2-3-4-5 and 2-3-4-5-6.
many ways these straights can arise ina roll of5 dice.
HE
Pr(straight) =
6>
It remains to count how
| as ssn
6°
12 @ SOLUTIONS TO EXERCISES — CHAPTER |
(d) A full house would have three dice be of one value (1, 2, 3, 4, 5, or 6) and two
dice of a second value.
1 Wh S chal,i hoe
PrFull house) = Hl46>0 5
10.5.
one
= 03858
(ec) The only way to get nothing 1s to have dice of the form:
1-2-3-4-6, 1-2-3-5-6, 1-2-4-5-6, or 1-3-4-5-6.
It remains to count the ways these can sequence can be arranged on 5 dice.
4.5!
480
Pr(nothing) =
a
eer
= .06%73
Bek
pag
3
Oa:
aticlie
ieee
as
aA
("a = So
(f) Pr(Three dice-of-a-kind) = =
aea aa
ZA
2a.
2
ACE
8 B 4 " b ge
(ey ePIC Eo pair ie ee
6°
(h) Pr(one pair) =
6°
a B
PA 5
ea a
aea"
geal|
4010-3:
a
63
= wee
= 15432
= 46296
It is neat that you are more likely to get 3 dice-ofa kind than you are to roll nothing.
Did you check that the sum of these probabilities is one?
1-61
1-62
Solutions tabulated in the text.
Expected Winnings
l
= 49,999,999 .———____—
120,526,770
4,00 006)
2
Sy, OSes
20526
oon
eet
sAlis
0)
ene
That is, the expected winning is negative 41 cents per play.
ee
t). 117,184,724
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 13
Section 1.5 Chapter 1 Sample Examination
L.
Order does not matter.
2
799-104
3:
Order matters here.
4
2.
15 P;
C2
8G
give slinky's
away to 2 of
the 15 students
give bop-on-nose
to | of the
— remaining 13
ee
Both solutions are 5Cs; = 65,780.
rei
=| 365 2— TY
14C2
give bop-on-nose
away first
give slinky's away
to 2 of the remaining
14 students
(a) 51Ci7
2 pair and 13 other cards without the old maid
two pair, the old maid, and 12 other cards
SSS SSS
(b)
6.
25
Gg Ga-9 C7" 33Cig (4G)> gh.
Gor Cy ya 7 93 Cp -(2C))
12
(a) 32Cs
(BD)
aC
ae
fo
which suit?
—S
Cin<
3500)
—— a
ways to select 4 the remaining
cards from the
selected suit
if
Ca Cog
51C17
card
(a) Cae 6. C2
:
162663
(b) Pr(5 kids selected included 3 soccer and 2 football) = acetate A408.
1305
8.
Using the multiplication rule gives 3-2-3=18.
s)
49Cs + 42C; = 80,089,128. The Jackpots were not getting large enough, so there
were more balls added to decrease the chances of winning and therefore increasing
cumulative Grand Prizes.
10.
(2x—y)>
= 32x> —80x4y +80x°y? — 40xy? +10xy4 - y°
14
LI;
SOLUTIONS TO EXERCISES — CHAPTER |
There are 15 terms in the multinomial expansion.
(x—2y+5z)"
= 1-x4+4-x3 -(-2y)+4-x3 -(5z)+6-x? esto -+1-(5z)*
12 cards have faces
—t
Pr(all 3 cards have faces) =
Az 7
= 00995.
3)
ways to match
ways to match
white socks
black socks
———
Pr(matching socks) =
8@
SS
mai 42
a 2
8-4
=1-Pr(one white and one black sock)=1— 7
There are a total of 100 birds, therefore j99Cs ways to take six birds.
Joe could take 2 Canada geese (and one from each of the other prey), or 2 ducks, or 2
eagles, or 2 cranes, or 2 flamingos.
This can be done in
20C2 + 25Ci- 401-101 5Ci
+ 29 Ci1* 95 Co = 49Ci 0 Ci= 5 Ci
+ 29 + 25C1- 4o€2 > 10C1 +5 Cy
+ 20Cy > 25Cy > ani «10 C2 > 5 Cy
+ 99C) + 25Cy + an Cy - 19 Ch - sC2
= 47,500,000
ways.
47,500,000
bird ofof each each variety)
£192,052. 400 = .033 985.
Pr(atr(at least leastone bird
variety) =~
ee
Pr(passing on first attempt)=
1oS)
Use the multiplication rule with the following steps:
Step 1.
Step 2.
Step 3.
Step 4.
Choose any 3 rows (order immaterial) - # ways = 6C; =20.
List the rows in ascending order for definiteness.
Choose a column for the first listed row -# ways =5.
Choose a column for the 2™ listed row -# ways =4.
Choose a column for the 3" listed row - # ways =3.
Answer = 20x5x4x3
= 1200
(C)
CHAPTER 2: GENERAL RULES OF PROBABILITY
Section 2.2 Set Theory Survival Kit
2-1
(a)
(b)
(Cc)
(d)
O=rl3
od.7-9} and, Pat 2 aio.
The odd primes less than 10 are OM P = {3,5,7}.
The set of odd numbers or prime numbers less than 10 is
OMT Pee 28 5.7.9.
The odd numbers less than 10 that are not prime is the set
Age
(a)
(b)
B ={az water}, “so N(B) =2.
(c)
AUB=(1,2,z,Jamaal,gum},
(a)
The Luisi family pets can be summarized as:
(b)
The number of pets of each type is summarized as:
so (AU B)’ = {water}.
Named
\
O— P = {1,9 Sf:
16 ® SOLUTIONS TO EXERCISES — CHAPTER 2
2-4
The shaded section denotes the indicated set.
2-5
From Subsection 1.4.3, we already have the probability of the winning prizes.
Pr(losing) = 1 — Pr( winning something)
> Rye
le
.
120,576,770
ae
120,576,770
1,712,304
probability of winning
the grand prize
probabilty of matching
only the powerball
_ 117,184,724
120,576,770 °
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 17
Doubles = {(1,1),(2,2),(3,3), (4, 4),(5,5),(6,6)}.
Sum is divisible by 3 = {(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),
(4,2),(4,5),(5,1),(5,4),(6,3),(6,6)}.
Doubles - Sum is divisible by 3 = {(3,3),(6,6)}
Doubles U Sum is divisible by 3 = {(1,1),(1,2),(1,5),(2,1),
(2,2), (2,4), (3,3), (3,6),
1)
a)
0,1), (024), (,.2),
(0;3),.020)}
Pr(Doubles U Sum is divisible by 3) = + = < + oa— =
Let K equal the percent of king size mattresses sold, O equal the percent of queen
size mattresses sold, and 7 equal the percent of twin size mattresses sold.
We are given: K+Q+T
= 100%,
Solving for the variables,
K
Q.= LK +7), ane Kae
3 2
=60%, Q=20%, and T = 20%.
The probability that the next mattress sold is king or queen size is K
2-8
A’ = {2,4,6,8,10}
B' = {1,4,6,8,9,10}
AUB ={1,2,3,5,7,9}
ANB =33,5,7)
A'UB' =(1,2,4,6,8,9,10}
A' 0 B' ={4,6,8,10}
(AUB) ={4,6,8,10}
2-9
A-B = ACB’
is represented by the shaded area.
A
B
+Q = 80% (C).
18
D210.
SOLUTIONS TO EXERCISES —- CHAPTER 2
“Cte eee
(4UB) =4'0B'
2-11
AU(BOC)=(AUB)A(AUC)
2-12
N(AUB) = N(A\B)+N(AQB)+N(B\A)
[Nv(A\ B) + N( AMB)|++[ N(B\ 4)+N(ANB)|-N(AOB)
= N(A)+N(B)—-N(AMB)
_ PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 19
2-15
(4 Bo
c) is the shaded area
AM(BUC)".
2-14
Method 1: Using the Axioms
N(A)
N(B)
6 = N(U)-N(A).
10 = N(U)—N(B’).
From the inclusion-exclusion relationship,
N(AO
N(4U B)+N(AMB)
B)=6+4+10-12=4.
Method 2: Venn Diagram
Method 3: Venn Box diagram (Subsection 2.2.4)
= N(A)+ N(B).
20 @ SOLUTIONS TO EXERCISES — CHAPTER 2
2-16
NM(AUBUCUD)
= N(A)+N(B)+N(C)+N(D)
—-N(ANB)-N(ANC)
—-N(AND)-N(BOC)
—-N(BAD)-N(CAD)
+N(ANBOC)+N(ANBOD)
+N(ANCND)+N(BOCOD)
—-N(AQNBOACOD)
There are 4C, =4 ways to select one of the four regions.
There are 4Cz =6
intersections of two of the four regions.
There are 4C3 =4
intersections of three of the four regions.
.
2-17
A four region Venn diagram could be drawn as:
2-18
There are seven male children without purple hair in this family.
Adult
Female
Bin
Purple hair
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
2-19
(B) 50% = 40%+10%
of the population owns exactly one of auto and home.
Auto_owner
2-20
(D)_
Home
There are 880 young Tharried females.
Married
2-21
(D) 52% of these viewers watch none of the three sports.
Gymnastics
aN
yy
Soccer
Baseball
@ 21
22
2-22
SOLUTIONS TO EXERCISES —-CHAPTER 2
(A)
5% of the patients need both lab work and are referred to a specialist. The
percentages in bold are given in the problem. Try this just using a Venn
diagram.
Lab work | No lab work
Referred to Specialist
| 5% 0 Bier
ae
Not Referred to Specialist
35%
60%
2-23
(D) Pr(A) =.60.
The key here is to note that Pr(AU B) = 0.7 implies that Pr(4’
B') = 0.3.
Similarly, Pr(A U Be)=0.9 implies Pr(4'n B) =0.1. These are versions of De
Morgan’s laws. We do not have enough information to find Pr(A A B) or
Pr(AB'), but fortunately we are just asked to find Pr( A) = 1—(.1+.3) = .6
A
(D)
B
We have two equations with two unknown variables x and y (see box
diagram below). The first comes from the fact that the sum of the probability of four
disjoint outcomes must equal 1. The second equation comes from the statement of
the problem.
22%+x+y+12%
22%+y
= 100%
= 22%+x+
14%
—.,—_——“~-
We
—-—-
probabilty visit
chiropractor
probabilty
visit PT.
exceeds
by 14%
x = 26%
and
y=40%.
implies that
x+y
imphlesthat
y = x+14%.
Pr(visit P.T.)
= 66%.
= 22%+26%
= 48%.
Chiropractor
22% + y
No chiropractor
x+12%
22% + x
100%
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 23
2-25
Draw the Mickey Mouse diagram, with Auto, Homeowners and Renters labeled as
shown below. Since H and R are mutually exclusive the subsets comprising their
intersection have cardinality zero. Statement (i) gives N [AUBU cy |Sa,
Statement (v) gives N(H \ A)=11. The remaining unknown subsets are labeled
eye Z AWN
From (ii) x+ y+z = 64. Also,
100 = 174+114+(«+y+z)+w
=>
we
8.
From (111), y+11
= 2(x+8), and,
From (iv), x+y
= 35. Solving the last two equations simultaneously
gives
Section 2.3
2-26
= 174+11+64+w
x = N(AXMR)=10
(B)
Conditional Probability
There are 3 levels of choices in the following tree. The first branch concerns whether
the professor put the batteries in an empty pocket or in the pocket with the two new
batteries. The second level concerns which pocket the professor withdraws batteries
from.
In the third level, if the four batteries are in the same
pocket, we will use
counting techniques from chapter | to determine the likelihood of grabbing either
two batteries that do not work, two batteries that do work, or one battery that works
along with one battery that does not work.
24
SOLUTIONS TO EXERCISES
— CHAPTER 2
Professor selects
Professor places
dead batteries
in empty
;
Select two
Bee
.
1
pocket with dead
eee
batteries
:
3
we
: Professor
pocket
er
Select two
selects pocket
1 __ good batteries
with good batteries
2
‘alt
Select two dead
batteries
Professor
places dead
batteries in pocket
1
with 2 good batteries.
4
oe
| gi ees
l
Fe
1
4
elect one
professor selects
beket with all
:attenes
dea
ie
and one good
battery
Select two
good batteries
(a)
It
Pr(get at least one good battery) = =.Bers
1
») eee
eee
(b)
Pr(two good batteries at least one good battery)
Peay
_
G
Pr(two good vat least one good battery)
Pr(at least one good battery)
2
Pr(two good)
Pr(at least one good battery)
;
mall
Tiga,
ae
4
24
(c)
l able l
play
m0)
4
Pr(batteries in same pocket|at least one good battery)
Pr(batteries in same pocket Mat least one good battery)
Pr(at least one good battery)
The
audio
device
will
play
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 25
2-27
(a)
This is the scenario where we will keep our original door.
You will win the car if and only if you selected the correct door originally (it
just doesn’t matter what Monte does). There is a one in 3 chance of this.
Pr(win auto|keep original door) = +.
(b)
If you originally select the door with the new car, then Monte can select from
either of two doors to show you a goat. If you select a door concealing a goat,
Monte only has one door that he can open. Consider the following diagram.
shows you
Monte
:
You pick a door
hiding a
If you change doors, ?
ei
the goat below
l
you win a new car
goat
ou pick a door
hiding a
Monte shows you
aA
goat
the goat above
If you change doors,
>
|
you win a new car
'
3
If you change doors,
Monte shows you
a door hiding
the first goat
you win a goat
vof—
shows you
the second goat
Pr(win auto change doors) =
~
i)[o
<)
ee
w+
a new automobile
If you change doors,
you win a goat
vi-
‘L-1+4-1-1 = veto
|=
Consider the following Box Diagram with the information in (i), (11) and (111) filled in:
b = 1000-14-22
= 64
From (iv),
> High] = Pr[High|[rregular]- Pr[Irregular] = +-15% — hi
a = Pr{Irregular
Similarly, from(v),
c = +: 64% = 8%.
Regular
Irregular
High
| Low | Normal
ee
Row Total
oy
SES ies
85
5
2
8
15
100
Column Total
Pr[Regular
~ Low] = 20%
Completing the Box Diagram gives,
(E)
|
26 @ SOLUTIONS TO EXERCISES
2-29
- CHAPTER 2
(C) Let x equal the probability that the male is a smoker, given that he does NOT
have a circulation problem.
Pr(circulation problem] smoker) =
Pr(circulation problem ~ smoker)
Pr(smoker)
oe ak
5
5 Oe
ee
4
ae
Smoker
2x
Smoker
x
25%
circulation
problem
2-30
Cyne
(C)_
2-31
(B)
Pe
oS
ees
:
Pr(smoker A died)
.10-.05
.005
ker|died) = ————-___—.
= —_——____—_____ = —— _ = 36°
Pr(smoker/died)
Ne
Pr(died)
<N=
oe
10-.05+.90-.01
iS
014
]
oe =
Pr(N <4)
=
; Ni : + ms + a
3 + - a +35 +96
a]
role
;
|
h233
2
6
et.
:
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
2-32
@ 27
(a) Pr(Penguins win series) =.55-.35+.55-.65-.55+.45-.35-.55=.47575,
(b) Pr(Devils win at least one game) =1—.55-.35 =.8075.
\
35% |Pittsburgh wins the series 2-0
Pittsburgh
wins
55% | Pittsburgh wins
New Jersey
Pittsburgh
wins
wins
:
ew Jersey wins 45%
New Jersey
the series 2-1
‘
ae
Pittsburgh
Pittsburgh wins
wins
Game 2 played
:
in
Spc
New Jersey
New Jersey
—.
Game I played
in
ne
Game 3 played in Pittsburgh,
if necessar VW
65%
eM CLSeD
2-33
New Jersey wins
New Jersey wins the series 2-0
(B) This is a standard Bayes’ formula problem, easily solved by drawing the
appropriate tree diagram. The data may also be displayed in a Box diagram similar
to the one in Exercise 2-28:
|
| Critical |Serious |Stable |Row Total |
Sie
a ee a
Lert >oa lr ec (SS el ae
10
Column Total
30
d
100
d = 1000-10-30 = 60
a = Pr[Die Critical] = Pr[Die| Critical] Pr[Critical] = (40%) -(10%) = 4%.
Similarly,
b = (10%)-(30%)
= .6%.
= 3%, and c = (1%)-(60%)
the box gives,
vai
fon
10
30
Critical |Serious | Stable |Row Total
Z
59.4
92.4
Column Total
7
Pr| serious
} survive
i
Pr serious|survive
=—=
L
Pr| survive |
60
J oro
a7
100
wal ayNaicaaeae
ha ey
Ee
B
(B)
Completing
28 @ SOLUTIONS TO EXERCISES
2-34
— CHAPTER 2
— Pr(exactly 1 hit in 3 at-bats) =.25-.6-.75+.75-.25-.6+.75-.75-.25 =.365625.
400
Hit
400
Hit
No hit
600
250
250
=F
Hit
No hit
-600
1
No hit
.750
Hit
No hit
Hit
.400
No hit
maul —
at
Hit
First at-bat
250
750 —<.
No hit
Second at-bat
No hit
Third at-bat
2-35
Pr(call “dad” ~ Packy is your father)
Pr(Packy is your father)
Pr(call Packy “dad” |Packy is your father)
e.
.01-.90
.01-.90 +.99-.05
One Car | Multiple Cars | Row Total
Using (iv), a = Pr[Sports Car > Multiple Cars]
= Pr[Sports Car |Multiple Cars]Pr{Multiple Cars]
= (15%) -(64%) = 9.6%
The complete Box Diagram is now easily calculated:
2
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
Thus, Pr |One Car 7 No Sports Car|=25.6%
Note:
@ 29
(C)
One may also display the information in a tree diagram:
Insure a sports car
Does NOT
.64
15
insure a
sports car
.85
Insure a sports car
exactly | ca
Section 2.4
Independence
511
Pr(Male-Male) =.511* =.2611
Female
gq Pr(Male-Female) = .511-.489 =.24988
First person chosen
511
Pr(Female-Male) = .489-.511 =.24988
499
Pr(Female-Female) = 4892 =.2391
489
Second person chosen™
30 ® SOLUTIONS TO EXERCISES = CHAPTER 2
2-38
| You are encouraged to observe that this is structurally the same as problem 2-37.
2
z
Heads
>)
Pr(Heads-Heads) = (2) ae,
3
2
.
3
Head
Tails
ae
Pr(Heads-Tails) = eae
2
Px(Tails-Heads) = 44 : =
3
é paresip
3
Heads
Tails
l
3
First coin flip
|
Maes
pyTails-Tails) = (+) =
3)
Second coin flip
Number of Heads
| 0 |
l
Probability
9
cathe thoagh the srobsbibiy
tree that have one
head and one tail
384
tee
Make
Makes his 3
point attempt
384
384
Miss
616
Nree
Misses his
3 point attempt
Pee
1
Make
: 616
Make
384
Miss
616
Make
384
eit.
616
384
John Stockton’s
_
Miss
616
Make
384
Miss
616
616 Ea
first attempt
second attempt
third attempt
Pr(Stockton makes exactly two 3-pointer)
3
384
.384
.616
ways to make
exactly two
3-pointer
on three attepts
probability of
making a
3-pointer
probability of
making a
3-pointer
probability of
missing a
3-pointer
= 2725
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 31
2-40
iy
Pr(B| A)=—.
2-41
(a) Since Pr(B)=b, we know Pr(B’) = 1—b. To show that4 and B’ are
independent, we observe Pr(4> B’) = Pr(A)-Pr(B’) = a-(1-b).
Pr(B) api
51
The events A and B are dependent.
52
Q
Ca [a RRS
w=)
2-42
‘Not necessarily.
2-43
(a)
(b)
B
B'
b
eoah)
C = {HHH, HTT,THH,TTT}.
Thus, Pr(A) = Pr(B) = Pr(C) = 4.
(c) Pr(AB)=Pr(HHH,TTT)=2-=1.
Pr[A
2-44
Similarly,
VC] =Pr[B AC] =Pr[HHH, TTT) =.
(d)
Yes, because for example, Pr(AM8B) = + = +.
(e)
Pr(4
(f)
ING Weiter Arie
Let
x=Pr(AM 8B). Using independence, solve (.2+ x)-(.3+x)
SGIUUONS, X=
=ePTCA)- PTC):
BOC) = Pr[HHH,TTT]=+.
Cyt
ee
= SPs
A] PLB) Pr.
= x. There are two
2 0lL =.
Pi
eet Obar 0.
Since the coin is fair (all outcomes are equally likely) and we do not need to
track probabilities, listing the sample space will be sufficient.
2-45
U =|HHH,
HHT, HTH,HTT,THH
A = first coin 1s tails ={THA.THT
B =the third flip is heads =
THT ,TTH,TTT.
TTH,TTT}.
{WHH,
HTH ,THH ,TTH }.
(b)
Yes, the events are independent.
32 @ SOLUTIONS TO EXERCISES — CHAPTER 2
2-46
(a)
Now is when tracking probabilities through a tree diagram is critical.
PrANB) =_ 4-.6-.6+.4-.4-.6 = 60%.
agPr(B| A)a = ———.
Era)
(c)
2-47
A
Yes, the events are independent. You can see this by computing
Pr(B) = Pr(third flip is heads) =.60.
Let O be the event of using orthodontic work, F fillings and E extractions.
Using (1) and (11) we have:
tele
eae
Ss
Gerth
Ol) Primal
1—21o—9
b=j12=1)2. By independence,
Completing the O and F box gives:
3
d= be=>c=a/b=213-
Row Total
1/3
Bes)
O | O' | Row Total
|
E
1/4}1/4
le2
E'
[1/4 ]i/4]
1/2
|Column Total | 1/2 | 1/2
We now use Pr| F |=
ly
|
|
anc Pri Efe l /2, plus (iv) to fill out the F and E box:
ip F
E
Row Total
LS
[ 1/8 |
al
E'
7/24
Li
1/2
Column Tota!
Note: F and £ are not independent.
Pr| EU F )=1-Pr| EV F'|=1-7/24=17/24 (D)
l
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
2-48
33
We take the statement that a “different brand of pregnancy test was purchased” to
imply that these test results are independent.
Pr(both tests are incorrect) =.01-.02 = 0.0002.
2-49
First we fill in the missing information.
ivajor | ee Grades yeas
| TAT BIC] DE |Totals |
|
Business|
2| 3 |3 [5 [1] 14 |
PSs
lem
ee
| Totals |9 |10}4[/5[3]
31 |
(a)
Pr(student assigned ‘B’) = =.
students - business majors
:
b
(0)
:
:
number of science majors — 17
Pr(science major)
= -———_-_—__>—_ = — =
\
yor)
number of students
on
(c)
Pr(science major ~ assigned ‘B’) = +.
(d)
This is most efficiently calculated by looking at the table. We are told that the
student is a business major. Therefore the student lives in the shaded row of
the table.
Pr(assigned *C’ b usiness
2-50
major)
31-14
at
3
et is
1
(e)
Pr(business major assigned 'A') = =.
(f)
No, because
(a)
Wewill use counting techniques (combinations) that we studied in chapter 1.
The number of ways to select 2 batteries from a flashlight containing 5
Pr(assigned *C’ business major) # Pr(assigned *C’).
batteries is 5C> =10.
Pr(both batteries lost their charge) = at eho:
(b)
Pr(exactly one battery lost its charge) =
:
=
(c)
Draw a tree diagram with appropriate probabilities.
>
Pr(exactly one tested battery lost its charge) = 2-
bo
in|
34
2-51
SOLUTIONS TO EXERCISES—~CHAPTER2
~
Let Pr(purchase disability coverage) = P[D]=.x, then
Pr(purchase collision coverage) = Pr[C’] = 2x.
0.15=2x? implies that x =.274 and 2x =.548.
The completed Box Diagram is:
ee KE
oa [ea
Sonn Tour]sasa2[1
Pr(select neither insurance) = Pr aoN D'] = 328.
2-52
(B)
Pr(select two consecutive face cards) = =-- ae or using combinations
Pr(select two consecutive face cards)
2-53
2-54
number of ways to deal 2 face cards
12C2
number of ways to deal 2 cards
52C2~
(a)
Pr(next roulette spin is black) = 4.
(b)
8
Pr(next eight roulette spins are black)= (48 = U02 53:
(E) Pr(no severe and at most one moderate)
= (0.5)? +0.5-0.4+0.4-0.5 =.65.
minor
moderate
severe
0.5
0.4
First Accident
Second Accident”!
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
2-55
(A) Let B denote the number of blue balls in urn 2.
Pr(both ball same color) =.44 = (0.4)-
16
16+B
B
+ (0.6):
_ 6.4+.6B
[6% Bael6e
Solve for B to find there are four blue balls in urn 2.
16
16+8B
16
16+B
a4
10
Red ball
Blue ball
B
Waa
Red ball
16+8B
Blue ball
16+B
16
Blue ball
2
ae
10
B
rm
Urn 2
Section 2.5
2-56
Bayes’ Theorem
Draw the standard tree diagram to track probabilities.
~
51
Ace
pais
on
52
Non-Ace
48
uy
3
2
Non-Ace
51
Sade
51
First card
Non-Ace
A
51
Second card
:
number of aces
-
Pr(first card is an ace) =
(b)
Pr(second card is an ace}first card not ace) =
number ofcards
4
a
(a)
52
35
36 @ SOLUTIONS TO EXERCISES —CHAPTER 2
(c) Pr(first card is an ace|second card not ace)
Pr(first card is an ace ~ second card not ace)
Pr(second card not ace)
Be
°
AGS
AS
ip
Se
FAT
Sil
2-57
40
11Face
50
a
40
eh See
5D
Non-Face
Face
sy)
First card
Non-Face
Face
11
Ae
50
oh
51
on-Face
Non-F
39
50
Face
50
51
Non-Face
=
39
Face
rr
12
=
Second card
&
Oo
50
12
ae
5
Non-Face
Third card
Number of Face Cards
Probability
.F¢
A
0
9,880
22,100
9,360
22,100
2
2,640
22,100
+
4
220
22,100
2,860
(b)
Pr(at least 2 face cards) = F100"
(c)
Pr(3" card is face|first two cards are face cards)
= Pr(all three face cards |first two cards are face cards) = e
PROBABILITY AND STATISTICS WITH APPLICATIONS:
(d)
A PROBLEM SOLVING TEXT
# 37
Pr(all three face cards| at least two face cards)
___Pr(allthree face cards)
220
— 22,100 _
220
~ Pr(at least two face cards)
2,860
2,860°
22,100
(e)
Solution Method |: Tree diagram and Bayes’ Theorem
Pr(at least 2 face cards| last card has face)
a
Pr(at least 2 face cards ~ last card has face)
Pr(last card has face)
this is the probability that the
first and third cards have faces
.
iP Mim = 1 0Nt B40 gio
5ST 50 Bes 50) Oh oD
Py i Pane
ol Th eee
3) GS
i i
SS
ee
=
This equals S Does it make sense to you?
— 11880 _ 3999
30600
Solution Method 2: Using Combinations
Imagine that you know that the third card is a face card. That leaves 11 face cards in
a deck of 51 total cards. Now imagine dealing the first two cards from this reduced
deck. The question asks us to find the probability that at least one of these cards is a
face card.
40 C2 se Hey
Pr(at least one face) = 1— Pr(two non-face cards) = 1— NE
Ge
5102
38 @ SOLUTIONS TO EXERCISES— CHAPTER 2
Section 2.6
Credibility
2-58
Ay
0.10
Good
driver
Bad
driver
Type of driver
A
0.50
Accidentor
not in year|
Accident or not
in year2
(a) PrA. As) =.8201)* 2-5)?
Cha
=.058.
Pr( 42 OA
SoG Lyhed)
SA bee
pete
mee
(c) Pr(G|A ro yy
Pr(4,)
28.5 (1) ba2ed)
O58
LEE
18
90
ese
ater
boa
Pr( A, C
SS
29
>. (1a. 2
<
(d) Pr( BlAt least one accident) = ae
Lee aaa
.2-(1-.5*)+.8-(1-.9°)
302
497.
PG
A Az)
As A
Pr(A, 7 A>)
mel. )s + .2-(.5)-
008 imi
O58
Ales:
(e) PROTAro AAS
A>)
8 .(.9)2
he)
_
.8°(9)*4+.2:(5)e
2-9
a)
Pr(1997|1997 or 1998 or 1999) = =e
.16-
2-60
(D)
Draw atree diagram.
_ .6482 — 9284.
~ .698
ae
(.05) +.18-(.02) + .20-(.03
)
Pr(ultra-pre ferred dies) =
= 455.
(0.10) -(.001)
=.0141.
(0.10) -(.001) + (0.40) - (.005) + (0.50) -(.01)
a
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 39
2-61
(D)
Drawa
tree diagram.
(.16)(.08)
Pr( young adult collision) = (.08)(.15) +(.16)(.08) + (.45)(.04) +(.31)(.05) —
2-62
(D)
Let xdenote the probability that a non-smoker dies.
Dies
Heavy
20%
Smoker
|.
Lives
1—4x
Dies
Smoker
4x
2x
30%
Lives
1-—2x
NonSmoker
50%
Type of smoker
l-x
Live or Die?
fi
fo ce
Pr( heavy smoker}died) = (.2)
+(.3)
(4
(2x)
x)
+(.5)x
lp
19
(.08)(.06)
it)1 lon)ios)
(B)
=
Pr(16—20jaccident) = (.08)(.06)
+ (.15)(.03) + (.49)(.02) + (.28)(.04)
158
40
SOLUTIONS TO EXERCISES
Section 2.7
- CHAPTER 2
Chapter 2 Sample Examination
1.
A
Z.
B
Pr(dealt a fullh
NE
Cus)
houses
eeber of full
ee
number of 5 card hands
me 13Ci C519
Ci aC a
3744
52Cs
3.
2598960 ©
You may wish to create a Venn diagram. The key is to note that that you must have
some insurance to be a client.
Auto Insurance
No Auto Insurance
Dismemberment | No
Insurance
Dismemberment
il7/
45
(a) N(both auto and dismemberment) =17.
(b) N(exactly one kind of insurance) = 65 = 2( a aw NM
(c) No, because Pr(auto ™
dismember)=
17
62
go * 95" 89
= Pr(auto) - Pr(dismember).
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
@ 41
(a) Venn Diagram
U =College class
Engineers
Female
Venn Box Diagram
(b) Pr(female > engineer) = a
oo
(c) Pr(femaleU engineer) = ca = pit
a
a2
by)
(d) Pr(female| non-engineer) = =.
(e) No, since Pr(male
Mengineer)
:
Ke
ea
aey)
# Pr(male)- Pr(engineer).
number of ways to be dealt 3 Kings
=
:
=
number of ways to be dealt 3 cards
ES
(b) Pr( NOT dealt 3 Kings) = 1
Cy
ce
Oy
22100"
9)
(c) Pr(3™ card King|I‘' and 2" card King) = av
(d) Pr(3 Kingsat least 2 Kings) =
4C3 ae
ancy
=ACTON
=
4C3
Z
52 C3
=
4
22100
e
42
6.
SOLUTIONS TO EXERCISES — CHAPTER A)
(a) There are 16 elements in the set.: The bold numbers denote a sum of 6.
‘a= {22,23,24,25,32,33,34,35,42,43,44, 45, 52,53, 54,55}.
(b) Pr(sum is nine) = =.
(c) Pr(at least one die is 5) =
(d) Pr(sum is 6at least one 5) = ws,
9°
(ec) Pr(at least one 5|sum is 6) = we
3
It
(a)
(b) Pr(sum is 4) =
(c) Pr(at least one 2) =
(d) Pr(sum is 4) ball from Urn 4 is 2)= Pr(ball from Urn B is 2) = i
st
(e) Pr(ball from Urn B is
2/sum is 4) =
~
=
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
@ 43
8.
Se] Coke
i |Ripple | Water |Red Bull|__|
eager [alte dt oaMeeSESEY
eee
[aes |e facia lela
Ma cl PE
arian SO Oe ese es Se 0!
|
b)
(b)
5
Pr(Red
Pr(Red Bull)
Bull) =—30
(c) Pr(female|
milk) = N(female
4 milk)
8
Namilk)
SG
d) Pr(male U milk) = 22.
307
(a) Pr) =07)
(b) Pr(A’)=0.6.
(Cy PH Arm.) =.255
(een
eyPr(AA B)
Pr(B)
(eyo
Pr AGB
ORCI
)\=HOs.
Pr(B
es
eerra A) _
(g) No, they are dependent.
‘40.
Pr(4 7 B) # Pr(A)- Pr(B).
10. Assume for definiteness that card with 5 credits is in the left (L) pocket and the 4 credit
card is in the right (R) pocket. She could either spend all 5 credits in Z, or spend one
from L and 4 from R, with the last one coming from R. These actions can be
displayed as the following 5 letter words using LZ and R:
LLLLL, LRRRR, RLRRR, RRLRR, RRRLR.
5
l
5
Each word has probability Gy , and so the answer 1s 5- +) ‘=
32
5
44
RL.
SOLUTIONS TO EXERCISES
—CHAPTER 2
Assume first that the left pocket is depleted leaving 3 credits in the right pocket.
This means a total of 6 credits are spent, | from the right pocket, 5 from left, and the
last one used is from the left pocket. This can be portrayed using 6 letter words with
1 Rand 5 L’s, ending in L. That means the first 5 positions in the word consist of 1 R
and 4 L’s. There are
5;C; =5 such words.
Next, assume that the right pocket is depleted leaving 3 credits in the left pocket.
This means we have 6 letter words ending in R with 2 L’s and 3 R’s in the first 5
positions. There are sCz =10 such words.
122
Thus, the total probability is
Consider what can happen to the team that is leading 3 games to |.
50%
Series won 4-1
Win
50%
Series won 4-2
Win
Lose
50%
50%
Series won 4-3
50%
Series lost 3-4
Lose
Game 5
50%
Game 6, if
necessary
Lose
Game 7, if
necessary
Pr(win) =1—.5° =.875.
13;
Pr(win) =1—.43 =.936.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 45
14.
Column Total
pes
b=1-1/3=2/3 and a=Pr| AN B|=Pr| 4] B|Pr| B]= Tee
Lk Complete
the Box Diagram to give:
Row Total
B'
eS
Beal
13
Column Total | 2/3
y
1
Then’ Pra B)=1—2715=13
715.
tS.
Begin by writing out the sample space of possible dance couples.
Betty’s
Danielle’s
Dance Partner
Dance Partner
Danielle
Dan
Danielle
| Boris
Dan | Danielle | Andy
Andy | Danielle | Boris
Pr(3 wives dance with non-spouse) = 2.
U = fabed,abdc,achd,acdb, adbc,adch, bacd, badc, bead, beda, bdac, bdca
cabd,cadb, chad ,chda, cdab, cdba, dabc, dach, dbac, dbca,dcab,dcba}.
Pr(all wives dances with non-spouse) =
Nae
tO
46 @ SOLUTIONS TO EXERCISES
We
-CHAPTER 2
There is one good young teacher.
U = math faculty at School University
Old
18.
Bad
Pr(first person wins)
Wei
Sea
Nae
‘On
5
Wy
AY
5
Red
bal
i
Se
4
Red
bal
White
4
ball
The
First ball selected
=|~
Red
bal
=
White
ball
IZ
Second ball selected,
if necessary.
sak,
l
White
ball
Third ball selected,
if necessary.
19.
(a)
i
2}
Pr(1* person wins) = Bama +|
6
6
7
-—+,..=40%
¢5
Use geometric series.
(Oy
> +
ers person wins) =.
G26
L978:
eos CT COeTera
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT ® 47
20.
To solve for x, we know:
a =Pr(A44 BAC|ANB)= PHA BAC) = ex
0G
3
Prodan yee ao ee
Pr(no risk factor| A)= =~ =.46.
Zzby,
(C)
(C) We have four equations with four unknown variables:
1
5
it
: a
, and w+x+y+z=l.
ee
4
Vg
et
e
12
e
Adding the first three equations together, we see
2D 22 =
=>
Xxytzy
=
|nN
—
Pr(no coverage)
=
w=1-(x+ y+z)=1 = = 5
48 @ SOLUTIONS TO EXERCISES —-CHAPTER 2
De.
Let A represent the event of heart disease. Let B represent the event of at least one
parent with heart disease. Then:
Pr( 4]B’) = —_ = 1712.
ZS,
(B)
P|
One Car |Multiple Cars | Row Total |
|
SPOnICACIeNs
|ae St en
ey
NSIS ports Cates |Seeea | Stae ee
eee
Column Total
a
70
Pr |Sports Car M Multiple Cars |
Pe |Sports Car |Multiple Cars |Pr |Multiple Cars |
(15%)(70%)
= 10.5%.
The completed Box Diagram 1s:
pS
One Car |Multiple Cars |Row Total |
Sports Car
9.5
No Sports
Car | 20.5
Pr(one car ~ no sports car) = 20.5%
24.
(B)
The Auto/Homeowner Box Diagram ts:
Thus, 35% have auto insurance only, 50%
both. Then
Pr(renew)=
(.50) +
auto only
(.40)
renew given
insure only auto
have homeowners only and 15%
~=+(.35)-(.60)+(.15)-(.80)
= 53%.
(D)
have
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 49
25%
Begin with the Emergency/Operating room Box Diagram.
Pr| E’\O') =1-Pr[ EVO] =1-85%
=15%.
Column Total
iso, by independence, 13% =" 25%7a)
=>
a = 00%.
The complete Box Diagram is:
Row Total
40
ee
Pr| O|=40%
26.
(D)
60
(D)
A tree diagram would be nice, but would have infinitely many branches. The
key is to figure out how many ways there can be a total of seven accidents in
two weeks. There could be 0 accidents the first week and 7 accidents the
second week. Or there could be | accident the first week and 6 accidents the
second week, and so forth. Let (a,b) be the pair representing a accidents the
first week and 4 accidents the second week.
Pr(exactly 7 accidents)
- Pr[(0,7) ]+ Pr[(1,6) ]+---+ Pr[(7,0) |
=
ie
gi
°
al
28
a
I
:
I
cay)
yi
prannerd
ee
ie
prob of ; 7
accidents in
Or i 2 o
irst week
second week
Sealy
feeet
28
.
ie
2!
=
eel
99
=
Saran
64
°F:
“ah
a
oe
=
ee
"y
——
’
et
:
ge <
ee
=
:
:
ee
Pe
(EE
ee ae
7
:
=
’
i.
tr
»
-
~
2
=
-,
=
af
CHAPTER 3: DISCRETE RANDOM VARIABLES
Section 3.1 Discrete Random Variables
3-1
Y denotes the number of starters among the three players that get off the bus.
Par)
LO phe se
eer
12C3
220
Z denotes the integer selected. “At random” means each integer has an equal chance
(probability) of being selected.
Equivalently,
is>)' ies)
Pr(Z =i) = 5 fora
6
1 0}
18
38
(b) The first two throws must miss, and the third throw must hit.
(a)
PriS$ =1)=
ae
Ps =.) -(#) (43)
y
(c) p(s
5—]
=s)=( 22] (&) [Oty
= l,2, 50°".
>
(a) This is exactly the same probability distribution as Exercise 3-3.
18
(b) For example, shoot a 3-pointer until you make one, where p= —— 1s the
38
probability of success (making a 3-point shot).
Di
52 @ SOLUTIONS TO EXERCISES - CHAPTER THREE
Section 3.2 Cumulative Probability Distribution
3-5
Value Of The Random Variable | Probability |Cumulative Probability
10/36
6/36
0
2
4
6
8
10
Sum of Dice
Cumulative Distribution Function
Sum of Dice
12
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
Bat
a) Pr = hyenG, (0.3) *-(0.7)* for #=0, 1.2.3, 4:
cpr |o0ost |0.0837 [03483 |0.7599 |1.0000|
3-8
Let
H be the number of heads.
=
Pri)
Pr[H >1] = 1—Pr[H =0] = 1-— = - * Then,
Pr sr
ST IRR
C3
|
t AY
SAS
7
it) ole
is
F(t)=Pr(T <n)
3-9
0.75 —0.40 =35%
of the homes cost between 4 and % million dollars.
Probability
Cumulative
0
200
400
600
800
1000
House Prices (in 1,000s)
Section 3.3 Measures of Central Tendency
3-10
E[W]=0-
5C2 +]- 30 +2: LS
11 C2
55
55
=
60 .
55
This is a little more than 1, which is
reasonable because the female party-goers outnumber the males by 6 to 5.
ii 28 Bee ED
Pere te as a 55
# 53
54
SOLUTIONS TO EXERCISES - CHAPTER THREE
Satie
I OMe
117,181a4,724
Toe —
;
ve
Seal
e
ee
Iv
18
SOc aah aST aeaes
VIR Verhsins ba“ IRROLIRYION
50,361,822 Oo
= in
en ;
Bea
3-12
Since the probabilities must sum to one, © =.04.
E[X]] = 32-(.16)+39-(.04) + 45-(.42) +57-(.10) + 62-(.28) = 48.64.
3-13
We need to determine the discrete probabilities from the cumulative distribution.
hice Xie [eae] <0 bol aaa Seale Ge
Perey Pe ptrfasPas 18|05
E[X] = —2-(.12)+0-(.11)+1-(.25)+3-(.28)+5-(.18)+6-(.06) = 2.11.
= =<© <43 o) =} eae II
iS=
Za = EM |
=
5
1(.2)+ 2(x-—.2)+3(.8-—x)
+ 4(.2)
sey
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
ay
F kas T
5
4
3
2
OQist
O0000 ree
15 7300 5 lela
15 738 2rye
l
pit
5
= 220,
3-17
(C)
(a) A starting salary of 45,500 corresponds to the third data point.
3
= 60” percentile.
4+]
(b)
Median =
43,412 + 45,500
= 44° 456.
2
(c) Find the 66.67" percentile.
=f =
in35. Then
wt
|
X3.33 = 45,500 + (53,
750 — 45,500) = 48,250 . Top third starts at 48,250.
(d)
The 60" percentile is 45,500.
The so" percentile is 53,760.
49,000 — 45,500 ~ 47 4%.
53,750 — 45,500
A starting salary of 49,000 corresponds to the 60 + (.42)-(80—60)
percentile.
3-18
(a) 25'"to 75" percentile.
(b) Linear interpolation using data points x, and x7. Namely,
15
— Xx).
25 —~
xy +—(
this will be
negative
= 68.48"
# 55
56 @ SOLUTIONS TO EXERCISES - CHAPTER THREE
3-19
We begin by deconstructing the cumulative distribution function F(x) to find
POS Pegabe |PRA 1)€9°6 By OG See Fee pees oh
FE | Re
(a) ELX] = 4-(.2)+6-(.2)+6.5:(.4)+7-(.1D+7.5-(.1)
= 6.05.
(b) The median is Q) =6.5.
(c) The mode is 6.5.
(d) The midrange is “min =*mx.
aS S575:
(e) The third quartile is QO; =6.5.
3-20
(a) See solution to Exercise 3-5.
l
2
l
= 2-——-4+3-—+---4+12:—
= 7.
b)-E
eet
a
Bee
eae
(c) Median, mode, midrange are all seven.
random variable.
(d)
3-21
O; =5
and
This is an example of a symmetric
O; =),
Let N be the number of distinct pairs of the form (a,b) , where a and 5 were dance
partners at some point. Then the mean number of distinct dance partners for a boy is
ine
ree
ne
ie
and the mean number of dance partners for a girl is
m
implies
mplies thatthat
ee
Pe
3.5=——.
n
This
ie
———=—
—— =—.
ey
In other words, the ratio m:n
2°|
ofgirls to boys is a2 ee
eT BR
po
Section 3.4
3-22
Measures of Dispersion
_ Lacrosse player heights (in cm) 1n ascending order:
(aj (¢)
1S 3erl64
51627-1624,
168, eed
181,
Ess
——
—
——
ene)
minimum
QO,
median=Q)
QO;
maximum
aS
mode
Mean height is (153+ 154+162+162+168+181+183)/7
Midrange is
183 +153
9
= 166.14cm.
= 168
“a
(d) The first 162 corresponds to the 37.5" percentile. The second 162 corresponds
to the 50™ percentile. Linear interpolation verifies that 162 will correspond to all
percentiles in between the 37.5" and 50" percentile.
(ec) Range = maximum - minimum
IOR = 0O3—
QO) = "270m.
= x7 —x,
= 183cem—153cm
= 30cm.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 57
5-23
1
3
3
1
12
(a) E[G]
E[G] = 0-—+1-—+2-=—4+3-—=
SaRehG Ge = age—
|
=: (7 -—
EIG?
[G*]
3
3
{\
Sie
oy ate
at iReese
gt? Ye gt
3
(b) VariG)=3=1,5"
=15,
3
= 0.75.
The standard deviation for the number of girls in a 3 child family is og = y.75.
3-24
We already found (Exercise 3-20) that the mean for the sum is E[S]= 1s =7.
Why compute it again?
l
2
oe}
ES
Ps ee Seibaaeh ee
nl36° gre [ee
36 254
VartS] = E[S?|-(E[S])" = <2 -49 ss2 E583
3-25
OF
——
min
alae)
—
—O7j
»
Be
me
median
een
ong Oe
a
O03
nn“
max
ae
mode
1
2
3
1
l
as
1
E[Classes] == 0 0-—+]-—+2-—
The Tt
ee +3. or a: “it 8-— Fr +24. 7 =
2
Midrange = at
56
1
= Ge
Inter-quartile range = 8-1 = 7
E|Classes“]earl
= 0°-——- +1 CoinaeGyeda” ue1]
eal
11
2,2
1]
es
ie
Var| Classes |= 42.45.
3-26
§=9E[X] = 32-(.16)+39-(.04)
+45 -(.42)+57-(.10)+62-(.28)
= 48.64.
E[X2] = 322 -(.16)+392 -(.04) +45? -(.42)+57? -(.10)
+ 627 -(.28) = 2476.4.
Var[ X]=110.55.
3-27
+23-(.01)
+22:(.05)9)
9-(.2
+20-(.38) +21-(.09)
(a) E[X] = 18-(.18)+1
= 19.57.
+ 222 -(.05) +23" -(.0D.
+21? -(.09)
E[X?] = 187 -(.18) +19 (29) +202 - (.38)
(b) Var[X]=1.2051.
58 @ SOLUTIONS TO EXERCISES - CHAPTER THREE
3-28
.(a)(-b).-
48
,59,
69, 69, 78,78,
SS
84, 87, 93,
Se
min
—
93,;..96,°-102
—>
ad
median
10
_ 784+84
eae
trae
5)
as
=169"
percentile
E({Exam Score] = 795.
There is no mode (3 values occur twice).
Midrange = ie
— a
(c) 80 percentile corresponds to the i = .80(12+1) = 10.4” data point.
X10.4 = 93+.4(96-93) = 94.2.
(d)
3-29
Oxam Score = 15.429.
E{Claim Size] = 20-(.15)+30-(.10)
+40-(.05) +50-(.20)
+ 60-(.10)
+70-(.10)+80-(.30) = 55.
E[{Claim Size2] = 20? -(.15) +302 -(.10)
+402 - (.05) +50? - (.20)
+ 60? -(.10) +702 -(.10)
+802 -(.30) = 3500.
Thus, oy =21.8. Pr[55-21.8< Claim Size <55+21.8|
= Pri 33.2 < Claim Size < 76.8 |
Pr| 40 < Claim Size <70|=.05+.20+.10+.10
= A5%,
3-30
(A)
The median is 50, so Pr(28.2 <claim size < 71.8) =55%.
The mode is 80, the percentage of claims between 58.2 and 101.8 is 50%.
BS
a) zg= AAI
1.5625,
(b) 10° =100—lh75(16)
3-32
= 72.
Ryan’s apple is z ed
= eee
2.848
een
eee ar}
Jen’s orange is z= Fag
ge
Both have impressive fruit, Jen’s orange being a little (relatively) larger.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 59
Number of standard
deviations from the mean
gee
ear
deviations of the mean
3-34
rede
elo oo 7s eee)
(oo984 | 9999
For the apple, the coefficient of variation is 100° 2.865 9,= 250%.
11.19
For the orange, the coefficient of variation is Seren =33.)%.
3-35
{sum of two fair dice] =7
and Ogum =2.415.
The coefficient of variation for the sum is Wee,
=34 5%,
Section 3.5 Conditional Expectation and Variance
3-36
Probability
Cubs win in 3 games
(.59)3
Cubs win in 4 games
3-(.59)3(.41)
Cubs win in 5 games
Cardinals win in 5
aACye(59)2( 41)"
6 -(.59)2(.41)3
Cardinals win in 4
3-(.59)(.41)?
Cardinals win in 3
(a) ELX]=31(.59)
+((41)5} +44ee
3(.59)3((41)
|
te +3(.59)(41)}
aa eee
probability that 3
games are played
probability that 4
games are played
" 5|6(.59)5(.41) + 6(.59)2(.41)3
mo ability that 5
games are played
3(.27430)
+ 4(.35109)
+ 5(.37461)= 4.1
EL X?] = 3° (.27430) + 47(.35109) + 5? (.37461) =17.45139.
Oxy =.799.
60
SOLUTIONS TO EXERCISES - CHAPTER THREE
(b) If the St. Louis Cardinals win game 1, then we have the following (conditional)
probability distribution:
Probability
|
Cubs win in 4 games | (,59)3
US379
Cubs win in 5 games | 3.(.59)3(.41)
| .252616
Cardinals win in 5
3-(.59)2(.41)? |.175547
Cardinals win in 4
2:(.59)\(.41)2_
Cardinals win in 3
(.41)?
| .198358
1681
E\X |Cardinals win game 1]
=3
{1681}
:
+4
{403737}
HO
E[ X? |Cardinals win
+5
{428163}
SSS
probability that 3
games are played
probability that 4
games are played
=4.26
See
probability that 5
games are played
game 1]= 37 -{.1681}if + 42 - {.403737}
!
J + 52 - {.428163} s
=. LOO ae
oxG ardinals win game 1 — * EAs
3-37
Let Hbe the event the husband survives and let W be the event the wife survives.
From the given information we have:
obs
es
W | .960 | .025 |andthe completed|W
w' | .O10
ae oe ee
Box Diagram is:
The conditional distribution given H is in the first column.
Let X be the excess of premiums over claims.
The conditional distribution of Xgiven His:
Event
xX
Probability
How
1000
OF
H AW’ | 1000-10,000 | .01/.97
1.00
E|X|H]
[X|A]
0.01
= ((1000-10,000) =
area
7
SS
excess if
wife dies
conditional
probability
wife dies
0.96
LOOO Wear
pais excess
=
"0.97
if
wife lives
690.91
;
eocn (5
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
3-38
@ 61
The 11 relevant outcomes from the 6 by 6 dice table are:
(4,1)
(4,2)
(1,4)
(2,4)
©G,4)
(4,3)
(44)
(5,4)
(64)
(4,5)
(4,6)
2
2
E{Sumlat least one 4] = 5 austere:
3-39
Let A be the amount of the loss.
06
03
E[A|A>0]
Tat 00
70
A|\A
== 500500-—+1000-——
a2
l
eae
2
2
82
AE ee
008
001
001
siaide
are+ 100,000 -———
70
+10, 000 - ——
+ 50,000 00- -——
= 2900.
3-40
(a) Let
x
=Pr(D=5000).
600 = 0-(.82) + 1000(.10) + 5,000(x) + 10, 000(.08 — x)
ganado
2
Eee | 225 ESTO 0G
(b) E[D?] = 02-(.82)
+1,0002 -(.10) + 5,0002 - (.06) + 10,0002 - (.02)
= 3,600,000.
op =/3,600,000 — 6007 =1800.
10
06
02
(c) E[D|D > 0] = "1000s (Wes, 000-[16) +10, 000: [
7
BO 3833,33.
are
10
06
a2.
(2
(
dy) EID* D> 0] = 1,000 (42 )+5,0007 (}- 10,0002
Mei
= 20,000,000.
OFp\p>=2981.42.
62
SOLUTIONS TO EXERCISES - CHAPTER THREE
Section 3.6 Jointly Distributed Random Variables (Round 1)
3-4]
Probability
Probabilit
Ets
ade
| fe
|
|
(b)
b) E[Y]
EY) = Oseot
a ee
Var[y
|= el
;
oael
ey
eel = 02 ee
Beer
|
:
Cola ee tants
ly
2
(ec) FIV+Y] = 0-41
+2 Ae 43 eA er hs, CNotethis equals say 1 E1Y)
haar
ECMEY)?
| 0 en
;
4
eee
pea
St
VarlX
+Y | = 3.5—15*
=:1.25,
oR
eee
(d) 1.25#.25+.50.
3-42
(a) We need to verify that
PrL¥ =xOY=y)
x =1,2,:--,6 and y=1,2,:--,6.
=
PriX =x)-PrY=y)
Yes, they are independent.
(b) By independence, Var|X
+ Y] = Var|X]+Var[Y]
calculate Var_X] = 1? eee
6
for
eee
6
= 5—.
= 2.916.
Of course we need to
bg SOby bg
ht
Se
bl
X+Y
Probabilit
4/36
12/36
13/36
6/36
E[X +Y]
RRSIES
> ><S
—
Se
Seby
—
be
(c)
~
|
l
l -—+2-—=].
~
4
(a) ELX] ={) -_—+
4
San
E
=
—
~
Probabilit
i
T
—
.
=FPTrarTryTr
—
ileell vr
Stes
a
SPTarT
——
ee
— Il
SN
3-43
—
Tare
ST
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
i fot° 2 3 ra {=
# 63
64
SOLUTIONS TO EXERCISES - CHAPTER THREE
(d) Yes, the joint distribution table below shows thatX and Y are independent.
The joint distribution for X and Y 1s given by:
Pie)
7)
at
0
1
2
1
pate)
3-44
(a)
1/8
e+
i)
e+
e
rt
bo
WI}
l
3/
3/
1/
(b) EX] = 0-5(oat
41-5
(c) EX, 4+Xo Xs)
| AU
ALIS es
=
|bo
to
Var Xt Xo tks).
“Su 5 Varlaeelt
rartxe lePareye) =
is
ee
+
5
4
by
independence
Or we could use the probability distribution for XY, + X> + X3.
Var[X;+X2+X3] = Bl(Xy+X2+X3)?]-(E(X1+X2+X3))
= 3-1,5? = 0.75.
+
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 65
3-45
Shoe Size
25
a
ee ieee
i
=
=
2
(a) E[Height] = 68-(.25+.05)+70-(.32)
+ 73-(.38) = 70.54.
E{Shoe Size] = 8.5-(.60)+12-(.40) = 9.9.
(b) E[Shoe Size Height = 73]= 8.5. =a)
(3) STs
(c) E[Shoe Size|Height =73]= 8.5- 3+£12. (:
= = 9.083.
= Shp ssOe (3s
(¥)
= 84.2083.
30
30
E[Shoe Size?
Var[Shoe Size|Height = 73] = 1.701.
Conditional coefficient of variation is 100-
NS
083
eur
Section 3.7 The Probability Generating Function
346
(a) ns) = (5+ = RtostSst ais)
3
ieee
g
innar =a
Vee
B-47
~
es?
g
(1+)? is
A's) = 2a}
8
243-(3| =k
(a) hytsy=5-C7) (3475)
EB lahe
(1) =3 5:
(b) A&(s) = 20-(.7)? -(.3+.7s)3
hp(1)=9.8
Var[B] = hg(l)+hp(l)—(hg()) = 9.8+3.5-3.5°
(c) hp(s) = (3+.78)> = (.3)°
= 1.05.
+5-(.3)4 +675) +10-¢.3)?
(75)?
+10-(3)2 «(.7s)? +5-(.3)' -.7s)4 +€7sy°
60
40
1.00
66 ® SOLUTIONS TO EXERCISES - CHAPTER THREE
So the probability distribution for the random variable B is:
LB
_| Probability |
a a ee
Note: This is the probability generating function for a binomial random variable
(wait for Section 4.2) with probability of success p=.7 and number of trials
Peale),
3-48
_ (-.21)-(8)-G8))-(2)
(a) hg (s) = aaa]
77 aaa
_-
8
Geant
E[G]
oe.
3
eee
ee
ORIG
Ye errs
559
CVA aoe
ars
a
Note: This is the probability generating function for a geometric random variable
(wait for Section 4.3) with probability of success p=.8.
Section 3.8
Im
(a)
Chapter 3 Sample Examination
Pr(S <s)
s, sum of two fair die
2?
2
3
4
5
6
7
8
9
10
1/2
2/25
3/25
4/25
5/15
4/25
3/25
2/25
1/25
(n
(b) E[S]=6.
(c) E[S?2]=40, o2 = 40-62 = 4, os = ,/40-6? = 2.
2
3/25
625
10/25
15/25
19/25
22/25
24/25
25
—-
4]
_|
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
67
1
2 eee,
3
4
5
Am
98
ee eae ee
re!
(Ou
alc meomme TOR 08"tySa
19 ~ jon
CM
eee
eon Sie1 a 2 a A 3 ee
lhl
(Oe
elon oe
546
eee 74
19
a4
Be SrIRSh
ao
Fe5
ig
4
le ot
10
Var[S
|S <8] = 28.74-5.16? = 2.13, ofS|S<8]=2.13 =1.46
1
with probability +
X =0
a
heads
=
1
De
X =1
with probability
heads
tails
@2
-;
heads
tails
rf)
2
tails
aii
D
(ate
nO
eo
OPO
S078).
MIN
17, fe 8, 5 O50 0,9509, 95.9; Dy 9, 10
SS
median
FT
mode
ees
max
=(7+7)/2
pyre
as
Midrange =
a
=e
(c) 80" percentile corresponds to the i = (80%)(24+1) = 20” data point.
(d) oy =25938:.
(c)
”)
CV cheeseburgers =
Z=
Z
11-6.5
Ahi
2.598
/
= 39.97%.
x29 =9.
68 ® SOLUTIONS TO EXERCISES - CHAPTER THREE
4.
(a) ELX] = 0-(.74)+50-(.12)+100-(.09)
+200-(.05) = 25.
(6) oF =3200—= 257 = 2575,
oy =V5200—5 650.74.
(c) Since the expected value after a year is 25 the prudent investor should not buy
the stock at 30.
(d) E[10-X¥ —300]=10- ELX]-300 =-$50.
(ec) Var{10-X-300] = 102-Var[X] = 100-2,575 = 257,500
25
49.3%,
(f) CVy = ae
—
ys
=
ae
0
Geometric series.
Expected
Claim
f=0
1000
1000(.95)
1000(.95)?
ae
f=2
f=
vee
1000(.95)78
f=29
1000(.95)??
t=30
The expected claim for month ¢ is 1000(.95)‘" and,
Expected total Payments = 1000 + 1000(.95)
+ 1000(.95)? +--- +1000(.95)??
as
e\30
awlegi 15,707.
7 000) 1—(.95)
S|S is prime
fe
(a)
E[S|S is prime] = 6.13
(b) 0 =2.5027.
cS
and
Var{S|S is prime] = 6.26.
(D)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 69
30
hit
30
miss
hit
;
hit
30
70
miss
—
4
>
hit
70
Y~
miss
HS)
:
DS
miss
hit
5
30
A
miss
TY)
.70
miss
5
hit
Probability
Chee
= 39375
C3) )Cs>)
= (25)07)7
(INCITS \25)
= 41125
(3200)
= (3)CC25)
+ (.7)(.25)(.3)
= 168
(.3)? =.027
(a) Efhits] = 0-(.39375)+1-(.41125) +2-(.168)+3-(.027) = 0.82825.
E{hits?] = 02 -(.39375)
+12-(.41125) +2? -(.168)
+32 -(.027) = 1.32625.
(Dy Ope
(a)
= v1.32625—.82825"
= 0.8.
22-(.05) + 23-(.01) = 19.47.
20-(.41)
+ 21-(.05) +
ELX] = 18-(.23)+19-(.25)+
Median is 20.
Modal class is 20.
Midrange is “ee
70 ® SOLUTIONS TO EXERCISES - CHAPTER THREE -
(b) Range is 23-18=5 years.
E[X2] = 182 -(.23) +192 -(.25) +202 -((41)
+ 21? °(.05)
+ 22? -(.05) +237 -(.01) = 380.31.
Var| X|=1:2291,
oy =1.1086.
Al
.05
.05
01
(c) E[X
|X >20] = 20-[)
+2198)
+22 [-B)+23-(-2)
= 20.346.
(a) ELX] = .10(0) +.18(1) +.12(2) +.37(5) +.23(8) = 4.11.
(b) ELX2] = .10(02) +.18(12) +.12(22) +.37(52) +.23(82) = 24.63.
oy = ¥24.63-4.1122 = 2.78.
(c) Q3=5.
Let p denote the probability that the damage is $2,500.
Solve 700 = E[ damage] =0-(.73) + 2500- p + 5000 - (.27 — p).
oN
(a) E[damage |damage > 0] = 2500-52 + 5000-2 = 292.59.
2
.26
(b) E[damage? damage > 0] =2 500° an
© damageldamage>0 a
(a) fy
te
5000- - oT
Ade ko:
= ELX] = 12.09) + 2(.32)+3(.42)+4(.13)+5(.04)
= 2.71.
The 50" percentile, median, is Y = 3.
E[X?] = 17(.09) + 27(.32) + 37(.42) + 47(.13)+57(.04)
Gy =8.23-2.717
(b)
£[Payment
= 8.23.
=.941,
at least
3 TVs].: = $75:2-2=
alee
e025 oe
GTeS
ines
iy
59
eS)
Nn
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 71
LZ
(aye
elses tel 4el42 15.91
=<
eS)
a
median
ae
mode
=(14+14)/2
O71 = X3,25 =13.
5153.15 9868 18
Sa
Hntebs
O; = X075 — 4ies
Midrange is 15.
jis
RY Wee
(b) JOR =15—13
=2.
1)
__ 10-14.416
— | 2.845.
eea0ee
Not an outlier.
E{merit money] = 0-(.09) +500-(.61) + 1000-(.15) +2500 -(.09)
+3750-(.05)
+5000-(.01) = $917.50.
E|(merit money)? |= 02 (.09) + 5002 -(.61) +1000? - (.15) + 2500? - (.09)
+ 37502 -(.05)
+5000? -(.01) = 1818125.
O merit money = $988.09.
Let p denote the probability of winning $10. Let W be the random value of your
winnings.
(CS
Ge
ae
E(W]-= 0-(.90—p)+10-(p)+100-(.10)
Var[W] = 100p +1000—(10p+10)?
= 10p +10.
= 73(10p +10).
Solving for p we have p=.2.
(a) E[W]=12 and Var[W]=876.
9)
(b) EW |W > 0] - 10-4100: = $40.
(Cc) CVy
=
12
876
= 40.544%.
Ty
72
155
SOLUTIONS TO EXERCISES - CHAPTER THREE
SEIR SSE A) = SCS
202 2) ane
Gi eo,
(48)
You might earn a raise if you informed your boss that
hp(s) = (.8+.2s)3
83 +3-.87 -.25+3-.8-(.25)? +(.2s)
= .512+.384s +.09657 + .008s7
So the discrete random variable R has probability distribution
eS
Eee
CC
16.
eee eo
ee ee
Let x be the common difference between consecutive probabilities, po, p}.---, Ds From (1) po is the largest, with the remainder of the probabilities arithmetically
decreasing in size.
From (i) and (11) together we have,
Mi
PossPe
CPE
Pe:
Then,
Py =(ParX;
and imeenetalsy
First,
ase
7, =
Dia ps =
Pp = Pix
Dy iis
Po
s,s
+ (Po
= (pox)— x — po
OL ee
—X) +++: + (po
P|
Second, from (iii),
2
= potp
2x,
—5x)
=
6 po
—15x.
Ds
= pot (po-x)
= 2po-X.
Solve simultaneously the two equations, 6p9 —15x = 1 and 2pp —x
5
Po aT
:
= 3 to get,
l
and x svat Then,
Pr| #Claims > 3 | =
pat ps
=(Po —4x) +(po —4y) = nee
fet
(C)
CHAPTER 4: SOME DISCRETE DISTRIBUTIONS
Section 4.1 The Discrete Uniform Distribution
4-]
(b) To prove i
= 14+24+3+--4n
n(n+l)
5
=
i=]
by mathematical induction, we need to
1
I. Verify that the statement is true for n =1.
Me = [=
=vk
i=l
II. Assume that the statement is true for n =k.
ew
tiie: eyseal
III. Show that the statement is true for
[+2+34ek
+ (ke
= 9)
n=k +1.
(en
= oe.
(c) The statement is clearly true for the case n=1.
Assume that the statement is true for
Pepe ay
n=k.
k(k+1)(2k+1)
6
iy aay
Then,
+ (k+l)?
(k+1)| k(2k-+1) + 6(k+1) |
6
(d)
showing the statement is true for
6
n=k +1.
The statement is clearly true for
N =0.
;
ataxt+ax* +ace+---+ax Rote ay
cee
6
Assume true for
k+l
ad—x*™")
l-x
+
l-x
a(l—x***)
l-x
showing the statement is true for
ee
1104
i =
Il
+
s|-Sei
—
+
s|-— Saw
—
We)
s|-
a
+
Se
to a ee) +
SS)
ars
+
N=k +1.
oo
“fe =
S
se!
Tal
—
———
\|
n-(nti)
wn
9
N =k.
K+
ax*™(1-x)
Shue)
9
Then,
74
4-3
SOLUTIONS TO EXERCISES - CHAPTER 4
(b) The total accumulation of rice is 1+2+27+---+2°
= 2° —1 (geometric
series), which is a bunch more than one billion grains of rice.
.
4-4
2
n?(n+1)*
We make use of the algebraic power sum formula | es =—_———..
| k=l
es aes
~
1
EV Xe | =o)
Uke
Ea
2
n
n?(n+l)*
ee
=
n(n+1)?
See
n
4
Then,
;
One may expand
E|(X-p)3| = E|X3-3uxX? +3n2X
-13|
(APG RVI BC APREIILIA
eye
II
= E[X?)-3uE[X7]+3u3 -pwe = ELX?]-3MELX2]4+2°
This can be evaluated using the formulas,
errr
vite
a
FLY?)
a
ee
and
3
(AO Ga i
n(n+l);
After a little algebra the result is seen to be zero.
Or, one may simply observe that the values of X — w (and hence (Y—y)*) that are
negative exactly cancel out the values of X — wu (and hence (Y¥—y)*) that are
positive. Thus, the expected value, by this symmetry, must equal zero.
Discrete uniform distribution on {1,2,3,4,5,6'.
(b) EX] =
= 3.5,
(c) There are 6 data points, the median is
(d) oy =
67 -1
12
3+4
=3
2
Lo)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
4-7
(a) Pr[(X =69) U(X =70)U---U(X
=85)] = a
(b) E[X]=
100 +1
2
= ehh,
(c) There is no mode. All 100 outcomes are equally likely.
ip eae 100? -1
4-8
(a) Pr[Y 25] =<. Let
X
=¥+1. Then X is discrete uniform on (Oop
(b) E[Y]=ELX -1)=ELX]- £11] =73h 124
(c) Chy
4-9
2
(7)
= — = 64.55%.
(a) Pr(Z>5] = 1-+ = z.
(yFIZ pe BI2 eX] = 2° RIX Ie 224) P10)
(c) Var{Z] = Var[2-X] = 2? -Var[X] = +
5] = 26.6.
oz =5.164.
4-10
+45] = 5E[X]+45 = 5(3)+45 = 60.
-
xk b>) = Pr|A>=
leer | = Pian
Il.
EY] = Ela-X +6] = a-E[X]+b
iy
Vary
|= Vvarfa-X +b) = a* -Var[| X].
—h
|
9)
# 75
76
SOLUTIONS TO EXERCISES - CHAPTER 4
Al?
a) PrLy is prime] ==,
(b) PrLX >11]=30.
Section 4.2 Bernoulli Trials and the Binomial Distribution
4-13
E[M] = 0-(.1184)+1-(.3681)
+2: (.3816)+3-(.1319) = 1.527.
E[M2] = 02-(.1184)+1? -(.3681) +22 -(.3816) +32 -(.1319) = 3.0816.
o3, = Var[M] = E[M?]-(E[M]) = .749757.
Ay WewePrMe=kh) ="35 Ce 8) 02)
fork = 0,2 roi
ELM |:=A2"(8) =.9.6
Var[M] = 12-(.8)(.2) = 1.92.
Pr[M <10 = 1—Pr[M =12]—Pr[M =11]—Pr[M =10]
= 1-.0687 —.2061—.2835 =.4417.
toa
10)"
19, = a0 Cie G2) sy ee
Pal Ye> 2 li = op Cond)
$8)" ona 2) G8)
E\Y| = 20-(2)i=4
Var\Yo) 92022). C8)
= O88:
4-16
Pr[ Yankees win 2 or 3 games] = 3C> -(.6)? -(.4)'+(.6)? = 64.8%.
4-17
Sy ueaa
aera
Pr[2 < X]=1-Pr[X =0]-Pr[X =1]
=1-3C,-(.9)!-(.1)? — 3p -(.9)° -(.1)? = 97.2%.
E[X]=3-(.9)=2.7 and Var[X]=3:-(.9):(.1) =.27.
PROBABILITY AND STATISTICS WITH APPLICATIONS:
4-19
A PROBLEM SOLVING TEXT
® 77
A twenty question multiple-choice particle physics examination written in the
language of Lower Lausatian, each question with four answer choices.
PrX=2] = 20C2-(.25) -(.75)
TEP ET Ny 00-025) = 5
Pr[X <p] = Pr[X <5] = Pr[X =0]+Pr[X =1]+---+ Pr[X =4] = .4148
OF
20825015
Oo; =3.75
4-20
Draw a tree diagram, where the first stage denotes from whence the shipment came
(X or X’). The second stage denotes the number N of defective vials (probabilities
from the binomial distribution), where we are given that
N =1.
Pr[X]=.2 and Pr[X']=.8.
Pr|N=1|X | = 30C-.1!-.929 and Pr[N=1|X"] = 30C;-.02)-.98.
Using Bayes’ Formula,
20+ (30 C, -.1! -.9??)
PrLX vee
s
20:+Go Gyeul! -.92?) +80:
4-21 = Pri[no shows =0
4-22
= 0957 (A)
or 1] = 32Co-(.1)°-(.9)?7 + 2) -C.)'-C9)?!
= .1564
Let N be the number of hurricanes.
Pry 02
4-23
-
66 G-.02'=.982?)
|= (.95)7° + a9C\ -(.05)! -(.95)!? + a9 C2 G05)" (.95)'®
5)-np=n- p-q=Var[X],
(a) (.25)-E[X]=(.2
7
so q=.25 and
p=.75.
ae
(b) PrLX =7]=10C7-(.75) -(.25) =.2503.
A24
-Pr(X >3.25) = 5Cy*(.65)* -(.35)' + 5Cs £(65) 7(bo)
="9245.
cette 8.703
(B)
78 @ SOLUTIONS TO EXERCISES - CHAPTER 4
4-25
hy(s) = (qt+p-s)".
hy(s) = n-p-(qt+p-s)"".
E[Y] = hy(l) = np.
hy(s) = n-(n-l)- p?-(qt+p-s)”*.
E[Y(Y-1)] = hy(l) = n-(n-1)-p?
Var[Y] lI= h'(y+h'ay—(h'))
n:(n—1)pp? +n-p—(n-p)?
= n?+p?—n-p?
+n: p-—n?- p*
n-p—np> =n-p-(l-p) = n- p-g.
4-26
Let N be the random variable that denotes the number of employees that achieve
high performance. NV has binomial distribution with parameters
n= 20 and p=2%.
The first few rows of the probability distribution are
E24
Biles
ad 7 Pe
as
Prv=n) |6076 |2725 |0528 |. |
Fo) [0675 [9401
|9929 |.
66.76% of the time, no employee would earn high performance, and C could be set
at anything.
Looking at the cumulative probabilities, 2 or fewer employees will earn high
performance 99.29% of the time, so the probability of exceeding two awards for high
performance is less than 1%. Thus, we can set
C=120/2=60. (D)
4-27
The only time a penalty occurs is if all 21 ticket holders show up.
Efrevenue] =
21-50
—-100-
(.98)*!
——
= 1050-—65.43 = 984.57.
(EB)
—
revenue from
sale of 21 tickets
probability all
21 people show
Section 4.3 The Geometric Distribution
4-28
Pr[D <2] = (.8)+(.2\.8) = 96%.
E\D]=-2=(25.
4-29
E{D*]= Var(D]+(E[D]) See
+{-
= 375.
The probability of success (not being rejected) is p=.10.
Pri R< 2) = GLC 9)GL) = 19%:
E[R] ==
|
9.
OR
=
Aas,
15
=
We are assuming that the geometric distribution applies, so in particular (1) your
advances in a pub are independent and (2) your probability of success remains
constant at 10%.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
4-30
Fi Macon failures] = 4 ais
a-31
Let px = Pr| X =n] = q"p. Then,
n
Pr| X >n| =
Pat
Pal t=
q’ptq™
pt
-=
aay
l-q
# 79
n
=
gor
P
=
Gs
Pr} (X¥ >a+b)n(X
=
Pr(X 2a+}|X >a) = Prix
2a+b)O(X2a)]
Pr(X >a)
P[X>a+b] = gt? Sg
=
Pr[X Da]
cage
iat cate
emesis eeepe
D
4
Pr{(X <1.5)] = Pr[X =0]+PrLX
=1] = (.4)+(.4)(.6) = 64%.
4-33
No. But you better have a reason.
(As in the previous exercise, the variance uniquely determines the value of p.)
peg
7—50 p implies g=(2)=.5848.
poet
Tey
\ gp
eee
ee
LX] =1.41.
stor y—1,2,3,---.
ee
el
p
p
p
Note: Var[Y]=VarLX +1] =Var[X]=—4
Pp
4-37
>°
This strategy will cause you to lose your bankroll if you have 4 consecutive failures.
4
Pr[lose bankroll] = (2 INCH
80
SOLUTIONS TO EXERCISES - CHAPTER 4
4-38
(a) For n=0,]1,2,---,48,
(b)
Distribution of X
=e
S
2
i)
=
i
10
20
30
40
Cards Drawn Before
(c) E[X]=9.6 and Var[X]=160-9.22 =75.36.
(d) If Gis the geometric distribution with p=
|
Nn
, then £[G]=
Aus
(Nnbho
Var[G]=156.
4-39
This is the alternate definition of the geometric distribution.
E[X]= =
iS
P
implies p=.08. Pr[ X =6]=(.92)5(.08)
=.053. (B)
4000
3000
2000
A
|
1000
E[ Benefit] = 4000-.4 + 3000-.6-.4+ 2000-.67 -.4+1000-.6% -.4 = 2694.4 (E)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT ® 81
4-41
Here Xis the trial of the first success, where success means rolling a 5. This is again
the alternative definition of the geometric distribution with p=1/6.
e[x]= Set (8-1 as
co
Thus,
n-l
Pp
Now, let
Z=X|Y=2.
Given Y =2, we know that the first roll was not a 6 and the
second roll was a 6 with certainty. Thus, either
Pee
Z=1 or Z >3.
Weeetsit2 3.4015) = 1/5,
Pr| Z = 2| =();
4
Pitz 3 Tei
3
>
a
I - .-( 4 \f 1
ie
(+4). Similarly,
TO
fistrol oer
rob
31droll
is not 5
rle~4l (56)
rz-s}-(s)es):
is 5
acs =gs203(3)e}-+(s)
aa)
-S(SPO+Oe- Od
and so forth.
Now,
|
of
we
n|— i
~S—Seee
This sum is by definition
the expected value of X +2
Ty
42)
5
5
1 (4 [eLx]+2|
= =+(=
= —+
2 (6 +2)
82 @ SOLUTIONS TO EXERCISES - CHAPTER 4
Section 4.4
4-42
The Negative Binomial Distribution |
Repeated differentiation of Aj (s)= p”(l-qgs)~” shows that
AO(s) = (rt+k-l(rtk-2):-r- p" -q* (-qs)-"*) for any integer k 20. Then,
A“)
(0) = (rtk-l)(r+k-2)---r- p” -q*, and the coefficient of s* in the power series
expansion of hy(s)= p’(1-qs)" is
WOO)
(reRaIKP+k—2
e r, ge
es
4-43
hy (s)= p’(-r\(-g)-qs)"",
PRI
= Pr] M=k|.
so ELM] =hy (I) = p’ (r\(q)-9)"* ao
hiy(s) = p’ (r\(q7\(r+)(1-gs)"7, so Ahy (1) = ae
and
Var(M] = hhy (1) +h) (Av)
_— OOH?) eg Pe?
p
és
rq?
+rq*
'
444
P
De
+rgp—r*q"
ts
p
=
=.
rq
oe
(a) 4, (35)2905)4(35) = 1087
(6) Cy 635) 065)2 Fa Cp C35)" 065)! 2 E35)? = 2352
(\ EI MV = yt iets ST and Pert]
Pp
(d
AAS)
rq? +rqp
—
Sie
p-
22)
E[Total Attempts] = £[M +3]=8.57.
(ay Pr y= 8) (Ce C8040)? = 00ed:
(bo) Ely] = 7-2 = 28,
(c) .oy-=/140 =)1,83
45
(d) Pr(¥ <45)= he
a
k=0
4-46
E[X]=6=—
2)
and Var[X]=12=—+
Pr(X <3) = (.5)o+4g
p-
implies p=.5, q=.5, and r=6.
Ci 35)" 05) 399 G3 G5) eat
a
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT.
4-47
Instead of px, =Pr(M =k) = +44-1C,4~p”-(l—p)*
[Ye
4-48
=
PrOh4-— Kk) =
ey er
aie Wes
for Kr
# 83
for k =0,1,2;---, we have
Opec,
Let success mean a month with no accidents. Then p= Be 4. Let E be the event
there are 4 successes before 4 failures. We need the probability of E.
Method I: Binomial Distribution.
in the first 7 trials. Thus,
Pr{E]
For E to occur there must be at least 4 successes
:
2, 7CEC4)* (6)
k=4
nA) 6) 94 Cs (4)> C6)? 4, Ce C4)° (0)
+(4) eno
Method II: Negative Binomial Distribution. Let
be the number of failures before
the 4" success. Then £ occurs if and only if M <4.
Pr[M <4] |= xyCO
(44)| 1+ (4)(.6) + (10)(.6)? + (20)(.6) |29
(D)
k=0
4-49
Let X be the number of hurricanes and let success mean damage occurs, with p=.4.
Let M be the number of failures before the second success.
Then X = M+2
problem calls for the mode of X. Mis negative binomial with,
Pr{[M =k] =
244-1C, -.6*" -.42. A partial probability table is:
k |Pr[ M =k]
iy eeakan team:
|
1920
jalan)
798
Ea
wey:
.
The mode of
M is | and so the mode of
X is 3. (B)
Note: It can be shown algebraically that Pr [M = k| is decreasing for k>2.
Section 4.5
H-300
The Hyper-geometric Distribution
“(aye Pr Dies) =
27Co + 14C5 F. 27Cy -14C4 Fi 27C2
+14C3
41Cs
41 Cs
4iCs
ae 2092
and the
84
4-51
SOLUTIONS TO EXERCISES - CHAPTER4
W denotes the number of women on the seven member committee.
Pr,
Se
(
4-52.
_
&
C6
a -34C
100 C7
100 C7
an
)
Cz
ga.
Cyee, ee
C
Oe
Oe
100 C7
Oa
100 C7
A denotes the number of administrators on the eight person committee.
the number of faculty. Note:
F =8-— A.
F denotes
E[A]=8(.75)=6. So, in a random selection
of committee members the expected number of administrators is 6. On the other
hand, again assuming a random assignment to this onerous committee work, we
have,
Pri Bs 4)
et
es
eae
= .1047
Comment to President Uptight: What do you mean by fair?
4-54
E[X]=E[Y]=£[Z]=2.
Var[X] = 5-(.4)-(.6) = 1.2,
vary)
gat= 5(22)
200)(53071)
49 1= 1.102,199 and
(3) [$F]
Var Ze 5-(1000
awe }{
on )(33] = 1.195.
) \ 1000 } \ 999
4-55
There are 10 defective modems, 6=30-20% from company A and 4=50-8%
company B. The probability that there are 2 defective from 5 selected is
MESES
he
WO
80 C5
Section 4.6 The Poisson Distribution
4-56
4-57
Pr(X =0)=.001007785
=e, so A =—In(.001007785) =6.9.
E[X]=2=6.9.
Worm
100Co (.03)°
| ,o0C} - (.03)!
Pr(Worm =k)
rcae
= .0476
-(.97)% |.2252 |.2275 |.1706 |.0074
= 1471
|
|
L
Poisson
ePes)"
e*:(@)E
Approximation
0!
1!
A = 100(.03) = 3
= 0498
= 1494
.2240 |.2240 |.1680 |.0081
from
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
4-58
@ 85
Let C denote the number of cars that arrive at the Wendy’s.
e 24. (24\3
0!
=.091
4-59
E[Z?] = 12 = Var[Z]+ (E[Z])" = A+A/?,
Solving this quadratic equation, we have 2=3 or 2 =~—4, but the parameter must be
positive.
3! e
Pr(Z <1) = Pr(Z=0)+Pr(Z=1) = e3 + = —~ 749. 0 AS9A
4-60
Consecutive Days of Rain|
E| Insurance Payment]
Insurance Payment
Probabili
2000
1219
= 0-(.5488)+1000(.3293) + 2000-(.1219)
= 573.1.
E(Unsurance Payment)” | = 07 -(.5488) + 10002 - (.3293) + 20002 - (.1219)
=9$16,893:
© InsurancePayment — NV 816893—573*
4-61
=699.
(B)
Number of Supreme
Court Vacancies, k
Actual number of
years from
(1790-1932)
with & vacancies
Poisson Prediction
of number of years
with k vacancies
14.3. p=4755
= 88.883
The number ofperiods is 1932-1789
= 143.
(87)(0) + (47)(1) +(6)(2)+ (3)(3) = 68.
The total number of vacancies is
The overall mean rate iS
A = 68/143 =.4755 vacancies per year .
/
Pel
(a) E[ Vacancies]
Pe= 0-775
87 a
Var[ Vacancies] = .4592.
47 ae 126
ae
ahve tO bes
lea
ATER
4755
and
86
4-62
SOLUTIONS TO EXERCISES - CHAPTER 4
(amet bom Hitetper sector=0 576
ee576
a
Var{bomb hits per sector] = 1.797 —(.92882) =.9342.
Bomb hits on
London, k
ee
see
Number of sectors
with k bomb hits
Poisson Prediction of}
576
eo©
93
35
1 97.6.)
30.2
ea ota
number of sectors
0!
with k bomb hits
= 220.3
210.2
The number of periods is 576. The total number of hits is
(229)(0) + (21 11) + (93)(2) + (35)(3) + (7)(4) = 535. The overall mean rate is
A =535/576 =.9288 hits per sector
4-63
;
e22:|
(a) SR
ha
=i}
oar
for f= Us2,7
(b) The mode occurs at T =2.
(CG) Var T= Aw,
4-64
(a) wc = E[C]=30-2.5=75.
Bale Se ies Ov
(b) Pr(C =70)=—|
4-66
e-"(n)?
A=n,
SOoPrl A=]
n!
—
en
=
(n-l)!
l
=9PriZz=(n—1)).
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT ® 87
4-67
Let
=P i
Z=n|=e" a for n=0,1,2,... The ratio of consecutive probabilities is
—A4Nn
!
fn = ioe ec
Pr
CA
Eto
ew(n—1)1
Uasincet enor ar integer, this ratio is
ln
greater than one for all n < 7, and is less than one for all n >A. Thus, the values for
Py are increasing for n< A and decreasing for n >A. Therefore, the largest integer
n less than 2 is the mode.
4-68
Approximate using the Poisson distribution with 2 = ae
e7 02026 (.02026)°
Pr(3 souvenirs) ~
=.000008. That lucky son-of-a-gun. Of course
3!
Davis had a nice seat directly behind the dug-out and was a cute kid, so the
requirements for a Poisson process likely do NOT apply.
4-69
Solve for 2, which is also the variance.
ma Vad
A>
4-70
eA?
i BigSha
implies
A~ =4 andsoA=2
ead 2 accidents
4 weeks _
accidents
week
4weeks
§ 4-week period ©
:
:
=o
Pr[ Accidents in 4-week period <5]= e° + se +
(D)
Q2
-8
a
83
4.0996;
E{ Accidents in 4-week period] =8.
4-7]
4-72
a
.
(a)
E[mistakes in 300 pages]
(b)
peter4-151 WN 4|«\0 " eet -15/16)l
\S)
0)!
Pr(Cusses = 2) =
1!
=
an
|
The variance is 8 as well.
mistakes
eee
=15 SOU pAnes
Gae5-15(15)4
(Loy
4!
ih
= .000857
—8
7 +
Q4
—8
A
88
SOLUTIONS TO EXERCISES - CHAPTER 4
4-73
(b) No, for the discrete uniform distribution all probabilities would equal 1/5.
(c) Pr[(S <4]=1/9+2/9.
snowstorms
ea (eo)!
0!
els
en 1.5
TeoP a
5)"
10,000 | 10,000
ae
10,000
ee (1.5)9
ee ees
10,000
ae
Note that the last column, Payment + ¥, is equal to 10,000-S.
20,000
i
30,000
40,000
Also,
E(¥] = (0)(0) + (10,000)
Pr[S >1] = 10,000(1—Pr[S
= 0]) Thus,
E{ Payment] = E[Payment + ¥]- E[¥]
= 10,000. E[S]—10,000- Pr(S > 1)
= 15,000-10,00011—e'?) = C.
4-75
Binomial Model,
n=80
and
p=.05
Poisson Model,
A=A4,
95°
= .0165|
.0695
1978
.2004 | .1603
0733
.1954
.1954
4
e+
= 0183
1563
1
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
Section 4.7 Summary Comparison of Discrete Distributions
:
AY
)
Uniform, n=5
(Oye
|
Ql 4
|
0
=
Prob)
4-76 (b)
an
1
> 0.2
.
=
2
0.2
a
3
0.2
P
Binomial,
i
~
n= 10,
4
0.2
5
0.2
[= =
eee ead
p=.3
3000
2000
-
4). 1000
0 0000
|
El
i)
a
1
BA
3
a
4
5
=
6
—
$3
—
=|
9
;—
Prob 0.028 0.121 0233 0.266 0.200 0.102 0.036 0.009 0.001 6.000 0.000
ee yr te
Hypergeometric
4-76 (c)
0.3000
0.2000
0 L000
OOOO
Prob
Ul
i
1
Z
3
rn
0.2166
04165
OT
0.0793
0.0086
a
5
0.0004
@ 89
90 # SOLUTIONS TO EXERCISES -CHAPTER 4°
4-76 (d)
|
|
ares
Poisson, Mean = 3
|os
t
_
|
|0.2000 |
;
| 0.1000 - ll
1Ih
» §.0090 1
i a
a
5
|
4-76 (c)
9
10
Prob0.049 0.149 0224 0m 0.168i 100 00.050 9.021 0.008 9.002 8.000
-
-
;
Geometric, p=.25
7
0.3000
0.2000
0.19000
0.0000
-
|
|
—
ii Hla
a
|
ra)
i
4-76 (f)
TET
TE URGRS SIT
10
0.033 0.025 0.018 0014
or
Nesaive Binaitia r= 3 and p=.5
0 3000)
0) 2000)
0 1000
0 0000
I
9
10
Prob 0.125 0.187 0.187/0156 0.117 0.082 0.054 0.035 0.022 0.013 0.008
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
Random Variable
[a
RS irromia] Sng |
pe
yee
4-77
aa
Pp
Negative Binomial
4-78
91
(a) E{coins}]=5<16=Var{coins].
distribution.
(b) 2-5 and ie
joe
P
yy
P
5Q
;
This information suggests the negative binomial
ple ep 16 andin= iG= wiCalculating pronapninies
with a non-integer r is not straightforward, although it can be accomplished using
the Gamma function, which will be introduced in Chapter 6.
4-79
Let S be the # of banana bunches purchased.
(ajeels|= 72. and, Varis|=1.26 {wba =.7416.
(b) Since the mean and variance are close, the Poisson model with 4 =.72 is an
obvious choice.
~(c)
We need to calculate the probability of at least one success in 1000 trials where
p=1—Pr|
S<4|=1-.99911=.00089.
Then the probability of at least one
success Is given by,
|
(.00089)°(,99911)!00 |= 5894.
92
SOLUTIONS TO EXERCISES - CHAPTER 4
4-80 The following table shows the specifics for each of the 6 discrete distributions.
Fy es Seat a
dd
je eed
Mean _|__ 5.5000, _6.0000,_ 1.2500,__4.0000|__3.0000|__
3.0000
2.44
:
pa area ae tee gs |Tenaya ere ee!
N
7
0.187
2
0.187
3
0.0793, 0.1954, 0.1055] __0.156
0.1000{ 0.1115] 0.0096 0.1954, 0.0791] 0.117
5
0.0593 0.082
| 0.1000, 0.2508)
| 0.1042 0.0445 =)
7 | 0.1000 0.2150
| 0.0595] 0.0334,__0.0352
0.1209
ss |__0.0298|
0.0250, 0.022 S
| 9 [0.1000 0.0403, ~~ |_—0.0132| __0.0188| 0.013
10 [| 0.1000 0.0060 ~~ |_0.0053|__ 0.0141] 0.0081
nu |
|[| 0.0019 0.0106 0.0048
re el es ea
ea
re
a
13
|
0.0002{
0.0059, 0.0016
14
|
L_|0.0001|
0.0045] 0.0009
fis |
eee eS eR hi code cost aces
sei —
~)
Nn[LW
SIN
ln|n
(oe)
i) lo
-
=)¥) - |
+o
w-2o
+20
u-30
u+3o
332
8.3723] 7.549
7.5492]
2.16721
| 8.3723|
2.167
| 0.2446
2.9016 _-0.5845|
| 11.2446 9.0984] 3.0845
| 3.1168] 1.3524 -1,5017|
| 14.1168 10.6476 4.0017]
6.0000, 6.4641|
5.4495
0.0000, -3.9282|_-1.
one 9.9282, 7.8990
_-2.0000|_-7.3923| 4.348
10.0000|_13.3923| 10.3
[in
(rr
| 0.8665, 0.730
(a)
0.6000}
0.6665)
0.6941
0.7977]
(b)
1.0000;
0.9817}
0.9900)
0.9786,
0.9437)
0.945
(c)
1.0000}
0.9983}
0.9996}
0.9972)
0.9822}
0.988 |W
I|N
co
>
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT . 93
Section 4.8
Chapter 4 Sample Examination
eras)
(7 )
rae
(4) i
pe
(as (3)-5
(b)— Geometric distribution.
6
(=) (Z| = 3.58%,
4
a
(c) Negative Binomial distribution. (2 )(Z| | = 3.35%.
—
i>)
© —
—_~
10gia
77-100
(b) on
mea 88.5.
(c)
Res =5 (2
ae12 = 1714,
(b) ELX]
o arty=rounrant? «(2)2)2)
—6X2+3] = 5-(3.933)-—6-(1.714)+3 = 12.378.
E[5X
94
Di
SOLUTIONS TO EXERCISES - CHAPTER 4
(a) A=2 > = 4 sneezes per 2 hour period
(b) o=V4 =2
Ce
(d) 1-e*[1+4+8] = .762.
Fl=OX 213+281
=~ 6E[ X72] 413E[X]
+28
- — 6{50-(.96)-(.04) + 487} +13-48 +28 = —13,183.52
6.
E[(3X +4\(7-2X)]
7.
(a) Pr(2 injuries next month) =
(b) Pr(5 injuries next year) =
eo (3).
7
= .0747.
ee 51Go)
==a
|N, Number of Claims [0 [7a]
23
,
pm
, 2
[om [SJ
, 3
(4) |
2
\|
-10 Nine
WE
/
mf
3
l
|
]
Pot (+)
Pot (+) Po + Gg Porte
a
- poyis(2)+(4) vf = rae
This implies that po =.8.
0.
Pr(N >1) = 1—.80-—.16 = .04 (A)
Let A be the number of people completing the study from one group of 10. Then 4
is binomial with n equal to 10 and p equal to .2. Then,
Pr| A >9] = 8194 19C, -.89 -.2! =.376.
Let B be the number of groups in which at least 9 people complete the study. Then
B is binomial with 1 equal to 2 and p equal to .376. Then,
Pr{B =1] = (2)6376)\1=.376) = 469 (B)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
10.
# 95
The scenarios where there are at least two more high risk than low risk include:
\hhhh, hhhm, hhhl, hhmm\.
Nae 2)
ale
n-p=8
Prex
Wer:
Cy (2) 03) £1, (25)
and n-p-q
10)
=
= 4.8 implies
20 Cio -(.4)!° =O)ue
=e
g=.6,
aCe (2)"23)2 = 0488 (D)
p=.4,
and n=20.
Ly.
Let X be the number of low-risk drivers with no accidents in a given month and let
- Y be the number of high-risk drivers with no accidents in a given month. Then X
and Y are binomial random variables with,
E{_X ]=400-0.9 = 360 and E[Y]=600-0.8 = 480
The total number of bonuses paid over 12 months 1s,
S = (X,4+:--+X12)+(% +--+ H2),
5S.
E[5S] = 5(12-360+12-480)
and the total bonus payment for the year is
= 50,400
(B)
a
oe
en
aca ee
‘
“oi
hp
)
7
tetas
Saray
shikai et Sard
Fe
hw
' 2%,
oe 4
|
neae
Si seen Gieg eeca
‘
CHAPTER 5: CALCULUS, PROBABILITY, AND CONTINUOUS DISTRIBUTIONS
Section 5.2
5-1
Density Functions
(a) F(x) has the properties of a cumulative distribution function.
G@)
0
itpex <0
F(x) = f@) = ,
eo
ih x
0!
so f(x)20 and therefore F(x) is non-decreasing.
(ii) lim Fy(x)= lim 0=0.
(iii)
lim Fy(x) = lim(—e~*) = (1-0) = 1.
x— 00
(iv)
Pr(a<X <b) = Fy(b)—Fx (a).
(b) See (a) part (1.).
(c) Using the CDF,
Pola 2) = FQ)-Fll) =U
—e°) (le 4) a lew ee’),
Using the probability density function,
Pree
[4e ditt ent ind = (e+-e78)
ene
(d)
ae S)
Prpx
2]
(b) fx(x)=
=
Pr| (X >2)(X
>))|
ls
Srv
Be eee
2)
» Tee Sh
ea
0
former
sxe
for 0<x<l
oe
2 & o-4
Ouror a = |
(c) Using the CDF,
Pr(.l<x<.5)=F(5)—F(1)=(.5) -(.1) =0.124.
Using the probability density function,
Pride
= 50) = Ve fe
CG
MEAN
Ne
oe 3
pePrleny (ik 5) bP rl a Wi<5)
ear ae)
Pr(X <.5)
wae Eri 5)
aimee
4124 = 992%
125
97
98
5-3
SOLUTIONS TO EXERCISES - CHAPTER 5
(a) {- "a dy =
ke”
he
=
k
=0-——= =. Since the total probability is 1, & =3.
aa
Ves
|” BSok ke 3 te F(0)=5=1 implies k =3.
(b) F(y) = [ke*at = ke
Silo
(c) Using the CDF,
Pr(2<x <3) = F(3)-F(2) = (l-e38)-(1-e3) = (e% -e).
Using the probability density function,
3
Pr? 2455)
D8)
i= I,36 “dv=—e—*
X=zZ
(d) Pr(Y >-3)
[fay
Ik3 (y)dy + IPSt(y)dy
[.0dy+ [-3ePdy = 041 = 1.
(a) Let f(t)=k for 82<¢<90. Then 1 = [kat = 8k, sok=—.,
E
80 |
IT] = leg
= 86.
(b) Zero. The probability of any individual value is zero for a// continuous
distributions.
87.5 |
(c) Pr(86.5<T <87.5) = [86.5ght =
5-5
5-6
PrlX <2|X >1.5]
‘
:
Prix alse as)
RS
PriX¥S16)
PrL¥ >16|.X >8]= ee
;
20
ek .1-—.005x) ax ee osaaa
e
:
TG
I, (1-.005x)de = =
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 99
5-7
4
F(x) = kf (W041)? at = k or
(VANS aa 10 oa ees
10+40
1
eP
: |Pls
10
104+x]-
icy uabas
Pr(X <5) = F(5)=12. 5|5: es
TER |p
10
(C)
12
Section 5.3. Great Expectations
5-8
Py
;
(a)
4
E
33
4 x4
ELX] = [x aac
=
>
<
5
—
——
5
le
—
—
4x
Seo ss
=
4x9
4)
=
x=4
wa
x°
45
—
x=4
(b) LX] = [x tatpele CAvk Ea CeGd ee
0
(Cy
ip
c= 0
756
1
0 = 4 = 4,
l
lie
4
co
Ey (7
=
oe
——“—
=
Sy ee
Ae
cy
aaa E
SS
SS
(dy Pron ane
<3) = [2
2 kea=ye ada =ee
256
:
5-9
>
k-(l+x)7.
kek) s
ibe
j,—*_(ax?
C dy =
Pe
———_}
(3140)
is
Lk3
= =.
Therefore the proportionality constant k =3.
ELAS
fa
=
5-10
ot
0 (14x)
ax
© 3(u-l) ,
[ free
=
E| VX | =
|v
ls
Letu=l—x.
AN eo ba
eae
2e4\) [—. a {tiued’
Sa Aa
4-x-(1—x-
ee
)dx
Pe
4.2Faoe
x
Aes
5
en5a x OD ¥
C)
(
ees
ill
100
SOLUTIONS TO EXERCISES - CHAPTER 5
2
5-11
Var[X] = ELX?]-(ELX))" = B18)
:
§-12
2
c=2.
= 426.
pr, dude =ee 4:
Ee 2xde PS
Bags
ELX]=Ban iF:
5. ELX PAB
1= [x
==.
peti’ =.0555.
ratx1=3-(3|
oy =.2357.
The coefficient of variation for the random variable X is Pt
A ipmet
5-14
5-(8) = 2211.
ier’| xdu=—.
Pao ELX Ai
eee
(a) ELX]==
le ai
(b) E[VX+1] = f° (eet a 2 2302,
]
5-15
1 = 35.36%,
0
(a) fry)= |
tor
Vez
0r 26
pa
Boe
ee:
Graph of the Density
Function
Ie:aS cei B-(3)
bo toe
= 2.027.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 101
5-16
Gaowie eo Cd
eS0 FOF Crew
|= 1.27
Var|Coig |= 1.44-260 = 374.4 (E)
f'(x) = 12x-6, so the only critical value is x =.5. Since f(0)=1, f(.5)=1.5,
and /(1)=0, the mode of this random variable is x =.5.
5-18
f'@ = 7.5-37.5t+45t? -1503 = (7.5)-(1-t)-(2¢2-4t+1).
P=
land tea 1+0.5./2 .
The critical points are
We need to use the 2"! derivative test and/or check the
values of /(f) at the endpoints and critical points. This random variable is bi-modal
with t = 1-0.5/2 = .2929, and ¢= 1-0.5J2 = 1.7071.
f(t)
(By el2)
Cie a7
Oe
jiew
ne
PO
1
Ne
I e702
<0
= el
ove
= =. Alternatively,
l
EZ] = 0+ |0 [I-27]a = ne
(c) The mode forZ occurs at the maximum value of f(z).
() (75-25 = .366.
The mode equals |.
102 ®
5-20
SOLUTIONS TO EXERCISES - CHAPTER 5
(ayre=43
0
(Dei
4) t=
for
eo ee for,
1
for
x<0
OS 1st
x>0(~~’
(c) py =aut[jx-4xe-(-x 2 )dx =coTe.
(d) The mode forX is jt:
(e) x4 = Jase Both parts (e) and (f) can be solved using the quadratic formula.
(f)
5-21
KO
= 33%
The answers are easy if you graph the density function.
wy =xs5 =z and the
random variable is bimodal, x = z/2 and 37/2.
Probability Density Function
f(x)=.25]sin(x)|
5-22
(a) No modes:
|
:
/(x) 3) on [5,8].
wiles
a.
:
Since the density function is constant there is
no single (or dual) maximum point. All values
5< x <8 are equally likely.
(b) One mode:
g(x) = 2x-—6 on [3,4].
The one mode occurs at x =4.
(c) Two modes:
sin z
f(z)= ea on the interval 0< z<2z.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 103
5-23
(a) f(w)= a for
0< w<
20, zero otherwise.
0
(b) F(w) = Pr(W<w) =
for
aa for 0<w<20.
l
(choppy
for
20<w
20
|= \, we dw = 10 _ using the density function f(w).
= E[W
-
w<0O
20
= 0+ if i
Ww
ass
P
‘| =)
using the CDF F(w).
Also, since the density function is symmetric about 10, the mean is 10.
(d) No mode since the density function is constant.
That is, f(w) =.05 for all
O< w< 20.
(e) The median is 10.
(f) we9 =13.8.
1
(a) Veer
31
te
ee7A
|,ise
ere
F(x)=
x5 =J/13 =3.6056.
5-25
tor
lay
The mode is x =5.
The mode occurs at the maximum value ofthe density function /(.x).
.
,
f(x) = F(x) =
for
x<0
‘
4x —— x
tote)
5
ee 3,
0
for.
ax
4-2x
The maximum value of f(x) occurs when f'(x) =—
T=
vis the moce..
(1)
A
hee
ee 0.
5;
104
SOLUTIONS TO EXERCISES - CHAPTER 5
$15
Jac pane 200)" Pe
= 93.06
(B)
8-27
BLX] = [x-flade = fn (Ele «(3 Jee BASRA
yy
1/2.5
1/2.5
oe 200 2) -200{12
Section 5.4 Mixed Distributions
5-28
f(x) = x—-1 for 1<x<2
and there is a point mass of p => at x =1.
We see this from the graph of the CDF F(x), which has a jump from 0 to 4 at x=1.
barns
ai ys ia +2
EX) PF
slyeared t[e
@Vae=
es
7
el =.
ee:
E[X2]apa
=1 sth? (Gl de el
= S47
5-29
E[Loss] = E[Loss| no accident]- Pr| no accident |+ E{Loss |accident] - Pr |accident |
———
(90)
- ele
0
probability
of no accident
expected damage,
given noaccident
I
5-30
(a) fas
ree ——C10)——Y |00 x-2e*5 de = (.10)-—2l = .05.*
probability
ofaccident
$7
expected damage,
given accident
50<x<110.
110 x
E{Payment]= 1000 |. Sp & = 80,000.
E{Payment*]
= 1000? ae x
ak= 6,700,000, 000.
Var[Payment] = 6,700,000,000—80,0002 = 300,000,000.
Grane slr a20a)
90
(b) E[Modified Payment] = [/1000.x. wt 000 a
lI
46,666.67+ 33,333.33
E|(Modified Payment)? ]= [,, 1000)
pe
O Modified Payment
p
= $16,966.
=
90
‘
= 80,000.
Gp dk +1000?
ya!
-
ae”
= 6,688,888,889
Ss
19
20000
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 105
Section 5.5 Applications to Insurance: Deductibles and Caps
SS
5-31
a
E[Benefit]
a
10
NE
y,
\ OR
= 1.8+10(.01) = 1.9
5-32
(D)
Note: The maximum payment is | million dollars, so solution choices (D) and (E)
are ruled out.
i x(4—-%)
3 x(4—x)
eee
a ed
E{ Payment]
1204+ .8148 = .9352
5-33
iy
er Xe)
Payment =
-
=
0
(C)
tor Vax <1)
ieee
ise fi OSC
.64 = Pr| Payment
<.5]-= Dipeaai
tee
= Pr Xx 225 tC)
5eCy.
(.5+C)=.8. Therefore C=.3 (B)
5-34
The loss NOT covered is equivalent to the loss subject to a cap of 2. Thus,
E[Loss not Covered] =
eho]
CUS
0
Big a
ae
2.5(0.6
ee
Alternatively, using the CDF:
2oy 2509"
ax = Oa
wee)
z
F(x) = i-(4)
Oe
k= Ol,
Let Y =losses capped at 2. Then,
aC [-£)
6 dx = 93.
E[Y] = 6+ |GNX
5-35
E{[X]=500. We want to set a deductible, D, such that
E[{Payment ]= 25%(500) =125.
:
1000
|
4
(rn))4
125 = 0/Pr(X<D)+| (e-D)- Baa Ute
Therefore,
D=500.
(C).
r=1000
a L000):
pean
-
106 ®
ye
SOLUTIONS TO EXERCISES - CHAPTER 5
Tapas [oy Eat 4 PLY > 4] =1.644-(2)=24.
FLY?) = fy dy +42. Pr[Y > 4] =4.26 +16-(.2)=7.46.
Var[Y]=1.706 (C)
5-37
E[Not Paid] = [x5 dr+10-Pr[x>10] =.066+10-.99 = 9.96.
5-38
f)=35 for 0<1<75.
me
_
10
E(Benchit)|= {- iedt +40: Pr[40 <T<75] = = ip
‘eit
Se
I ares
(100)°
‘ae,
eG
oe
eee:
160-507
100 ):
t=0
O<ses100
E[Payment
| X >10] = mre| sap 100(100—x
[ =| \" dx
i
ae
|
(100-10)
1-F(10)
3-100?
100-10
3-100
100
5-40
i
l=K
ili
tata
Net Premium
Insurance Payment, w/ deductible
60
137
ted| implies K=——.
oe OK hen ee
= .95-(0)+.05-
aed
[.
2. 15 4. 12iih =~ 0313
\ tes ogee
Ue ls Ae
a
5
iw"
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
5-41
The health care cost variable is represented by Y. The reimbursement random
variable is
0
TOU
R=,5X-20
pAes20
[Ore
100+ .5(X -120)
for
170
X >120
A reimbursement of 115 implies that the loss Y is more than 120.
Thus, 115 = 100+.5(X-120) gives X =150.
The reimbursement is positive if the health care cost is greater than 20.
G(115) = Pr(R<115|R>0)
Pr(X¥ <150| X > 20)
_ Pr(20<X<150) _ F(50)-FQ0) _ a9,
Feit 30)
ee
oer
ae
Section 5.6 The Moment Generating Function
5-42
In evaluating the following integral, we use integration by parts.
ne
Ge
=
EX]
IsNed d=
ELX2] = lex ede
5-43.
S44
545
(-xeek H
x=00
l
x=0
2 ;
j
= 0.5. oy =y.5-.52 =.5.
My(t)=.4e* +.6e™.
E[X]=My' (0) =3.8.
M(t) =.8e7! +3e*.
E[X*]= My" (0) =16.6.
M(t) =1.6e7" + 15e*.
Var|X ]=16.6 —3.87 = 2.16.
My(t)=.3e +.7e' =.3+.7e'.
My (t)=.7e'.
Ee leew
jae ee
M}(t) =.7e’.
Var(Y]=.7-(.7) =(.7)(.3).
(a)
My(t)=.5e7 +.2e# 436°",
(b) M'(t)=—.5e~ + 4e” +1.5e%, so ELX]= My (0)=—.5+.441.5=1.4.
(c) You could find the probability distribution for Y and the MGF from the
definition.
Mii
=
May_4(t)
=
a". My (3t)
l|
=
ew. (Sao +.2e° +,.3e!> )
4
5e7/t +.2e% + (3et",
® 107
108 @
5-46
SOLUTIONS TO EXERCISES - CHAPTER 5
(a) M(t) =.3e! +.2e7'.
M(t) =.3e! +.4e*".
EX
M7 (0) = 5.00 Van
Oo.
na
ey 25
= 45:
Re
Een) EIEN
(c) The mode is
5-47
x= (Me (0) 03)
x=0, the median is x =0, and the midrange for_X is =
=1.
4
(a) M’(t)=5(.8)e" -(.2+.8e')
3
M3 (t) = 4e! -3.2e! (2+.8e') +4e' (2 +.8e")
(aVara
E[Z2]=16.8.
4
Var[Z]=.8.
=e" (2487).
(b) My (t)=M3-7z(t) =e" -Mz(-Tt)
You may use
My (ft) to verify
Uy = E(W]= E33 -7Z]=3-7E[Z]=3-7-4
5-48
and op =|7|-J8.
The key to this problem 1s to find the probability distribution for Y.
the time (whenever any of the Y¥;=0).
and X; =1.
Y=1
Y =0 most of
ifand onlyif X; =1 and X>=1
By independence, this happens with probability =
Thus,
Yisa
Bernoulli trial random variable with p= + . Therefore the answer is (A).
Note:
All MGF’s have the property that M/y(0)=1.
So (B) and (D) are NOT valid
moment generating functions.
5-49
My (t) = My,-2x,43x,-4()
= My, ():My,
(-2t): Mx, 3t)-e™
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 109
5-50
Mt
Mae
fe
i
|= |e
as
ees
ion Ie
;
3
Mery.
x(t)
G-12
M‘ (t)= ah
B-
Dah:
y2]-2
er)
ea
ess ee
3
Omit
We use the shortcut method:
— A(t)=In
M(t) =—4In(1— 25002). A'(¢) = 10,000(1
—25002).
h"(t) =10,000-2500(1— 25002)*. oy =,/h"(0) =100-50=5000.
(B)
Alternatively,
M‘ (t) a
AU
(1- 25007)
ELX]=M'[0]=10,000
and M(t) — 125,000,000
(125007)
and ELX*]=125,000,000.
oy =5,000'= B.
5-52
Since X and Y are IID, we have
M()=MyirO=Mx(t):MyQ=Mr(t):
My(O=Mx(O’.
Thus, M(t) =0.09e~*! + 0.24e~ + 0.34 + 0.24e’ + 0.09e7’ must be a perfect square
and so My(t)
must take the form My(t)
= ae‘ +bh+ce’.
It follows that a and c
must both be 0.3. Since a, b, and c must sum to 1, we have that 4 must be 0.4. It can
now be easily checked that
+ 0.24e! +0.09e7"
M(t) =0.09e~*" + 0.24e7 + 0.34
Therefore My(t)
= (.30e7% +.40+.300e').
So we can reconstruct the distribution for X :
PrLX <0] = .30+.40 =.70.
(E)
= (.30e%+.40+.30e')?
.
110 @
5-52
SOLUTIONS TO EXERCISES - CHAPTER 5
fie sok
see ee >
MOQ)
P
Mx() == Oe
ElX]=Myo)=£3 a by
(1-q-e')
Dp
legal raheemlee)
2
Pp
(ere)
1G (oe
ee
P?-4-2p-q-P-(-4}
ae
ee ee_g(p+2¢)
Re »,_ eee
Pp
Vari x)=
P
2
‘(p+
42+ D2 isc
+2
ie
5-54
Pp
p
usto=| ie
1-ge'
Me
=Io(
M0) =o ayes
P
}=r ; P
l-qe
—qe'
(0
Teter
J=rincey rina-ge'
E[X]="@=—t.
}
meo=r (oe!)
5-55
euee
ke en
;
Dp
Pp
This is the MGF for a Negative Binomial random variable with r=4 and p=.3.
(b) Pr{V =2]= 5C3-(.7)° -(.3)4 =.0397.
5-56
0
Var[X]=h"(0)= Petre _
(a) My(t) een
a)
(b) My(t)=Mrx43(t) =e* ae
= ee" +31-6 |
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 111
oy
5-58
Yea
h(t)=In(M x(t)) = n(eHe) ale -1).
ELX]=h'(0) =.
Var[X]=h"(0) =A.
If Mz(t)= eles then the random variable has Poisson distribution with A =3.7.
The mode occurs at z =3.
Section 5.7 Chapter 5 Sample Examination
ies
(a)
ae fae
1=[
or
= x
== [fax
and Se
Solving for a and 5, we find the constants are
(ob)
ieee
cee
a=3.6 and
b=-2.4
Pr(X <.4)= iff(x) dt= 2368:
(c) -ELY2] = [G63 -2.4x4) de 229)
Var|X \=".42= (6)* = 06) so oy =0.245.
(ay meeey (t= .0%
2 Ay
(x)
9, 6 = 2Ax = OC
ye = 2/52 sincesthe graph
of the density is a downward directed parabola, we conclude the mode is x =.75.
2.
This is an exponential random variable that we will study in chapter 6.
tae
Pree)
= GS) 97709:
(bo)
5=F(x) = S=1l—e"" => the median is m=.2310.
ix) ae '(x) =3e°>* forx>0.
E|X]
= ux
(oS
xy
Thus, the mode is x =0.
=k [1- F(x) ee
|
= I.eo ax = =
65 = Le
mgs
= 3499.
112 ®
SOLUTIONS TO EXERCISES - CHAPTER 5
Re.
ere
= flee
wet) dy = 84375,
Fy) = (a
(6b)
= 13y-¥3 +2]for -l<y<l
The medianis_ ys =0.
The mode is y=0.
The mean is fly =0.
(CymlGsS=F Oy)= 43 y-y 342] = yes =.20278.
(Use a graphing calculator or Excel Goal-Seek.)
(ely
4.
Jape
A = = f,0? -y\)dy
E[T]=8.5 and o? ==.
The coefficient of variation is
§,
= =. Since wy =0, oF ==.
100-6 We =
100- 53
6 0% = 16.98%
Ell
Cee
E[(X-2))] = ELX? -6X2 +12 -8]= E[X3]-6E[X2]+12E[X]-8 = 0.
9-6-E[X2]+12-2-8=0.
7
OS
X*|=—.
[X7]==
E|
ps
|
Var|ar| X|)=—-4=—.
X|
r
r
6.
TheCDFisf (x)= [cencus ars pire —e?.004~)
Let Y be the benefit paid. Since
F(250) = 1—e~%*29
for X occurs to the left of 250, where
of X.
The
median
m=250
ln 2elio29
Te
ie = ae
004
(Ch:
of X
Thus, the constant
c=.004.
= 1-e7! = .63, the median
Y =X . ike the median of Y equals the median
is the
solution
to
és = ]—¢ 004m)
‘
250, the mode is x =0, and the standard deviation is oy = 250.
Which
is
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 113
The CDF is F(x) = f°c(l0+1)? at = a5 os | 0<x<40.
2,8;
a
2
Sones
(40)=¢
55° 80 ¢ = jesKore
l=F(4
Then Pr(X¥ <6) = F(6)=12.5- P | = 46875. (C)
l has
0
F(x)(Cele
=1.25
-—=
fio
for 0<
x < 40
7140)
iiaE —F (x)| dx
lI
("[1 F(x) Jd
40
The mean is
J
Sie
WAS
ae
08
+25] <
= (—.25)(40) +12.5In5 =10.118,
the median is 5=F(m)=>
m= S. and the mode occurs at x =0.
This is an example of Beta distribution.
(a)
Tra= 12;
(b)
Pr X>.80)= [sgh2x(x-l)? jie aiDae
(c)
l
l
Var[X] = ih
(x? )12x(x-1)? 2 dx -{[,
(12x01? 2 as|; = .2-(.4)? 7 = .04.
(d)
The mode of the random variable X occurs at the maximum value of f(x).
f(x) = 12x(2)(x-1) +12(x-1)? = 12(x-1)[2x+x-1].
The critical points are
;
;
|
x =1 (not in the domain of f(x) ) and x= =
The mode is x =>.
(Awe
PA
eS) eas)
(b)
The mode is x=0.
(c))
ix =:
(d)
1-—e-*
(6)
sox =I)
= le?
= .80 implies xg9
= 551,
= —In(.2) = In(5) = 1.60944.
114 ®
SOLUTIONS TO EXERCISES - CHAPTER 5
12.
(a)
EfDamage] = (.92)-0+(.08)-1750 = 140.
(b)
ETE
E[Ins. Pmt.] = (.92)-0+ (08)}say 0 —-
1000
1
3000
i
]
=10 0)- ee) as
meas
.08| 800 | = 64.
13. (a)
My() = Ele] = [oe% Sede
X=
=
Ctds= = ets] fess =
= 5 [e
:
14.
er
5—f
rcs.
: aay
(ye
15
G20)
(b)
Ae (ey
OnBr
(a)
Graphing the CDF allows you to see
that X is a mixed distribution. The
peer
point masses are PrlX¥ =2]=.3 and
PrX =3] =.2.
(2) dv+3-(2) = oie24543See = 5
(b) ELX] = 2-(3)+ fv
ct Miners cers
J (x)
1S;
If My (t) a
— 4e
then W has geometric distribution with p=.6.
(a) ow = Pe )
16.
(b) — PrfW =3]=(.4)
(6).
Since the man purchases the insurance at age 40, we are given that he has survived to
age 40.
E{Payment] = 5,000-Pr(40 <T <50|T > 40)+0- otherwise
conditional proability an
insured dies between ages 40
and 50, given he is alive at 40
= 5000-
F(50)—F (40)
1—F (40)
_
348.
(B)
CHAPTER 6: SOME CONTINUOUS DISTRIBUTIONS
Section 6.1 Uniform Random Variables
6-1
(a) f@== for 60<d<90.
F(d)=2—™ for 60<d<90.
6
:
:
(Db) a = “as = 75, conveniently located midway between the endpoints.
30"
o5=7 55>
(c)
9D = 5/3 =8.66.
d 69 oa 60
Solve ———
.
th
2
J
=.69 to discover the 69" percentile is d 6 = 60+.69(30) = 80.7.
ON=67-5
by Dee Ea
a
30
30
6-2 (a) fr(y== for -2<y<5, Fy) =
b
(b)
+5
wy = ——
a
=1.5.
=1.5.
ie Dees
Need:
oyo7 =—.
12
(c) The median for a uniform random variable is always the midpoint of a domain,
same as the mean.
Here m=1.5.
occurs for all -2< y<5.
The maximum value of fy(y) is +, which
Therefore there is NO mode.
2
(dierent ye< 0) ace
(ey 0Ptcy = 1) = 1),
6-3
We let 7 denote the arrival time of the cable guy in hours (after noon). So a value like
T =1.4 corresponds to an arrival time of 1:24 pm.
(a) f(t )=— and F() =~ for 0<t<5,
(b) sr =2.5 which corresponds to 2:30 pm.
eS)
o7 = 17 square-hours.
(c} Pr[3<T <4]=— =20%,
(d) Pr[3<T<4|T>2] = >
=
115
116 ® SOLUTIONS TO EXERCISES - CHAPTER 6
tae
Pfs
oy SX say roy] =
eb tater eth
7.
b=a
ne Pe
ae
tee
128
02
(b) Pr| wy -20y <X <py +20y
|=Pr Gi peheey 2ae 2c
ath
Convince yourself that
_,b-a
a+b
<
fi
ee eee
+
b-a
ie:
so that this probability is 1.
as
6-5
w4=10
and oy oily
6-6
This uses the methods of Section 5.5. Suppose D denotes the damage to an
automobile, D ~ Unif(0,1500). Then the insurance payment is given by
0
if if0<D<250
(with probability 4)
- |D-250 if D>250 (with density 7)
|
Et] = 0-44
[-”(x-250)-—41500 ae = 520.83.
6 | 4250
EUI?] = 0? +
e)
a}
[i (x-250)? zor de = 434027.7.
]
1500
2)
]
=
= 403.44.
5
(B)
Section 6.2 The Exponential Distribution
6-7
|
[ae
Sketch the graph of f(x) = lo
if x20
_.
_.
Itis clear that the mode occurs at
otherwise
x=0 regardless of the value 2. Using calculus, there are no critical points. Checking
the value of f(x) at the endpoint(s) gives
6-8
x =0 as the maximum.
Fy(y) is the CDF of an exponentially distributed random variable with mean +. The
>
variance 1s (4) and so oy = 4, You could also calculate Var[Y]= E[Y?] —(E[Y])
ra
using
eer ers
2
the
density
er
t
ion
function
ron
AS
ye
{0
for y <0
|2e “y
for
yp20
:
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
6-9
(a)
Oz = 23
(b)
(c)
The median of Z is 5-In(2).
Themodeof Z is’z=0:
(d)
Prl<Z<3) = i
—2/5
dz =
e > —e3/5
= 2699,
Oryusing F(z).= 1—e-¥*,
Pril<Z2 <3) "= F(3)—F
CG) = ee"
(ce)
Oal0
Sa)
= 2609;
Careful, f(z)=0 for z<0. Pr(-1<Z <5) = Pr(0<Z<5) = 1-e7,
7 (f)i=
t
1
4.0750
6350
ite
0
7 =0
otherwise
t
Also, F(Z) = le
07> 0.
—750
(b)
Pr(T <750)=1—e 70 =.6321.
(c)
m=750-In(2).
Conditional density
for X, given X>p
6-11 E[X|X>]
= [x
oo
(22
e749
dx
Substitute
V=X—-Q
CO
ss
= | TeOe
2
ged
“FF
Vie
fo@)
i,(y+): Ae?
dy
00
oe)
\,yp he A) dy+o | AeW*) ay
=
6-12
[47
dx+o-1
= E[X]+Q.
:
——.
) Q)
(b) The mode for an exponentially distributed random variable is always zero.
In(2)
(Cc) A=—.
7
—5-In(2)
(dyer
<S)panl—er
7
== 0.39,
6-13 B=a=10,s0 O;-Q, = —10In(.75)+101n(.25) = 10.986.
# 117
118
a
6-14
SOLUTIONS TO EXERCISES - CHAPTER 6
3
<
‘CO
E[X°] =
2 :
[,35
e
—x/7
7
=
av =) 2058:
The integral can be calculated using multiple (or tabular) integration-by-parts.
Also, we will learn how to calculate this using the gamma distribution in Section 6.5.
However, the easiest ee
is to use the MGF:
=
ELX?]=M0)
=-7
L20--1| 5 =C28)2° = H(29 4 = 2058.
6-15 Since
8? =9, wehave B=3.
ELX*]=M™)(0) =1944.
6-16 E[7Y2-11¥°] = TELY?]-116| ¥°| = —5,773,554 (using the MGF).
6-17
The median equals 4, so the mean must be £ = aa =5,//1.
PKG
5)
6-18
== Foy =e oo
(Dd)
First find the parameter { of the exponential distribution.
30=Pr(X <50)=1-e9/F,
6-19
= 429
B=140.18.
Let L denote the lifetime ofa printer.
E{refund]
Pr(X <80)
=1-e-80/140.18 =4349.
/
2. Then
Recall Pr(Z </) =F, (/)=l—e
200-Pr(0< L <1)+100- Pris L<2)+0-Pr(L
2 2)
= 200-.3935+100>.2387
= 102.56.
Since there are 100 independent printers, the total refund is 10,256.
6-20
(C)
Let X denote the auto loss,
¥ ~ Exp(300).
losses that exceed the deductible.
¢
OS Pt X=
ACA
:
ak 00)
=
(D)
Let 7 be the 95" percentile of actual
We need to solve the following for 77:
Pr(X >7 0X
>100)
Priv > 100)
=
Pr(X > 77)
PrX>100)
oat.
= e 300
3
7 =100—300-In(.05)=998.7.
(BE)
=~ a
3067),
100 ~
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
6-21
# 119
Let T denote the time from purchase to failure, T ~ Exp(10).
1000 = x-(l—e7/10)
4 2. (g-10 _ 93/10) — y..1772.
SS
SSS
————
SS
—
Pr(O<T‘<1)
Solving for xgives,
6-22
x =5644.
Pr(1<7'<3)
(D)
Method 1: Recognize the MGF as that of an exponential random variable with mean 2.
E[L00(0.5)*} = ["100(0.5)' ser? ee
=
50 [eno
eo X!/2 dy 25
i) [ eX (1/2+In(2)) hy =
0
lee.
0
=
41.9.
+ +In(2)
Method 2: Using the definition of MGF (with some slick work).
Recall, My (1) =Ele*]= os
C2)
0k
10ex
Ie= 100E[(0.5)4)
FALOOCO:5)2
II
6-23
:
100
100M
x(In(1/2))
= iain
= 41?
F(x) =1- e 0: x>0Q. Let Xy be the payment per loss with a deductible of d. We
use the CDF method of Subsection 5.5.3 to obtain,
x
BX gh =
[ [1- F(x) |de =
oo
IkC0 dy
oct
= Gee,
100
Then 2000=6e
? . Also,
E[Xs00]
6-24
=
(200021
Ge500/0
F(x) =1 -e 8 -x>0.
|e-5009
= 2000e4?
Let X¥% be the payment per loss with a cap of d.
We use the CDF method of Subsection 5.5.3 to obtain,
| [I-F(a)|dv=
im |e —x/0 dx-= 4 O(1-e74/),
ei)
ELX¢]aby =— I,
1
d
/
8),
Then 20 = @(1-e7 190
Thus, EL)
§ (C).
(1 — e7500/8)
= 60 =e") = 205(1 ——_*
ae = 100/€ ) )
(8).
(C)
120 @ SOLUTIONS TO EXERCISES - CHAPTER 6
Section 6.3 The Normal Distribution
6-25 (a) Pr(-2.98<Z<.99) = &(.99) -D(-2.98)
lI @(.99) -[1-D(2.98)] = .8389-1+.9986 = 83.75%
(b) Pr(-1.75<Z<-1.04) = [1-.8508]—[1—.9599] = 10.91%.
(c) Pr(Z >~.23) =59.10%
(QePHIZ 12073) ei P13
27 73) = 1-2 {lO
= 1-2-0907 = so
(ec) Pr(|Z|>2.05) = 2-[1-®(2.05)] = 2-.0202 = 4.04%.
626 (a) 10s Pr)2227 |31 —Py|
Z sz,
Z
9
That is, one needs to be 1.28 standard deviations above the mean to be
(b)
in the top 10%.
Zo5 =1.645.
(c)
P49) = Pee
(dy
95 = Pr 222,|=I Pr| Z<7, |= Pri 22, | 05 =e
05
\|
ss
NR —
=
II
Oana
1
O(=2Z,) — 29)
Se
=
Oe (5)
Sn
e705
ea OF O
Thus, 27 ='Zos =—1.645.
(e)
Paar ® 122
(f)
Probabilities are always numbers between 0 and 1, so such a zy does not exist.
6-27 (a) 1-.2119 = .7881 and
z78) = .80.
(b) ®(z_)-[1-D(zq)| = .9030
= @P(zy) = 9515
=e
6-28
2 = 255;35=
(a) Pr(X <84") = Pr| Z
(b)
X >79"+(.842)-2.5"
<
156
84-79 )= Pr(zZ <2) = 9772.
2.5
—————_-
= 81.1 inches.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 121
6-29 IO > 100+2.055(15) = 130.8.
6-30
Let T denote the completion time of a competitive marathon runner,
T.~ N(41=200min,o? = {24min}?)
(a) T = 200-—.955(24) = 177.1 minutes.
(b) T = 200+2.17(24)
= 252.08 minutes.
6-31 (a) Pr(F>16) = Pr{Z> ent) eau
(b) F > 11.7+2.33(3.9) = 20.8 inches.
hee
Pr Ae 60) = Pr[Z< 2-862. 764
(b) Pr(48< A<59) = Pr(-1.64<
Z <.56) = .6618.
6-33
(Gc) A = 56.2—1.28-5
= 49.8.
(a)
Similarly,
me
= —.33.
A test score of 79 corresponds to a z-score of z =.5.
A test score of 96 corresponds to a z-score of z =1.92.
(b) G = 73-1.62(12) = 53.56 and G = 73-.47(12) = 78.64.
6234
6-35
#(3) F295
= C3005)
= Pre
i 7 <7 1) S52.
(b) C 2 30.00+1.645-.07
= 30,115 inches.
(a) Bob would be assigned a *C’.
(b)
Course Grade | Percent of class
Range
|69.5,81.9)
|46.5,69.5)
|34.1,46.5)
| (-co,34.1)
122
SOLUTIONS TO EXERCISES - CHAPTER 6
6-36 Pr(50</O0<70) = Pr(-3.125<
Z <-1.875) = 3.0%.
6-37 (a)
1.22
E[Y]=e*2 =e°82 =336.97. Var[¥]=e26DH2 .(el2’ 1) =365,710.
(b)
Pr(l50<¥ <250) = Pr(150<e* <250)
= Pr(In(150) < X <In(250)) =Pr(-.07 < Z<.35) = .17.
(c)
The top 10% of the normal distribution corresponds to z =1.282.
In(Y)—5.1
19
=1;1.282 implies
impli
Y == 763
6-38 (a)
leet
:
This implies (e?* -1) = 3 and oy =,/In(4).
(b)
6300)
(b)
Has
2 =f" eit tells us that wy =0.
Pr(l<¥ <3) = Pr(l<e* <3) = Pr(0<X <In(3)) = Pr(0<Z<.93) = 32.
ee
bi] som 2e=e! 1197,
Vary] = ee te 1) = eo.
Pr(.9<L<1.5) =Pr(.9<e* <1.5)
Pr(In(.9) < X <In(1.5)
. py{Be
4
(c)
:
E[Y]=e "2° =2 and Var[Y] = e2#**%% -(e% -1) = 12.
See lise)
4
)
ere
The bottom 5% of the normal distribution corresponds to z =—1.645
aoa = ~1.645 implies Y =.5724.
Section 6.4 The Law of Averages and the Central Limit Theorem
640) BUY eG ance or ee eee
The exact binomial probability is Pr(¥ =8) = Cg -(.3)®-(.7)*
= .1144.
Our approximation using the normal approximation is
Prix =8) = Pr{778
7
eo —|
=
Prgsee e122)
= .12.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
6-41
# 123
(a) Let H represent the number of heads on 100 flips of a fair coin.
H ~ Binomial(n =100, p =.5). We will approximate H with a normally distributed
random variable X ~ N(u=n-p = 50,0 = J100-0.5-0.5 =5).
Pr(H =60) = 100Ceo «(.5)™ -(.5)* = .0108.
This is the exact binomial probability.
Pr(H > 60) = 100Ce1 -(-5)® «(.5)°? + 100 Con + (5) -(.5)°
+++ 199Ci99 -(.5)! = .0176.
Pr(H = 60) = pr(2550 By < 05-50) = 4821—.4713 =.0108,
60.5-50
(b) Pr(H > 60) = Pri ZS aoe A
6-42
Mis
Pretty nice approximation.
The number of no-shows is NS ~ N(u=9.3,0 = 2.95).
PrOvs =
2) 5) Se ET(WA
6-43
Pr(Z > 2.1)=.0179.
4.5—9.3
= Pr(Z 2 -1.63) = .948.
795
Let S be the number of sales. Then Sis binomial with n =500 and p=.1.
hus;
6-44 (a)
5 99:5 =G00)0:1)
Pris
= 56[°= m2
~ 1(500)(0.1)(.9)
“|el
939 =e 20/6/ae BY.
E[X;]=75 and VarlX;]=l| 35.
(b)
E[X]=75 and Var[X] =a) =(3) .
(c)
-Pr72<.4;<77)\= ay =.1.
That is, only 10% of the time will a randomly selected
number will be between 72 and 77. On the other hand, for the sample mean,
Pr[72< X <77]* Pr)Uo <Z< |
6-45
(a)
Let X=
then
(Dyin
X, ae X?2 Fama eet X 100
100
Pr(98<X
wet X=
= Pr[-1.8 < Z <1.2]=84.9%.
=
<102)
Xy + Xp +--+
400
—2
2
pref2 <7 <2
=
attttow
X00 , then
=
Pr(98
< X <102)
5
ap
= pr{ 2 <z < 2 = ,9670.
C
(c) There is a larger probability that the sample mean is within 2 of the mean.
124
SOLUTIONS TO EXERCISES - CHAPTER 6
6-46
Theerror is U ~ Unif (—2.5,2.5).
Let
U = al
PH 252
and Var[U] = 2.083.
be the sample mean.
<5) = |ee
ee
E[U]=0
J -0434
fe
er
= see ALE
5000/25
X;~N(u=3,0=1).
and Var[U] =.0434.
(A tes )= 77. .@)
/.0434
NTN
Ligh THO
6-48
os +U 4g
E[U]=0
T = X)+X2+---+X,,
is the total lifetime of n bulbs.
E\T|=3n and Var{T
|= n.
ape mT > 40) = Pr(Z > Ie }Since @(2) =.9772, we have
nf
2
Ee Oran
ee
Ore
nh
x=Jn. Then 3x2 —2x—40 =(3x + 10)(x—4)=0 implies x =4 and n=16. (B)
6-49
Let
S=X,+-+-+ X2925.
Then ps =3125-2025 and os
o=n[sss]-| 254 ee |
ae
:
=
250-/2025
Sel
= 6,342,548.
integrate f(w) = 12w? -12w°; 0<w<§<1 to find:
eer
[2m
=
2
,
E(w]
Ps
==
oi
[2m
2
|
lawi)dw = Aaa
2
a
4
‘
Ot
-_—
Psi
== = =
=
law? dw = ee
9) i)
=
tps
250:,/2025-
6-50 You will learn in section 6.5 that W ~ Beta(a =3,8=2).
E{W]
. Thus,
Elin abe EEE
@)
g§ = 3125-2025 +1.282-250 4/2025
= 250-2025
=
=
=
=
2
i = =->eee
= 5e.
Then, Var[W]
(a)
E[S]=225.>= 135 and Var[S]= 225. =9.
(b)
Pr(S<141) = pr{z < HS 9 = 773
(C)
In the meantime you can
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 125
6-51
You should recognize the density function as exponentially distributed with mean
1000. Thus, each policy is sold for $1100. The variance of this exponential random
variable is 10007.
If X; denotes the claim of the i" policy holder and T = X; +-::+ Xjo9 represents the
total claims, then we are asked to estimate Pr(T > 100-1100).
E{T] = ELX, +---+ X10] = 100-1000 = 100,000.
Var[T] = VarLX, +-:-+ X00] = 100-VarLX;] = 100-1000
= 100,000,000.
or =10,000.
Pr
OrodOy hyena)
mene Zot
0 S100 DOU Ise oe Sy
10,000
SoD)
Normal
Approximation
6-52
N ~ Poisson (A=2), so un =2 and oy =./2 . Let
S = Ni, +---+ Nj250.
Then “5 =1250-2 = 2500 and os = 1250 we Si 2200FUe
Pr[2450<S < 2600] = P| _ 2450-2500
50
<7< 2600 — 2500 _)
50
= .9772 —(1—.8413) = .82
(B)
Note:
Technically, since N is discrete this application of the CLT should use a
continuity correction. But, (i) the statement of the problem makes it indeterminate
which way the correction would go, and (ii) os is sufficiently large that it would make
little difference in the final result.
Section 6.5 The Gamma Distribution
Geom
ye
i
al(275)
= 22 Dal sell)
(b) 1(5)=4!=
24.
(c) (2)=1.
GAT)
6-55
= [, xe td “to mar md
pester dees TitA\s = 30:
ee
(5) =
= 2.071197,01
15
vr
8
126 ® SOLUTIONS TO EXERCISES - CHAPTER 6
6-56
(a)
M ~
Poisson(A = 4) is the number of insults
permonth.
23.34
Pt 3) =
‘ik
1680.
Och.
YP
Gv
is
spell je
a
he
~
AGS
(b) W ~ Poisson(A =1) 1s the number of insults per week.
-1,40
Pry=0) ee 7 = 3679.
(c)
(d)
PrW>1) = 1-.3679 = .6321.
3679 - 6321
—
Vi
= .2325,
—-———"
not insulted
at least once
first week
next week
(ec) Let S be the time in weeks until the 4" insult. Since insults are occurrences in a
Poisson process with mean rate
4=1 per week, we have S~ /(4,1). Let Ys be
the number
of insults in a 5 week period. Then
Ys; ~ Poisson (As =(5)(1) =5).
Pr[S >5] = Pr[4” insult arrives later than 5 weeks]
= Pr[at most 3 insults in a 5 week period]
Then,
=o Priye=
6-57
(a)
0, 2.3)
Z
3
os(ies+ 4S45 |WG
Dis
Soe!
2ES:
W ~ Poisson(A =1) is the number of failures per week.
=1 42
PrW =2) = <=" = 1839.
(b)
The mean rate of failure is one per week, or 1/7 per day. Let T be the time in
days to first failure. T is exponential with mean {=7 days. Let D be the
number of days such that .5 =Pr| F< DI. Then D is the median of T and so
Doin
=F ine = 4s
6-58
(a) ELX]=10=a-f
Ly
es= an
(D) efx)
q
BS
(C)SWSerrat
ee
yee
Srdays.
and the variance: V[X¥]=20=a-f?.
SE re erenoe
oe
ee
An
edt
esi
ene)
lay
oat
aS
=. lee
d
2
Lheng Prix
= 6) a (6)
So B=2
(Ax)
;
7
3
as
without integrating?
Rte
24
= .1847,
4
(d) E[4X -5X?] = 4E[X]-S5ELX?] = 4-10-5-120 = —560.
how to find EX]
I
with A= sn =5,5=6.
= Ine|tsaedrs a om
yan
]
and a=5
Do you remember
E{X?]=Var[X]+(E[X))
= 20+102.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 127
6-59 (a) E[X] = a-8 = 12 and the variance: VLX] = a-f? = 24.
(b)
Property (6) of the Gamma
following integrals:
Distribution is quite helpful in evaluating the
a=9, B=2
x
EV Ye
BEA
oo)
]
i,)x 3: see)
=
5, >. pp li2x
xe
z
dx
|
esi
4
nn
ioe)
See 2%
ihXYre
dx
= AY
(c) E| VX | = Se
= PRG
= 2688
ER
a=6.5, B=2
= a=
he
erent dy
=acinsall
neeee 6.5 -I"(6.5)
2
Aa 6.5): (4.5)- 3.5) (2.5): (1.5)-(5) Vr = 3.393.
6-60
(a) D~ Poisson(A = 2) is the number of business trips by your dog over a
2-day weekend.
Pr(D <3) =
Ce)
01
echo
Seaiy
he 0
a
H
er
5
= .857.
(G4 1) 80 f= ate"; t>0
(c) Pr[T <2] = Pr[4” trip occurs before end of weekend] = 1—.857 = .143.
In other words, the event 7 <2 is the complement to the event that 3 bags last
through the 2-day weekend.
6-61 S~I(50,2)= 72(100). E[S]=100 and Var(S)
=200.
(a) Pr(S<5-J2 -o5)=Pr(S <5-J2 -J200) = Pr(S < 100) = Pr[Z <0] =0.5000.
(b) Pr[S' <100]=CHIDIST(100,100)
=0.5188.
(c) Pr{S <100] =GAMMADIST(100,50,
2,1)= 0.5188.
Section 6.6 The Beta Family of Distributions
6-62a B(4,2)
= [B(l-x)lde
= ~5 = {0 c'(l-x)3dx
= B(2,4).
.
he
canes
:
Of course we would not integrate in general.
Mae
LCM) oa
ee
stl!
ee
1
rr
128 @ SOLUTIONS TO EXERCISES - CHAPTER 6
sete RAs
HOO ens
ay
EB
1215 3960.
Tyo)
25}
B(3,1.5) =
6-64
TAS)
235225 15.0805)
= 1524,
X ~ Beta(2,6).
SINCE I
ee
FB) eeTOTO
(a) °=Fa)
=
2
ee 42.
oer?
(b) Kase
Vary x |= OEOs
substitute w=I—x
() PX> 5) = [42x18 de
& f242(1-w)u5 (du) = 42 [G8 -u®) du = .0625.
‘(O) Oe a 42x(1-x)>dx “implies Q, =.138.
6-65
X ~ Beta(2,2).
(@) fy (x)=
0
if x <0
43x°—2h
ei
0s):
1
Ne
ames
(b) ELX2] = hee Someta
(Cy Van xa 3-(4| S105
al ee
ee
25 fo fa 2236.
med
iE 6x(1—x) dx implies med =. N
(e) The mode is the maximum of f(x) = 6-x-(l-—x).
6-66
X ~ Beta(2.5,3).
Aer
EGS)
;
(Seats)
i
eee
(hae
a
Ole
(d)
This occurs at x =.5.
ease os
2
eek:
ae
eS
(5.5) (Gr5)
ioe
Pr.4<X<.9) = [, 19.6875-x° 2. (1-x)2dx = .582.
iy)
ie)
>
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
6-67
E[TX-5X°] = [,(x-5x6)(60x2 (1-2) dx
= 60.|7[8-2
27
de—s i;x8(1-x)3 as
LAOS
r(8)
oro
T(13)
81:3! ey=
ii Peeskst
he,PE
7!
a
a
6-68
faye
Y= Betala = 1,B= 5).
(Dye =o)
(cy
y=
PY ay) =] 1=1= yy.
(d)
y=1-.25'> =.242 thousand dollars. That is, the students needs $242.
(fe)
Y= =. On average the students spends $166.67 on beer per month.
(f)
Note that Y lives on the interval |0, 1].
L
]
Pr(.8<Y¥<1.4) = [,50-»)4ay = 00032.
6-69
f=
30x! -(1—x)* on the interval 0<x <1.
ig
F(x) = Pr(X¥ <x) = 41-6(1—x)? +5(0i—-x)®
r
Here
socal)
for
O<x<l.
fOr
ead
Use a graphing utility to solve for the requested percentiles.
5 = 1-6(1-x5)> +5(1—x5)°
implies that the median is x5 =.26445.
No) \| 1—6(1-x9)° +5(I—x9)°
implies that the 95" percentile is x9 =.5818.
# 129
130 ® SOLUTIONS TO EXERCISES - CHAPTER 6
Section 6.7 More Continuous Distributions
6-70 (a)
2
5
5,
Pr(3<X <4) = F(4)-F(3) = (2) HH) |= 1004.
(BELLY =2.5.
(©)
pA
2
2=B[X?]-2.5?
752
GB _{_@B
_ 20(10)
Var[X]
= = CE
(2)) -4
(0).
This implies ELX?]==2.
FY ReB | = 7-2,5-3--2 205.
6-11 (a)
On
EX] geope
=F 14
= T2 = 56,
abo
= = egy Sone
3.5
Q
(
yiesolving.> = 1-(
(dye
gives median =4-27/7 =4.88.
median
ekccallihatey (x) =O fora
0 =4.
Pr(3<.X <9) = Pr(X <9) = i-(4] = 9415.
aph ege= i-( >
08
x+5
3
(a)
9 = -(.|
(b)
Let Y, be the payment with a deductible of2.
x+5
Givesco. = 571
ee
OL
E[Y2] = [ -F@)] dx = hl = =)dx
6-73 X is Pareto type II with @ =3, 8 = 2000.
(a) E[X] = an = 1000
(b)
nae
F(x)
_
2000)".
1-200)
7x >0.
ee
Thus,
; 2000 wages
Pr |A > 1000] =|
1000+2000}
27
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 131
(c) Se)
2000’
| [
eee
im ne
ae
i
We= 519.84.
== 2000(2"°-1)
givesxs5
(d) Let Y be the payment.
3000 [1-F(x)|dr = iF3000/2000)’
bs {,
a
dx
= a
G +2000)
3 2000"
2006" (+2000)-
5 {3000
i.
(00
ne
Lee i
-2 | 50002 20002
_or
2000], 1(2000) ||=
_ Ho
in)
6-74
X
is Pareto type I with a =3, B=3.
aruba a = =.
(bo EG)
Pr
3
= 1-(3) ByeeeLIS
E(6)=—F (4)
X <6|X >4| = Sar
3
(cy
= 1-(3) Sives Kh = 321 = 3/8
t
6-75.
X is Pareto type I with
@ =2.5,
8=200.
Thus, F(x) I|
|
—
~
EES
pos)
paee 1-(22] gives, X25 = 2004
2/5
= 224,39.
2.5
13 = 1-(22)
re
gives x75 =200- A'S = 348.92,
The difference is 123.83
(A).
SN
i)S fn)
SSS
V
~ S Se
132 @ SOLUTIONS TO EXERCISES - CHAPTER 6
6-76 X is Pareto type II with
a =3, B=1 and so E[LX] = see = < (C)
DS
4p)
6-77 2X 1s.Pareto type Lwith o@ = 2.5, 6 =0.62
Thus, Fix) — 1-(4|
3
he cabs
Let Y be the losses not paid, 1.e., losses capped at 2. Then,
.
25
E[Y] = 6+ [1-F(x)]de = anf,
ieee
6-78 (a)
(b)
25
LS
dx
—
| 9343
(C)
Pr(X>1) = 1-F(l)) = e2" = 0.1353.
FLX] =
are
me
er(i+2] age r(¥] = 70876.
Note: The Gamma function can be evaluated in Excel using the built-in function
GAMMALN (natural log of the Gamma function). Thus,
r{¥) = EXP(GAMMALN(4/3)) = .89298.
Section 6.8 Chapter 6 Sample Examination
1.
(a) E[B]=100.
The mean, median, and midrange all equal 100. There is no mode.
(b) Range is 200, JOR = Q;-Q, =150-50=100, Var{B]= 200° , and OR = 57.74.
12
2
Let F denote the number offailures,
Pri’ $66) = mz
3.55.
25 ede
< 66.5 — 70
ef
F ~ N(70,0? = 45.5).
|= Pr(Z s$—.52) = 3015.
ee
ee Tb,
In(.75)
The third quartile is found by solving .75 = 1—e-*/*476!,
Thus,
Q3; = —347.61-In(.25)
= 481.88.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
Ames
oO
Galerie
Se
ClO
anceiu
ad
aL
6? -8!
0
inact
78.
—cl6
mere; Cag
Ld
ato
@ 133
iy
arene
this expression equals 1, since
integrating a density function
5.
The mode of D occurs at the maximum value of f(x) = cx? -(l-x) on 0< x <1.
This maximum occurs at x =.75.
6.
L=e*, where Yis normal.
=
Priam
Note: It isn’t necessary to find c.
Let m be the median.
Then,
ml = py]zs
S|
= Pre
=F
Mie Ol,
So, m=e>"!.
ie
8.
9.
Let N denote the total sales revenue.
E[N] = 64-7 and Var|N] = 64-9.
The 90" percentile is 448 +1.28-/576
= 478.72.
PrO>120) = Pe Z>0 ie
14/./19
= Pr(Z >2.49) = .0064.
Let L denote the individual loss covered by the insurance company. We will use the
Central Limit Theorem to approximate total losses, S = L, +---+ Lago.
20,000
E[L] = ee (x-5000)
-
i]
7 555 dx == 56 5625.
20,000
ELL?) = [.., (%-$000)? -5—
|
dx = 56,250,000.
So the standard deviation of a single loss is 4961.
S —(200)(5625)
Then E[S]=(200)(5625)
Os = (200 }(4961) . Thus,fe Z
oe
es
wire
§ <1,200,000)
00,000
Pr(1,0<
<1.07)
< Z 78
= Pr(—1.
AN
= .82.
(D)
and
134 @ SOLUTIONS TO EXERCISES - CHAPTER 6
10.
The loss random variable X has a Pareto type II distribution with
2100
———_
Thus, (x) = = |-| (in|
EY]
4
ae) If Yis the payment then,
for x>0.
ik L ‘ F(x) |iis
4
_.
®
2100
oe fl ao
|a
ae
(x+2100)3|"
00E
3
a
= 468.95.
24003
=
11.
@ =4 and £ =2100.
300
Let
X and Ybe the two errors and let W = 5 (X4Y),
Then gy =0 and of = —-[(.0056h)?
+(0044)? ] = ow =.00356A.
Then,
Pr|W |s.005h] = Pr Z|
Osh aa = 2@(1.4)=1 = .8384.
(D)
CHAPTER 7: MULTIVARIATE DISTRIBUTIONS
Section 7.1 Joint Distributions for Discrete Random Variables
ae
=|
a) Pa)
a
ee
tors
z
=
2 PC7i)
3
00
EY
Te
ee
=
d x! y!
“
=
oe 00
3x71
=
3e 2) ) eam
x!
.
ee
Bes
—
3”
x!
.
l
=
e3
x!
= 0F))235:
9) ELX |] = ais :
fe
fs
aes
a
s
=
ee) 3
3e <i a——==
x
:
3 y e--e
i
ek
Sh
[eesye ap(x,y)
= 0?- p(1,0)+1? -p(1,1) + 4?- p(1,4) + 0? - p3,0) +1? - p3,1) +---+ 4? - pS,4)
Using marginal probabilities this is
= Oe(30)1a (25)44°>C45)1= 7:45
(b) E[J¥ |= -32+ 2 22 =1.253,
136 @ SOLUTIONS TO EXERCISES - CHAPTER 7
7-4
ae vpsari 6apts
Sp Sige
pela
+2. 1
qed
te rae
te pay
1/4 Pty
7-5
I
teatRA By7ndi. eae ees)aay (8mest
aero -—+1]?.(g+4)+
Curae 12
Pe
07
Var
= “t-(3) eS
E
2
gas)
(a) and (b) The individual probabilities are determined from the hyper-geometric
distribution.
ee
po
G
s
(he 0,
Ces | oe
eee
See apenas
220 | 220 | 220
123
36
220
48°
220
nOUR
220
ile
220
120
(c) E[C] == 0 0: a0
90
SYN
Che
ee
TI
ea)
10
oDs 590
= -
et
=.
20
0
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
® 137
Section 7.2 Conditional Distributions — The Discrete Case
7-6
(a)
(b)
ELX |¥ =3]
On
LY
=
yes]
EAE)
=
Ven
0.
ee
ee cec:
Var[X
|¥Y =3] = 5.5—(2.2)? = 6173.
(dq)
ELX] = 1-(037)+2-(.42)+3-(.21)
= 1.84.
BUX
3221)
1
C37) ees C42)
+
3 94:
Var|_X ]=.5544.
0(0) + 1(13) + 2(23) + 3(6)
=
(f)
ior
,
= ge
=
&
E[V¥
1x=3] = SOND LD a2 C)S A Save)
Let T equal total accidents and W unreimbursed accidents. Then
Since we are given that_X is zero, the possibilities are:
EIEAPAA CANOE
S
wr]
nlp]
Nl
oO;
S|
Oo|o|&
|
Se
W = max ie = 2,0].
138 @ SOLUTIONS TO EXERCISES - CHAPTER 7
Then the distribution for W (given_X is zero) is:
7-
EQ ILX =1)
05
= 0:
ORES a)
pl2e
|-
Er
Si OE She
aad SG ile
La)
EU
3
We
have
p(m,n2)
1
Ay
= +(4|
E
oe
9
(E)
= .741286.
Vary (X= 1) = 741286741286"
7-9
ae7
= .741286.
a
05
JOE
2 211
= 204"
“*(C)
l
e™(l1-e-™)"@;
m,n) =1,2,3,...
and we
need
to
calculate
E| Np |N = D | Let p;(2) be the marginal probability that N; =2 and let
P2(m|2)
= Pr| N> =nz|N,= zi = oe
Pil4
be the conditional probability function
Na
for ny
4
dl
all
pL
>
-2(J-e-2
ym =|
Al
Aen
5 \21
2
no
Wn
| +elea (1-«
—-e- ary
=1,2,3,.... Then p(2,n2)
3(
=].2
INEXt. 571 (2) <= zen)
no=1
3
]
A
4
4 Je Ce
geometric series with
:
>
ratio l-e~*
1)
4x
ee
3} Ze)
AED,
2” <=
n=!
Thus, p2(n2|2)
pP(2,m)
= ———
Pi
0 (2)
jl—-e77)m@-)
=
oie
ia}
- 5
—2
+—
= e7*(1-e7)@71;
7
Giese«=
l
ea
Ww =1,2,3,...
;
=
TL:
374
)
.
.
This is a geometric
‘
distribution in the form “number oftrials to the first success,” where the probability of
.
9
~
success 1s e ~. Therefore, the expected value, B| N2 |
= 2 |, equals
——=e?.
|
aie
(E)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
@ 139
Section 7.3 Independence — Discrete Case
7-10
Method 1: We will complete the joint distribution table (using independence) and
compute the expected value using all joint probabilities.
Be
ale
4
ta) 2). 2
Method 2: Since X and Y are independent,
marginal probabilities.
(5) (3) COTY
15
E[LX-Y] = ELX]-E[Y]
We can use the
ELX] = (-1)-(.4)+(0)-(.5)+(5)-C)) = .1.
BLY] = ()-(.6)+(2)-(3)+)-CD = 15.
E(X-Y] = ELX]-E[Y] = (.1)-(1.5) = .15.
Section 7.4 Covariance and Correlation
7-11
Cov[X,aX
+b] = E| X[aX+6]]-E[X]-£[(aX+5)|
= aE[X?]+bELX]-E[X](aE[X]+5)
aELX?]+bE[X]-a(ELX]) -bELX]
= a(ELX?]-(ELX)) | = a-Var[X].
7-12
E[F]
-+3-4-—+3.5-—+3-6:—-—
E[T)
(2)-(4)
=
w|r
140.
SOLUTIONS TO EXERCISES - CHAPTER 7
7-13 Var[X]=10(.2)(.8)=1.6.
Cov(X,2X) = 2-Var[X] = 3.2.
II 25Var[X]+81War[Y]+2(5)\(-9)Cov(
7-14 Var[SX —9Y] =
X,Y)
= 25(13— 9)+81(20)-90(7 -(-3)(4)) = 10.
7-15.
17,000 = Var[X+Y]
= Var|X |+Var|[Y]+ 2Cov(X,Y)
= 5000+10,000+2Cov(X,Y)
Var{(X +100) +1.1¥]
=
Cov(xX,Y)=1000.
Var X]+1.12Var[Y]+ 2(1)(1.1)Cov(x,y)
5000 +1.21(10,000) +2.2(1000) = 19,300.
(©
7-16 8 = Var[X+Y] = Var[X]+Var[Y]+2Cov(X,Y)
= (27.4-52)
+(51.4-72)+2Cov(X,Y)
=>
Cov(X,Y) = 1.6
1.6 = Cov[X,Y] = ELXY]-ELX]E[Y] = E[XY]-5-7
=>
E[XY] = 36.6
covariance formula
Cov| Cy,Cz |= E[(X + Y)(X +1.2Y)]- E[X +Y]- EX +1.2Y]
E(|X?]+2.2E[XY]+1.2E[Y?]-(£[X]+ E[Y])(ELX]+1.2£[Y))
= 274+ (2.2)(36.6) + (1.2)(51.4) —(5 + 7)(5 + (1.2)(7))
= 3.8:
(A)
7-17 Cov(X,Y) = 1-0-(.17) +++-+3-3-(.08)=[1-(.37) +2-(.42)+3-(21)]-[1.45]
ELXY]
te
aa
SE
= 2.8-1.84-1.45
= 132.
t
718) Cov3Xx,—1Y |= E121 XY SE Rete
—21{ELXY]- ELXJELY]}
l| ~21Cow(X,Y).
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
7-19
(a)
The joint probability distribution with replacement is:
ELXY] = F(U+2+3+24446434649) EY Cov)
(b)
4220)
The joint probability distribution without replacement is:
E[XY] = (24342464346) =F. conx,y) = ee
36
(|(1.2)2)5 30.
(a)
6
5)
2
Tae aeWe
= 0e
Var[X
er ]ee
(b)
Var|Y | == 02 fea B47 =e
PEE
COVEY.) = 2.=(1.2)(.8) ease
(a)
Var[X+¥] = E|(X+¥)?|-(ELX+¥])
10
Alternate solution:
Var[X+Y]
10
9
aa3
2 al
a3 eis)
>
—
36
10
= 4-(2)? = 0.
= Var|X]+Var[Y]+2Cov(X,Y)
= 0.
Also, note that X¥ + Y = 2 is aconstant random variable, so the variance 1s zero.
@ 141
142 @ SOLUTIONS TO EXERCISES - CHAPTER 7
7-21
The joint distribution for X, Y, and Z is summarized:
So ea
(a)
Cat
ra
Since the joint probabilities are the products of the marginals, it follows that Y
and Y are independent.
(b)
Var[X+Y+Z] = E[(X¥+¥+zy |-(B[X+¥+z]) = 3-(3) ~ 75
()
SEXY Zi m7
()
gy =.
ea
AX]: EIY]-HZ] = ie5 a-5-=
(e)
They are pair-wise independent but not all three independent.
Bae Te Sh
7.07
S
4
85 —(1.95)(.6)
ee
2° 2-2
= Prix sl)
Pry =1)- Pr
ayy
4.35-1.952 -/3.4-.67
7-23
CovLX,Y]
=Cov[_xX,-3X]
Cov[X,Y]
a
For example,
= -3-Cov[X,X]
—3Var[X
|
\Var(X WarlY], 7 Var x1(9VartX))
= —3Var[X]
|
=i)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
7-24
Because daily sales for B are more spread out,
Var(X) <Var(Y),
and
Var(X+Y)
= Var[X]+Var[Y]+2Cov(X,Y)
negative
< Var(X)+Var(Y)
(E)
Section 7.5 Joint Distributions for Continuous Random Variables
7-25
ed a aii “2 dy de
Prox
Y <1)-=
(
Ne
7-26
(ah
shaded area
total area
=
1/2
1
6
ie
f(x,y) 1s constant (because joint uniform). The constant equals
l
areaofregion
(a)
(b)
ee
ee
4/3
4°
fx(x) =
f° fvay
ely)
rt.
ar
s a
Lem
Bysymmetry,
= a
3,/l—y
= Sys
E[X]=0.
for -1<x <1, zero otherwise.
:
for 0< : y<l, zero otherwise.
You should do the integration to check this.
ia
E|Y] =
0
} (11) -40'0?
& = +f) y-(l-y)!"* dy =
[iy Ir(y)
vo
|to
substitute w=l—y
fw? Ho
me
ie Ou = A.
(—du)
—
# 143
144
SOLUTIONS TO EXERCISES - CHAPTER 7
7-27 The device fails if either component fails.
The correct answer choice is (E). There are other correct ways to write this including:
1- ie [ fls.ds dt
or
lizff(s,t)ds dt + he lief(s,t)ds dt .
7-28 Pri(X+¥<l) = {,[ede dy= fl-ee?] * dy
= [[cet+e)] dy = [-ye!-e"] .
=l—2er*.
(D)
y=0
7-29
7-30
The joint density function is constant on the triangular region with vertices (0,0), (0.L)
2
and (L,L), whose area is =. Thus, f(t,,t2) ==
for
Biel
(C)
p= { f,(e +13) = dt; dt) = 2E
0<4, <b <L.
Both components are still functioning in the given triangular region.
Mm
40
x=50-y
Boundary Line
Region of Integration:
~~ X,Y greater than 20
i)
10
20
30)
{0
SO
6
30 -50—x
pres
Thus, the probability is the integral, 125,000 We iF IU)) Gy)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
7-31
Pr(X +Y 21) is found by integrating f(x,y) over the shaded area. (D)
ee
Pr(X+Y = 1)
1 x?
eee
ie
eee
2
Seen
lt17
aA:
0.75
0.5
y-values
0.25
x-values
Note that Y takes on values from 0 to 1.
g(y) = PY Pony) ax = JP IS yde = 1Syx my
oy
gly) =
Jisy? *1—y")
0
otherwise
|
7-33
7-34
No.
f(1,2) = 1.3-.8—.8
20-30
for 0<y<!
,
= —.3.
(E)
Density functions cannot be negative.
= 30%.
ey. 50 9
7-35
The area of an ellipse of the form
Pr(safe
(safe rabbit)it)
=
(xh)
i
(4 1-20-30+-4-407]
|
40-80
:
lat
bh?
= 3973.
5
=
1 is Area
= z-a-b.
@ 145
146 ® SOLUTIONS TO EXERCISES - CHAPTER 7
7-36 The joint density is constant on the
shaded region whose area is
rae
= 34
Thus, f(¢,f2)=—— on the shaded region.
By symmetry,
E\T, + T)] = 2£[T]
2E[T;]
148)
9
6 pl0—1
\45 8
2
| J,jptaran+ [fa asa = A34 (488
7-37
The joint loss lives on the region 0< x<1land 0<y<1.
(x+y-—1) over the sub-region where
E{max(X
+ Y -1,0)] =
(Est
Je 0@)
x+
y21.
_
= 512540)
The policy has a payment of
So,
{,2x: (rt y=) dedy = + (A)
leif(v)6e*%e?” dxdy
II
Flee” cof 0 dy
II
he{6ye?” .(-e-? +) dy
J,(Sve?
6ye )dy
Don’t get scared now.
a)
IF.2ye*” dy -2
il.3 ye"dy
SE
this is the expected
value of an exponential
ry. with mean 1/2
= 3(1/2)-2(1/3)
10
—
this is the expected
value of an exponential
rv. with mean 1/3
Cees:
5
= —=-= ==.
+i,
6
|
lo
(D)
x?
wh
7-39 ELX = [ fia
7 (10- xy’) dvdx = ali [10x- : Ja = 5.77.
(C)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
7-40
The failure time, /,
eas
A
# 147
is the maximum of Y and Y.
if SY
i ae
inital = ake (35)dvde + [wee
, (%) dxdy = 2.54+2.5 = 5.0. (D)
I(x)
f(x,y)
if y>x
if x>y
Section 7.6 Conditional Distributions — The Continuous Case
FAV fimyy=44n
0
(a)
sets
x*+y* <4
OO
f
otherwise
fx (x) = | ‘ th iy ee
4-12 4a ~
20
for —2 <x <2, zero otherwise.
Similarly for fy(y).
(Dyess
(ilno=1)
=
*
edhe
AMESOY,og
fx)
aa,
for=/3 2 y<./3. That is, given
Tae
et
X =1, the conditional distribution for Y is uniform between —J/3 < y <3.
7-42
3
Bie8 = 2691.
folx=Day
= oe
(c)
Pr(¥>.8|X=1) =
(a)
f(x,y) =
(b)
eer
l
Given X =<, the conditional distribution for Y is uniform on o.3|
:
1
Ib
fry |X =x): fx(x) = aaa =i
as
x= 1.
fomQ sve
Therefore
l
|
ahaa
We nae amr”
[x9
PY > An = 7 =
vea5
148 @ SOLUTIONS TO EXERCISES - CHAPTER 7
(c)
eee
We have to find the marginal density function for Y, fy(v)= [= dx =—In(y),
foreU=y =< i So
fy (1/4) e= In(4).
Sa
(1s)
al)
F(i/4) ~ taf)”
4) - =f
fx{v1¥=
Then,
i pesekal a iste ees)
Ne ”
Pe x>Hy=t atin hoe
(d) ~~E(Net Loss] = ELY—Y]= i,[o--y)t PGi {de -+.
(e)
Since Y given X =3/4 is uniform on ne/4],
elyix=3, _ 3/440 _ 3
2
(f)
7-43
8
“Off with her head!”
Fy (y) = [fy dx = 4y—3y’, for O< y<1.
ELX|Y=y] = [x-fe(xl¥=y)ak
A
{poe ;
1
(2—x-y
% Y)
ees
es
4y—3y?
Sy(y)
for 0's yal. So, for example,
HE]X |Y=
l=
sdk
2.2
Ce
=
7-44 fr() = ibS(x,y)dk = hes a=
Lye.
25—2y
=
,
4-—3y
= 1622.
2
é 2s tor 0< pao
E[X2|Y=y]= lice fr(x| ya = [ox ev/y
y
e =
Alternatively, note
that f(x,y) 1s a function of Y alone.
This shows that f(x|Y = y)
is uniform on the interval [Oy] , Hence,
E[X?|Y=y] = Var[X|Y=y]+£[X|Y¥=y]
Since the joint density function
- eal
aba
is constant on the region bounded
os.
by y=0
and
y=4-x’, we have that, given X =1, Y is uniformly distributed on 0< y < 4-1? =3.,
3-0)?
Lheny |Var|y
1
|x =| ce ca)
-
3
>, and so oyj\y-(oh
.866.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
7-46
# 149
The event that the device fails in the first hour is represented by the L-shaped region
ys lf
(Ose SOs
The complement to this region is the square (1SxS 2))(iy
2),
Therefore, the probability that the device DOES fail is
ff
1A es
— dxdy = 1- fp Gy = 625" (D)
(Ole
ake
tr (y) = [24x =
y
+2 Bo) tor Gaye eo.
yeesO f(x
|
y= Ly) yor
Sy (y)
lay
Oye
(E)
Section 7.7 Independence and Covariance in the Continuous Case
7-48
By independence, we have f(x,vy)=e
~* <y on the rectangle [0<x<o)x[0<y <3].
We need to determine the region in this domain for which John and Mary meet. In
order to meet we must have Y < X < Y+1( see diagram).
Pr(shaded region) |=
SF
a
|t
—
Se Q.a S
[-e'(v+0]
=5(I-e") Se
ey
= <(-e(I-4e") = 1125
using integration by parts
x
X denotes John’s arrival time, Y denotes Mary’s arrival time.
150
SOLUTIONS TO EXERCISES - CHAPTER 7
TAQ (ay
LD1
ofc) =e 2
Porereniy ee Sa at cy 2
a
ar
sees
0
otherwise
(6) fr(y)= [)fy) ar=1, for O< ysl.
But we already knew that Y was uniformly distributed on the interval [0,1].
()
7-50
£E| VX -¥3] = f,foe pha = 6928.
The region where the joint density lives looks like:
The shape of the region makes is easier to conduct all integrals in the form
i [C --) dydx .
EX
epee
ee
|=SEA
3 I,I. Sree
X-V dyde
= 3 i.
5% Vee
as Meee
5 = 800,
(eo
EY Bed
mall
I xy* Se
dyaxhe =
Gel
_
Cov[ X,Y]
8 pl
ex.
3 j,i) Ayo
5
a3 iheeee
OF1.244
Bats
ri
fica
Oa Cal ieckaeaeta Se
dvax = a liea=
=1.037 —(0.800): (1.244) = .042..
a
oT 7 1.037,
(A)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 151
3
7-51 Weare asked to find Pr(1<
¥Y<3| ¥ =2) = I1: f(v|X =2) dy.
fe2)= "fd
= [Syed = 4
:
yk
2)
ibaa
NE
Vey
Fx (2)
Se
|
etry
=|e
3
*
Pr(l<¥<3|X=2) = f'2y3
dy = =. (E)
7-52
Given X =1, the conditional distribution is uniform for 0 < y <1.
EY? |X=1]= [\y?- QIK =Ddy = fy? -ldy = +.
2
1
3)
1
3
I
for
7-53 (a) f(xy)=4n
0
o:
x-
vege
9
»)
ye <
ro
otherwise
2
©)
b
Prl(X24¥?) >.5]=1- 203) =1205) _
(ei
Pix<Y] =>.
@
fe@=|
se =~
fal j2
Ly
(e)
?
Ree
1
:
The random variables are dependent since the joint density does not live ona
rectangle.
152 @ SOLUTIONS TO EXERCISES - CHAPTER 7
7-54
Var[Z]| = Var[3X-Y-5]
= 3? .Var[X]+(-1)?Var[Y]+ 2(3)(-l)Cov(X, Y)
= 911) +1(2)-6(0) = 11.
7-55
(D)
LetX be the time to the next Basic claim and let Y be the time until the next Deluxe
claim. The joint density is given by f(x,y) =
eot!2
ev/3
Des
. The event Y< X has
probability given by,
= x2,
PY <X) = i =
—x/2
i} € :
=
TEN
Ne
1
¢l
ew
5 bs
(1—e-*3) ad =
Hie
\ [kx aeay mer
Sie
—¢-
6) dy
a
2 (C)
:
implies k =2.
ELXY] = {,[22x dx dy = 7
EL
=
{,[x 2x de dy = 2.
1
el
Ele If[jy 2x de dy = >
Covey) = s-25 S70,
We could also observe that X and Y are independent because f(x,y) can be written
of a function of x alone times a function of y alone, on a rectangle. Since independent,
then Cove.),) = 0. (3)
7-57 VarlY|x] = E[Y? |x]-(ELY | x1)
Ix (x) = [Fe
of Ya
tex
ees
x)dy = Finede fy mde
f(x,y)
is ax
Ix (x)
2x
== |,
That is, given a particular value of x, the conditional distribution for Y is uniformly
distributed on the interval [x,x. +1].
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
x+l
SY see)
2
EL
Pete] = (oy erence 3 eetnioers
=
3
and
ais
E[Y |x] = [ l ydy — (x4)? =x?
Z
Vanhy \x=
ely
2
3x? 43x41 =|aes al
4
”
se
We could also have recognized this immediately since the variance of a uniform
random variable on an interval of length lis 1/12.
.
Ese
(A)
l—x
f(x) = Is f(x,y)dy = i, 24 xvdy = 12x(I-x)?,
Jee
ae
i
yO)=
9
r{ .
WV
~~
Ix (x)
l
2y=
t) ee.
=
24xya
SS
ay=aEe
SoS
Vege egy
=
Oy
2
==>,
O<
0<x<1.
(l-x)*
4
<a
ay 2 priy<ty=4]
bl promt meadj(eenan
Pr <4 sl ioe:
x=4]
= nein]i
7-59
(C)
The joint density lives on the region shown, with
iver aex)
fy(xle lo
b
Of xsl ys ysrtl
Since the joint density is uniform, we can use
simple geometry to complete the problem.
The event that Y is greater than 0.5 is the
complement to the triangle with vertices (0,0),
(0,1/2) and (1/2,1/2), whose area is 1/8. Hence,
Pr| ¥ >.5|=1-1/8=7/8.
(D)
Xx
# 153
154 ® SOLUTIONS TO EXERCISES - CHAPTER 7
7-60 F(x) = [,30-9? dt = 1-(1-x)3: O<x<I1.
In this instance F(x)>x and |x—F(x)|
= F(x)-x
= (1-x)-(I-x)
and G = 2[ [(1-x)-(-x)'
Jar = 4-4 =5
Section 7.9
(©)
Bivariate Normal Distributions
7-61 Var[X|¥ =y] = 02 (-p?) = (32)-.84 = 7.56. oyy-6 = V7.56 =2.75.
7-62
Given Y=-2,
X is normally distributed with
Pep
3
= 8+=(-3)(-2-(3))
= 7.82
Var[|X
|Y =-2] = 9(.91) = 8.19.
Pr(X <9Y¥ =-2)=Pr(Z < O78) ) =Pr(Z <.41) =.659.
V8.19
7-63
(a) ox =1, uy =0, oy =2, wy =—1, and 1— p* =.36.
So
p=+.8. We know that
p=.8 by looking at the sign of -1.6x et5 }
\
2
(b) Pr(¥ >.5)=Pr(Z > oS: ONS PNZ > 5) = 3085,
7-64
Let
A~ N(w=10,000,o = 2,000) be the random claim amount from company A, if
there is aclaim. Let
B~ N(u=9,000,o = 2,000) be the random claim amount from
company B, if
there is a claim.
If X = A-B, then E[X]=E[A]— E[B]=1,000 and
Var(X] = Var[A]+Var[B] = 8,000,000 so that oy =2,828.43.
01,000 )
:
Pr(X/ <0) = Pr(A<B)
r(
) = Pr}
i Z < <————_
7898 43 | = Pr(Z
( <-.35)
5) = .362
6
given claims
from A and B,
Pr( d<B)
——_,,
PrA<By=le
G60)-(30)
————_
"=
33625
oa)
G0)
a
probability no claim from
probability claim from A
A and claim from B
and a claim from B
= .2234.
(D)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
# 155
Section 7.10 Moment Generating Function for a Joint Distribution
7-65
X = J+K+L.
Since the losses are independent,
Mx(t) = Mj()-Mx(@)-Mz() = (1-22)? .-28)?° - (1-28)49 = (1-24)-”
M(t) = 2001-20)",
M¥(t) =4400-20)-!2, and M%(t) = 10,560(1-21)-33.
E(X3) =M% (0) =10,560. (E)
defn
7-66
Myz(ttr)
aa
= Ele”
+22|
2 El et(x+Y+n(¥—x) |
E| eft +t.)X+(-t; YY |
=
EI ett +12) X o(ta—h yy |
Since U and V independent
SE
ge
aa
II es Lei +f x |B ele yy|
(t) +t)?
= My(tit+h)-My(m-4)
=e
2
(t2-t)?
-e
2
= ef th’,
(E)
Section 7.11 Chapter 7 Sample Examination
1.
0 = E|(X-2)3] = ELX3-2X?+4x-8] = 9-2E[X7]+8-8>
Gh +
AP
Zz.
(a)
ey
TAG.
= 3-4 = +.
The area ofthe triangle is 8, so
;
I(%y) =
{t
for
lo
otherwise
O0<Y <2
and 2y-4<x<4-2y
see
Wy
(b) By symmetry E[X]=0 and E[Y = [ ike g ddy =
2
(c) The event [Y >1] is a triangle (see drawing in text) of area 2. The subset
consisting of [¥ <1/2][Y >1] has as its complement in [Y > 1] the triangle with
vertices (1/2,1), ( 1/2.7/4) and (2,1).
Its area is 9/16 and so the area of
[X<1/2}A[Y >1] is 2—9/16 = 23/16.
a)
73
Thus,
PrL¥ <1/2|¥ >] = zone = = = 71875.
(d)
By symmetry, E[X |¥ =.3] = 0.
156 ® SOLUTIONS TO EXERCISES - CHAPTER 7
3.
tt
for Osa Sh
0
otherwise
and fy(y) =
yt+.5
for0<y<l
0
otherwise
(a)
fx(X) =
b)
ALX]=21Y]=5.
(c)
X and Y are dependent. The region is a simple rectangle, but f(x,y) cannot
be written as a product of a function of x alone and a function of y alone. Or
realize, for example, that f(1,1) = 2 ~ €.5)-(:5) = fy@:- AO.
x+y
(dq)
f(x|y)=——
yt+.5
for0<x,y<l.
(e)
ee
9
EIX|¥=.7] = =...
f
Pay
(f) E[7X -3Y]=<.
‘gap gia we1/144Re
oe
a naa
(h) Varl7X -3Y]==>.
a
1
rela)
(b)
2
(c)
ay 8 Sa
(lee
el7
so.and ee)
y= 2s 33.
(e)
X and Y are dependent.
p(x| y=3)
(g)
(h)
(i)
Gj)
(k)
(1)
E[X|¥=2]
= 19/11.
E[_XY]=3.7575 =124/33.
p= —.03108.
V[X]=.19835and V[Y]=.6336.
Cov(X,Y) =—4/363.
Var[7X —3Y]=15.88.
ae
ee]
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
(A) rE |
(b)
ey
# 157
a47 = 92EXE BY 4 bz pas:
Var[2X-Y +4Z] = 4Var[X]+(-1)?Var[Y]+16Var[Z]
—4Cov(X,Y) +16Cov(X,Z) —8Cov(Y,Z)
= 4-(2)+1-(4)+16-(5)—4+16-(-1) -8(2) =58.
(.5)
Se Cover
2)
time Covix4)
OneION:
25,
Cov(X,Y)=-6.
mis
on
:
)
Cov(X,Y)=82/3and p =2.768. Since the correlation coefficient must be between
negative one and one, this made up problem can not exist.
From the statement of the problem,
rts)=|2ix+y
+9)
0
for O45
0< Y< X¥ <1. Thus,
leand 075
otherwise.
fr) = [fe yav = [204+4
hen
=) =
jy)
= 3x2
elon
2 21+ y)
fx (1)
03
~
05
2
my
05 2G jpe Ye
| = \, fr(y|x=.)dy = | cia
Then Pr(¥ <.05
¥ =.10)
> ~ fy 6.1) = 03.
416. (D)
The key step is to note that,
= on the triangle 0< y<x and 0<x<12.
whee y)= fr(y|x)- a)
X
Since X is uniform on [0,12], ELX]=6.
You should do this without calculation since
X ~Unif (0,12).
Vp
fi[,v Sl.) dvdk = ie[=
ee: dydx = (45a oe = eh
=[ j; a dyd.
12°
=" >5 dydx= (E x de==e
Cov(X,Y) = ELXY]-E[X]-ElY] = 24-63 =6.
(©)
24..
158 @ SOLUTIONS TO EXERCISES - CHAPTER 7
10.
Let O denote the number of tornadoes in county Q. Let P denote the number of
tornadoes in county P. First, Pr[P=0] = .12+.06+.05+.02 = .25. Then,
jy Ve
ee Se
ee
it
Pry 22)
eS
09, 32 SS (.88) =. 980.
(D)
6 i hk (l-x-y) dydx
_
2 (l-x-y)?
= 6[/ere
l-x
x dx
_ 3 [ex de = - (ex)
Bri
12.
Let
let
8a= 488
G~ Exp(A =6) denote the random wait time for a good driver to file a claim and
B ~ Exp(A =3) denote the random wait time for a bad driver to file a claim.
PG
<3( B= 2) =P
nG = 3) Pile2)
I
(een?) (le)
(=e).
DB
syle"?
13.
(C)
d-e
e's
7?)
e-7
(C)
Let P=premiums ~ Exp(A =2) and C=claims ~ Exp(A=1).
joint density function is f(p,c)=+e-?’7e*
for
p>0,c>0.
By independence, the
We define X ==
We
want to find the density function for this ratio. We will use the standard technique of
finding the CDF first, and then taking its derivative.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 159
FGe)e =: Prea)
= Pr
S
= |
Rite
< ae
_
i, WP
=f
=
ere
i>
—p(x+1/2
))dp
—ePl*
(e-P/2
= pie (l-epsa
ey
ee
~plo 4
I
[ee)
—p(x+l/2)
p=0
SS
IS
eecie 4
2
(x41?
Th ell. be)
‘@) =F F(x) @)=———.
The CDF for an exponential distribution with parameter 8 is F(x)=1—e~/. Then,
Ene8) 0
0+5-[e% — e895] 476 [e48 _¢-#i| = 71223 (0)
peeks
————
Sa
N=0
Let 7;
N=l
N>I
=X; + Y; denote the total number of hours watching movies or sports of the i"
randomly selected person.
E[T;]=50+20=70 and Var[T;] =Var{X;+Y;] = 50+30+2-10 =100.
Let T = 7, +72 +-:-+Tjo9.
Since the people are chose randomly, it implies that 7; are
T; are pair-wise independent, for all i # /.
E|T | =
E{T; +T) +---+Tigo | =)
7,000)
Var{T] = Var[T, +T2+---+Tioo] = Var[T]+---Var[Tioo] = 10,000, so that
Or
= 100.
Pr(T <7100) = pr|Z<
7100 — 7000
00
= Pr Zi< |) = S413:
(B)
160
16.
SOLUTIONS TO EXERCISES - CHAPTER 7
Sketch the region where the instrument will fail.
X fails
in first
hour
Y fails in first hour
Pr(fails during first hour) = 1—Pr(both instrument work after 1 hour)
-=
x+y
gohka
dx dywee= > = 407
(B)
AI
17.
Again, the key step is to find the region for which the total payment is less than 5.
Since we have ajoint uniform distribution on 0 < x <10,0<
y<10, we can calculate
probabilities using the ratio of the area of the event divided by the area of the square.
Region
Ee
where t al be efit
is less t an 5.
Pr(event)
2:61
= 10
Sess 5
ae
=
;
529),
(G)
PROBABILITY AND STATISTICS WITH APPLICATIONS:
18.
A PROBLEM SOLVING TEXT
@ 161
Since the joint density is f(x, y)=2e-¢*?”) =(e*)(2e*”) on the rectangular region
[0,00) x[0,00), it follows that XYand Y are independent.
Thus, the value of X is immaterial and we need only consider
Var[Y|Y>3].
If
W =(Y —3)|Y>3, then by the memory-less property of the exponential, W is also
exponential with a mean of 1/2 and a variance of 1/4.
Then, since Y| Y >3=W
+3,
we have
Ee Yi3) e735. o-and VarYY > 3)
Var[W +3]
Var[W]
= 5” = 25,
(A)
cae
atn
aif Sao
fr9rn Taw iy
m
Ose
if
CHAPTER
8: A PROBABILITY POTPOURRI
Section 8-1 The Distribution of a Transformed Random Variable
8-1
For l<y<e,
Iny2=0.
[a dy = (Iny)? "=
So ei
12-0? = 1,
y=l
See
ror 0 =x <1, Fy (x) -=—> =x.
=
l<yse.
Since Y=e*, the random variable Y lives on
Fy(y) = Pr(¥<y) = Pr(e* <y) = Pr(X <In(y)) = In(y).
(a)
$0) =O)
=— for lay e.
(b)
Method 1: z[Y] = ["y-tay
= [~ e-1.
1
y
Method 2: E[Y] = a Mere.
Deer
Ore.
0. 7(x)=
~ooince ) = ase. the random variable Y lives on 0< y<o,
Fy(v) =Pr(¥ <y)=Pr(JX <y)=Pr(X < y2)=1-e
(a)
f(y)=Fi(y)=2ye” for
(b)
Method 2: The key to computing the value of the integral of the following type
is to use Subsection 6.5.2, property (6) of the Gamma distribution with
parameters
=
(c)
SA
a =1.5 and #=1.
1
\ Jx -e7*dx
=
Also, r[$|-vr
2/5
j,x22.
a
ody =
3
r(3| =
X ~exp(l), E[Y7]=E[X]=1.
Since Y2=X, and
Ore.)
0< y<om.
Y
fy=pen"
=
nC)
I
‘2
]
r(4) =
aie
ve
Z
— Aa
x = ce" = 1 and a =e.
ay
(a)
fyrQy) =
fx (x)
for
y<o.
OS
(by eel | = [yee
/
a
— 1
ete.
6d Pe rs
/4\e |=
ae
a mee
oe? Se4
mk
Ody = 1.3409 (Using a graphing calculator).
163
164
SOLUTIONS TO EXERCISES - CHAPTER 8
8-5
CDF Method:
Fy(yy
Fy(x)=1l-e™~*, so
= Pry)
= Pri0- ee <-y)
1/8
SS
Tt
1/8
a
Pree ea
Fy
V
(2
a
=
1/8
/
(yoy’®
_
(y/10)'> 1.25
_(v/10
Saf
(a
l-e
a
1.25
Foyt Ce 1352 eNO = 9.125(0.1y)25e OY)”. (B)
Transformation Formula:
y=10x8 = x=(1p)'” and & =1.25(.1y)” (1) =.125(.1y)". Then,
-125(1y) 125). ly)” eth)” (B)
LO) = feo} eS
We
sofia) Wee
ee S for 1<x<5.
Let the random variable Y = a
(a) Fy(y) = PHY sy) = Pr[ps y}= [x24] =1-2
lore
(b)
te
see
aes
2 ye:
fora ves:
=p
fO)=F
ve
=|
() EY) = | y-
= ~In(.2)/4 = 4026.
>
8-7
(diem
ely | =
(é)°
Noand no,
| yd
2£[X|=3 and Var
x |=
For X, fx@)=51 <x<4.
x=Iny and ae = i
dy
Fry) = fx (x):
eee
y
“
= Atle
oy melon
(5-1)?
12
The random variable Y lives on Eaak with
Then,
-_
r
ries
-
<a
Te
ay ee.
iiss.
ne
So, fy
(8)
Q) =
5
Loe)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 165
8-8
Fy(y) = Pr(¥<y)
= PXT2<y) = Pr(-Jy << Jy]
= eipeirs
ip) = SG 2 a
Oj One =e for
y>4.
(A)
Alternatively,
Sr(t) = F'(t)=8t%; t>2.
)
8-9
at
The transformation y=¢? for
i) gid ee =
1
.
F(v) = PrV<v)
= Pr(10,000e% <v)
¢>2 means
t
=
ys Then Pi) = fr =8(y"} ao =4y™. (A)
= Pr| R<In——
r
In--4,
10,000 —.04
.08 —.04
_
8-10
ee
(E)
10,000
R= ai customery eal customers per minute.
T minutes
dK
Ferry = Po
10
Sr) = tPri <r
di
:
;
#
fir)=Fn=2>.
oe
8-11
mad
Pri T =
;
‘ab
gerald
=
r
ae==
-
(E)
X is the profit of company I. Y =2X
fr(v)= SOE
[
fsa
:
is the profit of company II.
l
Since Y=2.X, we have ay:
ebcme
Sl
and ios
Then
ivalently in terms ofx,fr(x)=+-f|oad
Lx]. (A)
fei V— hh Gi oe or equivalentlyin terms of x, fy (x)= pa
166 ® SOLUTIONS TO EXERCISES - CHAPTER 8
fs(s)=Ae~** and fr(t)=Ae™.
8-12
f(x) = [fs(s)-fre—s) ds = (Ap): [er 2 A) ds
= A? ee
8-13
x
ee -e ds = A*.x-e,
The joint density function is f(s,t)=¢ on the rectangle [0,2] by [0,1].
t-axis
0
l
2
S-aXiS
There
are 3.cases, O<
x <1, 1<x<2,and
2<x<3.
The event [S+T <x] hasa
different shape, and hence different limits of integration for each of the 3
dashed lines in the diagram correspond to the boundary line s +¢ = x for
Casel:
O<x<1,
Fy(x)=Pr[S+7 <x]= igi 3tds
dt =
Case2: 1<x<2,
Fy(x)=Pr[S+T <x]=(le tds dt = aa
Case 3: 2<x<3,
(integrate over the triangle in the upper right corner to obtain the
complement of F'y (x))
1—Fy(x)
=
La
p2
[se ares dt =
Fy (x) =
Finally, differentiating,
=
who
= —_a
;
i)— |
:
A(x 2)+ Ltx-2)°.
—-
| S) ~~"
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 167
Note:
This can also be solved using the convolution method for each of the 3 regions,
but care must be taken to choose the correct interval of integration for s:
Tx (x)
[, BO): fries) ds
for
O2xs1
[_ fs(9): fr(x-s) ds
for
lex
[_ fs) frlx-s) ds for
2<x<3
Bel
iP=m-2(x-s) ds
4
[Olen
[os 2(x—s) dseior
2
es
<%= 2
za 2d X—S)-AS = 10F = 2 oS
IL: Z
=
=
fore
Ose <1
Le
5
[Olea
ce<
AEG eae
8-14 (a)
(Oj
fx (x)=e** -e™* for x20.
Fl Xe
co
ifie
(oe
\ ar
(See Example 8.1-9.)
3.
Of course, E[X] = LS +7) = 2/8
|+ £7] = 241 =:3.
8-15
Let J, ~ Exp(A =10) be the time to failure ofthe first electric generator.
Let T) ~ Exp(A =10) be the time to failure ofthe first electric generator.
Var[T, + T,] = Var[T,]+Var[T,]+2Cov[T,,T]
= 200. (E)
=0)
Alternatively,
Let_Y be the time until failure on the first generator and Ythe time until failure of the
Ve eter
ye then
§ =-1(2,10), $0 Var|S]=28* = 200,
(BE)
168 ® SOLUTIONS TO EXERCISES - CHAPTER 8
8-16
If S denotes the sum of the roll of two fair dice.
io 51362] =e15]sb TOR
The random numbers .28918, .69578, .88231, .33276, and .70997 correspond to sums,
S, of 6, 8, 10, 6, and 8 respectively.
8-17
Suppose X represents the outcome of the roll of a single loaded die.
Pr(S) | Cumulative Probability
1/441
4/44]
1/441 =.00227
5/441=.01134
10/441
15/44 =.03401
35 / 441 =.07937
70/441 =.15873
126/441 =.28571
35/441
56/441
70/441
76/441
196/441 =.44444
272/441 =.61678
36/441
441/441 =1.0000
73/441
345/441 =.78231
60/441 | 405/441=.91837
The “random” numbers .28918, .69578, .88231, .33276, and .70997 correspond to
simulated outcomes 8, 10, 10, 8, and 10.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
8-18
X ~ Uniform[—5,25].
Prix)
# 169
We want to simulate outcomes x such that
te () aeee
25-(-5) ~
30
= de
solving for Xyex =.30u —5)
The random numbers .79936, .56865, .05859, .90106, and .31595 correspond to
simulated outcomes 18.9808, 12.0595, -3.2423, 22.0318, and 4.4785.
8-19
B~ Binomial(n=10, p =.7).
LB | Pr(B)__| Cumulative Probability
ON]15:9049E-06)
NG 5.9049E-06 5 |
|6 |0.20012095 |__—0.35038928
|8 |0.23347444
| 0.85069165
|9 [012106082
| _—0.97175248_ =
The random numbers .79936, .56865, .05859, .90106, and .31595 correspond to
simulated outcomes 8, 7, 5, 9, and 6.
8-21
By the inverse transformation method, w=
(6)
3.756
.690
ZZ
998
BM:
(b)
{—In(l
care
6.248
2.083
14.39
2.093
The mean ofthe simulated outcomes Is fy = 5.2034 and the sample standard
deviation is s, =3.939.
170
SOLUTIONS TO EXERCISES - CHAPTER 8
(c)
The mean and variance of the Weibull distribution are given in Subsection 6.7.3:
With parameters 6 =2 and a =.03,
ELX] = par(is aL)
= (.03)-"2 (3/2) = (.03)-/2(1/2)F(0/2) = (.03)-!2(1/2)Jz = 5.12
Var[X] = pl r(u+2}-r(: A
|
- «asy'|ra-r(3) |= 70 ear PAG,
8-22
(a) c=12.
This is a Beta distribution with parameters
~@=2and £=3.
(b) F(x) =6x? —8x3 +3x4.
(c)
Erie
(d)
(ec)
Fora Beta distribution with p parameters
Lx Se
=.40 and oy =
@=2 and fB=3,
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
8-23
@
171
Here we had two different assumptions about Bert’s age of death. Both models
(uniform and exponential) imply that Bert would live an average of 20 more years. We
do not know when Bert will actually die, so we simulated his age at death.
Note: Under the uniform assumption the future lifetime is uniform on [0,40].
Simulated years
Simulated years
Given
“random”|
numbers
until death —
Present value
until death —
Present value
uniform
assumption
of $100,000
if die at age
exponential
assumption
of $100,000 if
die at age
16.8167]
36458.25
ae
ee!
x=-20In(I-u) | 100,000e~%*
100,000e~°*
x=40-u
1.207531
46847.15
12.638
boa
et
Pees )
Sa] 9 95.2,946:72 a
(d)
Bice
93011.05
Under the uniform assumption,
_
oer
ORS
PV (ea= 100,000 |.
Clay
Pataa 00
e7 06x
140
531.60).
Under the exponential assumption,
e Le = W0e dx ly. _ 5000 [-e! dx a
PV &= 100,000 |eGo
e
(soe
R245
(a)
Lets
= mint
5000 x.
,y-):
F(t) = Pr(min(X,Y)
<2) =! -
es
(b)
P
PriminCx,Y)
<3} = FG)
10-3
10
(c)
&{[min(X,Y)]
=
= 1-
10, 10-f
10
iF t- 50 nies mae
:
oe
(d) E[min(X,Y)*] = j, t
50
ae
at =
o0
ra
K
So
Omin( X,Y) — 2.397.
172
SOLUTIONS TO EXERCISES - CHAPTER 8
8-25
(a)
SuncesY=2X,
YS
Xvand max(xX,Y)=yY.
Primax(X,Y)>7)=Pr(Y >7)= ee= 65%,
(b)
min(X,20—Y)=min(XY,20—2X).
This number is always < = =6.6.
Therefore, Pr(min(X,20-—Y)>7)=0.
26543) we Get 7 minx.)
3Z,).
We 30en:
FO =~ 000
E[min(X.Y,Z)) = foe
3
at
10
|10-18 (10-14
~ 1000
3
12
es
ayy
4
ei
using integration by parts
10, i
300-2)
YZ) |
1 == [
(b)byeewminx,
E[min(x,¥,Z?
a
__3 |_ 20-93
1000
_
200-9
3
=10
2010-9) |
iz
60
|
t=0
using integration by parts
10m
Then,
(c)
Omin(x.Y.Z)—
Vat) = 2.
1.93649.
_
-\n-l
Let T=min| X,...,Xio
|. Then f=
E|T]
=
1 © n(lOmt)"=
i, area
O<r<l0.
*:
10
eo hal
107
tO =9)*
eee
Be
n
n(n+1)
10”
using integration by parts
a
10
n(n+l)
nl
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
10
10-1)"
;
pein |ef Urs
£08
n
10
2t(10-1)""! e 200—7
n(n+l)
n(n+l)(n+2)
‘=0
using integration by parts
ke
10”)
a
eeAlte
n(ntl)(n+2)
— (n+1)(n+2)
Then,
2-10 2
holt D é
(n+1)(n+2)
(n+l)?
VariT] =
-
2 -]-
(n+l)|n+2
107
n+l]
(n4l)2(n+2)
n=)
Next, let S=max|
10
EIS]
X,...,Xi0 |. Then fs (t)= ae Os
ns?!
A
es
tO"
10n
=
-——dt =
gers!
care : eye = men
ee
0
10”
10”. (n+2)
n+2
n—|
n+2
2
(These integrals don’t require integration by parts!) Then,
Var{S]
ae
=
107n
~=107n?
_
n+2
(n+1)?
3
= 10°n
]
n
=
=
|=
n+2
(n+l)
107n
5
(n+1)*(n+2)
Note that S and T have the same variance, which might have been anticipated
from the symmetry between the two random variables with an underlying
uniform distribution.
9)
8-27
(a) E[max(X,Y,Z)]
Ke
(b) E| max(X1,X2
=
Sate
ae .—Q
175 dt
|=
X Xa) )|
ai
—e|
3
Zs
m See = iy
=_=--
5,
(c) Pr| min(X,Y,Z)<3] = |—Pr[min(X,Y,Z)
>3]
1=—Pr[X 23(\¥ 23(|223]
= ]—PrLX >3]-Pr[Y¥ >3]-Pr[Z >3] = 1-(2] Sa
# 173
174 ® SOLUTIONS TO EXERCISES - CHAPTER 8
8-28 Let X ~ N(70,37) and let “success” be the case that XY > 78. The probability of
success is given by p = Pr[X >78] = plz>
78-70
=2.61 = 1—.9962.
Let
n be
the number of trials. Then,
60" = Primax( 44,495", 4,) > 76)
Pr[at least one success] = 1—(l—p)”
THT
O20
In 0.4
Soeee
0.006)
= 1—(.9962)”.
= 240.67, son=241.
Ge ct(10-—t)dt == c |
Gl [0r-1 2 |adte== c 5x
pate
Ey (x) Se ik
=
i
a
1000:
Then, 1= Fy (10)
=c| 500- “36
Psy é =10:
>
and
so c =.006 and
Fy (x) = 0.03x? —0.002x?.
Fy(y)= 453 0<y<10, and F7(z) = 1-e)?; z>0.
(a) Pr[max(X,Y,Z)>6] = 1-Pr[max(X,Y,Z)<6|
= 1-Pr{LX¥
<6]o[¥ <6] o[Z <6}!
= 1—PrLX <6]-Pr[Y <6]-Pr[Z <6]
= 1—|(.03\(62) -(.002)(6°) |Sha -e6/5}
—[.648]-[.6]-[.6988] = 1-.2717 = .7283
(b)
Pr[min(XY,Y,Z) <3] I| 1—Prl| min(Cee
ut
ese
& Z)>3]|
data
ies 3]
1—Pr[X 23]-PrlY 251.2 1[Z >3]
=
ae
a)
1-|1-((,03)(3%) -(.002)¢ ERA en]
BS)
= 1-784}
=
10
aa
i
5-3/5
|5488] -= 1.3012 = .6988.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
8-30
# 175
Let Wbe the maximum of the three future lifetimes. Then,
Fir(w)=2 3 and EW] = [30 es 4 Jem2 0a 3 = +30 is.
8-31
Let Wbe the maximum of the three exponential random variables. Then,
Fy (w) = [1-e-#3]-[1-e-4 ].[1-e-5 ]
II 1-ew3
Jw (w) = Fy(w)=
peut
—e
ws Sr ee wil2 = e 8w/l5 ai e 2 w/20 — e 47/60
sews tet
ae
fi
12°
—Tw/12i
ews
8
i
—8w/15
y,
eee eee
20 ~
—9w/20
47
—
60
—47 w/60
Note that each term in this sum 1s the density function for an exponential random
Xx:
variable. If XYis exponential with parameter # then f(x) = oe , and
E| Xx, = 26-8 Now.
es Variexd
|= p and EX
Ei be Vor
1a
Le
oe. eed i.Wwhale
id wie ee iedw+e-+
i.why (w)Re dw i = j,
a0 ° an
= 34445
DB
_ 9
8.
VAT
dw
= 7.4651.
Similarly,
2
E|W|:=
47 —47 ti w/ Wied dy ,
ee
aie!
we2 ——e
we ae —w/33 dwt--+ I,
\eew+ fy (w) dw = j,
Sa
mei
rs | a
20-60
Pieilig ate? (2) (= ye
= (2)(40.2370) = 80.474.
Thus, Var[W] = E[W?]-E[WP? = 24.7465,
Note: The CDF for the minimum VofX, Y, and Z is given by
Fy(v) = 1-[1-Fr(v)
[I-A 1-20)
1-[e-” ak [e-” cit Le calee
eet
Consequently, the minimum V is an exponential random variable, with par
This would make the calculation of the mean and variance of V much easier.
60
Ay”
176
SOLUTIONS TO EXERCISES - CHAPTER 8
8-32
Let
V=min[ X,Y]. Then the CDF is
Fy(v) = 1-[1-FyQ)]1Fy ()] = 1-[e?
]-[e-r"’7] = 1-e 25; v0.
Thus V is exponential with £ = =.
(ayn
(vy = Te 2) and fu asses v>0
(by)
Pry s3i
(c)
py “By
(d)
Lear
oy =e
1— et)
= 164248)
28)
Let W = max [x ; ia . Then the CDF and density are,
(6)
se
ey (ie) = [l= a2] [Fe PT] =-1 ee
os Peony
Vell aeBese
35°
e 12were
belw/5 Ga+ze
oe
=
tw lw)
and
5 ee
= i573
(fy GC) Meee Ole)
(g) EW] = J,whi(ov)dw
Se w—e 5
=
0
Bia
th)
5
dw+ |aN w—e™ry
0
7
dw
owt
|oe
w—eo-’-?
0
35
dy
45 ern O84
72
oe
FEW?) = Ihew fw (w) dw
=
OO
we] Mee cure
ay ee
} Wore e wi? dw+ I wze
= aysPe
eo,
04 Reet bs
wil dy — j, wr
a wI/S
el2w/35 dw
-(3) |
= (2)(65.4931)= 130.986.
Var| W|= E\W? |= E[w] = 48.4792 and ow =J/48.4792 = 6.9627.
E[max(T,2)] = 2-(Pr(T-<2))+ [°r-—-e
No
= 2-(l=e-*\+2-¢
PROBABILITY AND STATISTICS WITH APPLICATIONS:
8-34
@ 177
Pr(max(storm, fire,theft) >3) = 1—Pr(max(storm, fire, theft) <3)
II
1—Pr(storm < 3) - Pr( fire < 3) - Pr(theft <3)
fe"
8-35
A PROBLEM SOLVING TEXT
!")
—e
"5.
(1)—e 3/24) =
414.
(E)
The pay-off time for the first product is PR, = X.
The pay-off time for product 2 is P; = max(X,Y).
ETR]=ELX]=20 and E[P)) = +40 _ Now,
pee Xe
Oey
7
ote
aX
= Y,
Thus,
P, =X
on the region 0< y< 40, y< x< 40
and
P, =Y on the region 0< y<40,0<x<y.
Thus,
8-36
40 -40
40
EUR -Py|= i. il Ta * drdy + | I aaan wtedy = a+
= 600.
Cov(B,P:) = E(B -P)]—-ELB]- EL] = 600—-(20)- (2.40]= 66.6.
(C)
cet)
=
=[F(y)}4
Pr(max {%, ¥>,¥3, Ys} < y)
Uae (y)
=
Eas (y)
a
nl. [itsinay
m-cosmy-(1+sin wry)
4
E{max|= an [ieycoszy(1+sin zy)> dy (E)
Let_X greater than 50 be a success.
Letp be the probability of asuccess and let S'be the
number of successes in 10 trials.
Then,
p=
Prix > 20)= 1 Fy (50)= Ce
eh] BO
The second smallest claim is greater than 50 if and only ifthere are at least 9 successes.
The probability that the 2" smallest claim is greater than
50 = Pr[S =9,10] = p'°+10p2q = 3152.
(B)
178 ® SOLUTIONS TO EXERCISES - CHAPTER 8
8-38 Let Y less than 1/3 be a success. Then p=1/3
is the probability of success.
the number of successes in 3 trials. Then Y) <1/3
if and only if S =2
Let S be
or 3.
Rentoe
Pr[S =2,3] = p?+3pq = (+)
+35) ee:
(2) = 2593. (D)
8-39
700-300 = 0154. Lens ve
Let X be aclaim amount and let p = PrLX¥
> 300] =
700-50
the number of claims that exceed 300 in three trials. Then
Pr[S <2] = 1—Pr[S =3] = 1- p? = .7670. (E)
8-40
Let Xbe uniform on [0,10]. Then the expected value of the 4" order statistic out of a
sample of size 5 is given by wd
= 6.67.
(B)
Alternatively, the density function for Y4 is,
S05)
i/peut
ead
js eevivieeryahb:
f 1 at arte ~ 105 [do-y)],
and
ElYs] = 20“~ [“poy*-y5]ay
= 2 (B)
0
10°
8-41
Fora single observation Y, Fy(y)=l-—e7” and p=Pr[Y >1]=e7!
of success.
Pr[Y5 >1]=Pr[at least one success] = 1—qg? = 1—(1—e7'!)> = .8991. (E)
2 se fetta eT
8.2 The Moment-Gencrating Function Method
n
=AD
=
n
y
i=l
|
Qa;
=
|
Q, +Q2+---+Qy
_
ot
M(t) = [[“« (t) = I Je
2
och
;
|)
j=
1
8-43
.
= ev
:
This is the MGF of a normally distributed random variable with
mean
is the probability
n
n
f=
i=l
"1; and variance }'o? .
Y"
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
# 179
8.3. Covariance Formulas
8-44 CovfU,V] = Cov[2X-Y,-X43Y]
= E|(2X-Y)\-X+3Y)]- £[2X-Y]- E[-X+3Y]
= E[-2X?+7XY-3Y?]-(2E[X]- ElY])-(-E[X]+32[¥))
= —2E[X?]+ 7ELXY]—3E[¥2]+ 2(ELX]) —7ELX]BLY] +3(E(Y])
= -2(ELX?]-(ELX])?)
+7(ELXY]-£LX]- ELY]) -3(21Y?]-(E1Y))?)
—2Var|X |+ 7Cov_X, Y]—3Var[Y]
—2(2)-3(3)+7(-1)
8-45 Cov[2X -4Y,X+7Y]
= —20.
2Var{ X]—28Var[Y]+10Cov(X,Y)
= 2(5)—28(2)+10(-1) = —56.
8.4 The Conditioning Formulas
8-46
Using the conditioning formulas directly yields,
E(B] = Eg {E[B|G]} = .4(50) +.3(70)
+.2(65) +.1(24) = 56.4.
Var[B] = Eg \Var[B|
G]} +Varg {EB
|G]}
are ae }:2(652) +.1(0)
+|.4(50-56.4)? +.3(70-56.4)? +.2(65-56.4)? +.1(24-56.4)?|
Here is the complete calculation in tabular form, modeled on Examples 8.4-1 and
8.4-2, where_X is the sporting event attended and Y is the quantity of beer consumed:
Pr{X] | E[X|¥] | Varly|X1 | ELEY X]] | £[veri¥ |X1] | Eley xP]
6.4
20
Cubs
aa
or
=
—
1470
845
845
E(Y]= a X]]=56.4
E[Y|X] = ELey | XP |- ely) = 3372.6—(56.4) = 191.64,
Var
E|Var[Y |X]]}=1101.4, and
Var[Y] = E[ Var[¥| X]]+ Var| E[Y | X]] = 191.64+1101.4 = 1293.04
=
Oy =55.96
1000
250
180 ® SOLUTIONS TO EXERCISES - CHAPTER 8
8-47 (a) ELX] =Ep{E[x|6]} = Ey 2 2-5 mifhion:
(b) Var[X] = Eg {Var X |6]} + Varg {ELX |0}
Fo
62
0
D |Yar {8)
I
3h
h+ qo
To Bet?
I
Seer
2252 nis: Soi Fa (million)° =>
=
Prev
> he) =
poe
18}
Me
oy = 3.323 million
0
fig Ot
ta
Pees
6
Pree ee joe
d0+ ,
x -e-5d@ = 5742.
5
8-48 (a) E[S] = Ey {£[S|N}
= E[X,+X2+---+Xy]
EL X,]+ E[X2]+---+
E[Xy] = N- py = un: Mx.
(b) Var[S] = Ey Var{S
|N}}+ Vary {E[S| N]}
= Ey{N -Var[X]} + Vary {N-ELX]}
Var[X]-ELN]+(ELX]) -E(N] = E[X?]-E[N].
Note:
8-49
See also property (2) for random sums.
If we let T denote the total losses, then
Sern
l
12,500
Pp
E(T |N]= ux N =12,500N
15,0007 /12
and Var[T|N]=o2N=
15 ,0007
12
E(T]= E[E(T |N)]= EluyN]= wx ELN]=12,500p
Var[T] = En \Var{T |N}} +Vary {E[T |N]}
2
“
é)
7
=Var{T
|=Ey \o} N}+Vary {uN} =03 pt ui pq =
Ss= +12,500? - pg
lis
)2
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
@ 181
8-50 In the case of partial damage the expected payment in thousands is given by,
by parts with w=x-1
and dv=e-*/?dx
reoOF
J, @-1)-[.5003-e~*?dr] = .5003 f“ete ak
15
5003[-2e"(x+1)]
II
15
= 1.2049
]
Exe
ReymeeelDI
1.2049
pele
E| Payment] = E [£(Payment |Damage) |
= .94-0+.04-1.2049+.02-14
8-51
= .3282 thousand
This is an exercise in detail. As an example, we calculate
(B)
Pr(S =4). Think about all
of the ways that the sum could equal four:
Pr(S =4) = Pr(X¥ =0,Y =0,Z =4) + Pr(0,3,1)
+Pr(1,1,2)
—.
FH
denoted
Pr(0,0,4)
+ Pr(1,3,0) + Pr(2,0,2) + Pr(2,1,1)
= (.4)(.25)(.55) + C4)C4)GC15) + (3)¢35)(.2)4+ (3). 491)
+ (.3)(.25)(.2)
+ .3)(.35)C15) = .14275
S=X+4+Y+Z
0.0365
0.07025
0.0965 |
0.14275
0.20125
0. 141 |
N]rRe
|]
BI]
NH}
DA]
CO}N]
OO}
EUS] = 0-.01+---+9-.066 = 5.2
0.16975
0.066
0.066
182
SOLUTIONS TO EXERCISES - CHAPTER 8
8-52
EY] =
y-2-(l-y) dy = 7 Let N be the number of claims, a binomial random
{,
variable with n=32 and p=1/6.
= Ven
Let X,...,Yy be the policies with claims and let
Ae
[S] = E[ES|N]| =wy -sy
8-53
Let S=¥,+¥%)+---+Yy
8-54
ae 68
16
denote the random sum.
RN
oD= 32, p=).
ee
N ~ Binomial(n
Val S\=00
Leal
mite
sy =+ and o; ==.
_16 and oy
Fa
NIELS 37.1
eS We Prarie
37
a
=32 2h
a:
ale
Opal
On iy + Lino far = LORE
E eae
The random variable T|.X ~ Unif[X,2X].
Soper
fe a)
ee
0x
ef
0
otherwise
2x-3 yp
Pr(T
(i) = — :
——_
5 dxeS
id
=
64
x>15
ere
Sal.
Let N be the number of people hospitalized.
The unconditional probabilities for N are,
Pr[L <1|
(A
(A)
Then N is binomial with »=2 and
N=0] = @ and Pr[L <1|N =1] =.
p=.3.
Given N =2, let.X be the driver’s loss
and let Y be the passenger’s loss. The joint distribution of XYand Y is uniform on the
square [0,1]x[0,1] andso PriL=4+Y<1|/N=2] = 0.5..Now,
Pr[L <1] = E[Pr(L<1|N)| = (.49)(1)
+(.42)(1) +(.09)(.5) = .49+.42+.045 = .955.
We use Bayes’ formula to calculate the conditional probabilities for NV given that the
loss L 1s less than 1.
Pr{N =0|L <1j=-%,
955
Then,
2 2V |ie<1|
1 <1] = Oh.
By =l|L<1]= err yand Pri Vi=2
20: $e
(B)
=
+1--92.+2--983 534
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
@ 183
Section 8.5 Chapter 8 Sample Examination
I:
Ha
Vinee
x. for Osa 61,
0
otherwise
Alternatively, using the transformation formula:
Se
V=e
2x
=
]
dx
1
X=—
=
x 5 In ny and —ose
SO,
trf (y)p=
2.
1
3
ai
=i y|
)
, ay
c
3(In y)?
=
5
<<
<p)
San : Ieyece
=
:
(a) F(z) =Pr(Z <z)=1-Pr(X =z NY =z) =1-(e-25) (e773) = 1-825,
(b) Z~ Exp(B =>).
jes
(c) pz <r
note
using independence
3
242
(2 =e (Gr
2-e 142507
(a) Mot) = Myzy() = Mx(t)-My() se
a
sea
a
(b) The MGF for S is that of anormally distributed random variable with mean
variance 34.
(CyAPiCs —4 ett Z =< 7 Veer 214) =.097:
4.
Let Ny represent the number ofclaims from class 4,
N, ~ Binomial(n = 500,p=.01) so that E[N4]=5 and Var[N4]=500-.01-.99.
Similarly, Ng represent the number of claims from class B,
Nz ~ Binomial(n = 300,p =.05) so that E[Ng]=15 and Var[Ng]=300-.05-.95.
| and
184
SOLUTIONS TO EXERCISES - CHAPTER 8
The claim amounts, X 4 and Xz are constants, so wy, =500 and wy, =300,
with
ce =O7, =0. Thus, if S4 represents the total claims from class A and Sz
represents the total claims from class B, then,
E[S.s] = Mw, «Hx, = (5)(200) and E[Sg]= pv, x, =(15)(100), with,
of, =o} Hy =(500-.01-.99)(200)? and 6}, =02, - u2, =(300-.05-.95)(100)’.
Let
T=S4+Sg
represent total claims.
E(T] = (5)(200)
+(15)(100) = 2500.
Var[T] = (500-.01-.99)(200)? +(300-.05-.95)(100)? = 340,500 = of = 583.52.
The 95" percentile is 2500 +1.645-583.52
= 3460.
_
k 3460
=< =1.38.
(EB)
5.
Let f; denote the number of pensions provided by the city for female recruit i.
The probability distribution for f; 1s
0
l
6
(.4)(.25)
=.1
E[ fi]=.7 and Varf f;]=.81.
The total number of pensions is
T=
E[T]=70
fitfre+fot+-++fiooe
and Var[T]=81.
Using a normal approximation of the actual distribution of 7 (with correction for
continuity), we find
Prt 390) = Pr{Z
. 90.5-—70
N
———
os
—
= Pr(Z $2.28)
= 99.
(B)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
(aye
Pt] Max(XY)
@ 185
<-6)'—=" Prix <6 <6]
II
i]s
Prey
)26]
(ietemye(3)
1354
(b)
Let Mbe the maximum of X and Y.
Fy (m)-Fy(m)=(l-em |, 0<m<2
Fy (m) =
Fy (m)-Fy(m)=(I-e")();,_ 2<m<oo
E[M] = lp[1- Fay(m) |]
dm
. an (2-m+me") dm+ |e” dm
: [4-24
d+m|
er
= 5[2+(1-3e%)]+e? = $G-e%) = 143
Var[N] = E[Var(N |A)]+VarlE(N |A)]
fee,
ee
= BA|
=),
=
Vari A| = 15>
(3-0)?
= D5
rv)
5000
Fy(x)
=
‘vy(x) = 1-| ——
We
Pr[.X > 2s25,000]
and d so Peixe=
000}
Pr[Y; > 25,000] = i Pe
(BE)
12
5000 |) —
)
ea
= (.2)!2,
Ss4r (Ey
Let XYbe a single claim severity in units of one thousand.
Fy(x) = 1-e~* forx>0.
Py Gj
E[W)
=
Let Wbe the larger of two claims.
a l= ge te
[, U-Fr()
oboe
Then
[2a
4
Then
iorx> 0.
ee"
| Hie
Dn
+ =1.5 thousand =1,500. (C)
2
186 ® SOLUTIONS TO EXERCISES - CHAPTER 8
10.
Let X be the smaller of two observations and let Y be the larger.
Then the range is Y —X.
The joint density for X and Yis given by,
fii
via cere
tor0aycoe
Usrays
and the expected value of the range is given by,
2[, [deter drdy = 2], Lyffexae— fjxetarle dy
2D:i |y(l-e” )+te* es
|
e dy
= 2f [pre-e ay
l
2=
21+ 5-1)
(D)
CHAPTER 9:
STATISTICAL DISTRIBUTIONS AND ESTIMATION
Section 9.1 The Sample Mean as an Estimator
9-1
(a)
Weare to find v—Z05° te. Ly <W +2Z05 |
Vn
Dre GAS.
Vn
2350-1.645- 2 < pay <2350-+1.64522 |=(2144.4
<py < 2555.6).
J16
(b)
2==—
ia
jz=
J16
S
Jn
ro?
Ao mek h sarees
J16
9-2.
Let
H ~ N( uy =70,0H =5) denote the size of arandomly selected hamburger.
a=
stim
att
denotes the sample mean.
4
Pr(ff <77) = PZ < a Ber 8) = 9974,
Sih) 4
9-3
The confidence interval for the population mean is
[ghee
poss
Ke
(a)
The sample mean must be in the center of the confidence interval, so X =11.00.
(b)
Theerroris ¢=2.50=Zo4
ero
rar
MESas IRY . The sample size was n=49.
oar
187
188 @ SOLUTIONS TO EXERCISES - CHAPTER 9
9-4
yw=100 and o=16.
(a)
With asample of size n =100,
ro7 = 100" Te
| 167/100
Pr(97 < X <103)=
103-100
16/./100
= Pr(-1.875
<Z <1.875) = 93.9%
(b)
With a sample of size n = 200,
:
Pr(97 < X <103)
=P
(c)
97-100
16/./200
r(—-2. 65
2
<Ze
103-100
16 //200
2.03) = 99275
The likelihood that the sample mean is within 3 of the population mean increases
as the sample size increases.
cae
leit
So
Section 9.2 Estimating the Population Variance
9-6
X=
L434 9,.67 15,04 1254106
;
:
5
cd
_
= 12.36 is the unbiased estimate for su.
ie):
*
Zz
S? = (14.3 12.36)" +--+ (10.6 =12.36)"
5-1
= 5.363
is the unbiased estimate for the
population variance.
9-7
CERO
Xe
(n-1)
;
v
Gas
:
Via
4)
3239 se or eranly Leas
So the 95% confidence interval for the population standard deviation is
HERE
vey es VIAL
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
9-8
n=10,
so there are 9 degrees of freedom for the chi-square distribution.
We are to find the value @ so that .025 = Pr(S <@).
OZ
5e— 9 PTGS <a)
= POS <a
.§2
= Pr a
OY
an
}= Pr(9-.
SS? <9
-@
WAL:
ae O|
= Pr| 729)< Ais
OY
9:
7)
a Oy o
= 72 »s(9) = 2.7
41.67-
ere
So =, jeTaT _ 92.92,
9-9
Begin by calculat ing the sample mean and sample variance.
X =147.5 and S? =629.1.
a
Hohl)
Lean)
(5)-629.1
11.07
(5) -629.1
ipl
16.86 < ay <52.30.
9-10
.95 = Pr{S2 <c}
y
(19)-c
5
as
Wiel)
“9.
woe
Om
= rr Ain
Bade
=
———
-
O-
=3.17
ie)
Ou
Zs
@ 189
oy =41.67.
190
SOLUTIONS TO EXERCISES - CHAPTER 9
Section 9-10 The Student ¢-Distribution
=
9-11 Frente
a
| Sy$C]eee Ftgate-|
Sy
$e]
Vn
Vn
92-2650
17.9
12 Bee -.2+2650/
=A. 22a
y=
12
Ji4
43.2 +127
—55.9 < uy < =—30:5.
9-12 (a)
OQ
298+ 2.93 $027 +2,50
X = alee
eee
2.619.
si
Se +---oeee + (2.50—2.619)
=
2
(2.58-2.619)*
Gee
— 092353.
(11-1)
(ey
eX =F (n-1)- DIT
S
Wie <= X #1, (nl: oe
Vn
2
vn
2619-2, 298:|022293.) 2 i < 2 6192,098 4oe
|
28!
vi,
2.619—.062
2.0)
9-13
(a)
< tty
aye
S
Ji
< 2.619+.062.
Obs
1
fy(y)
= ———
yp te
Bo -I(a@)
/
)
ana .r(2)
u=,[2-. Then, y=nu? and ole
n
]
WAY = —_____.
du
>
yl"2)-1g-)
Nw
9 V5 0, Let
ae SO,
dy
VO) ee OY)du
z
=
=
=
I
_ ylriD)-1p-(W2)¥
aera a
I
renee
9
NE 2 )\(n/2)—1 e ,-(1/2)(nu*) -2nu
oe
ste
ie wee
an? F(a)
apy
n
y”
2 ‘yew
2)(nu? )
=
2)
: Sax
2
(4)
yl
l
enn
2
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
(Hea
ae
ae? for so Aes
AG
Then, f7(t) = iL.fu(u): fz(tu)-u-du
of
=
Oy
n
n/2
4
3
tel
ete
ae
2
IP
n/2
»)
= (+) :
gt
]
:
5)
v=ure
-
>
U=vV
_ nu
2
fda
20
4,
and
2
yy)
2 -@
at
2
aa yn -eM2Y(nte"\u* hy
n/2
fr
r i) zh
5)
1
je
1
2B
n/2
gi
=
)
lI
N|s
a
Ee,
eS
9-14
Sx :
S?
Foatmeln—-l)
ls~~
(
1
|
>; —o<f{<0o
y
81 a eee ne oya _ —8i - F'95 (12,7)
ee tA)
ce
era
a
Ep
foae2.
ere
OF
et 357
Ips
a)
erin
i)y
Ox od Bas -Fyj2(n-1, m-l)
oy
2/n
© ee
(a)
).
=
eon
eg
yey
IF
Ben =b set Ly |
2
ee
. [ave ue eoW/2)y(nte)v . hy
0
r( )Noe (p
r( orl
=
fell
2
1
+
—
2
Z
y4
:
n
© nl, p-(l/2)(n4t?)
[2
n—|
2
—
-y-du
a ) the
2
eae)
-u-du
B(1/2
l
- [)
ie
2
du= svn ) dv
n/2
(c)
_tu
(Qo
[ Uy'-@
V2
py)
1
ie
1.86
< CX < 19.28.
OF
5
ee
@ 191
192 ® SOLUTIONS TO EXERCISES - CHAPTER 9
9-15
5.42Bl)reel
EG
meen1
5,42
ee of
ROai ia
AGH Taye
Ae
TEEN
ce
or
2
2
eb 2 lS) ee
-3.80
5,07 4 05m
ioe
5,97
Taking square roots,
9-16
Ue y
SP hs < 1.7846.
Oy
For your bowling scores denoted as Y, n=6,
X =147.5 and Sy =25.082.
Your sister’s bowling scores denoted as Y,
n=8, Y =194.375, and Sy =14.272.
25.0822
|
oe
ee
ee
(a7
Fe)
25.0822
1
(OT
<<
25.0822
of.
Ones
<A
Ge
i428
082"
44272"
5
4.88
BR) = 10m 3 889.
Sister
9-17 (a)
U~y2(m) -7($.2) Soleene ae 2), using Property (5) of the
2:
.
-
m
2 m
Gamma distribution in Subsection 6.5.2. Similarly,
1SCy eapoy
Y
emma Teer)
ky
ed Ds
n
ZR
5)
fs(s) =
Foror
he
SS URIS sean
2
or (tee
S
W =—,
+
(4)
Dae
m/2
n/2
(4)
2
\ (ip(n/2) =I oe = ti2 9 (wt)
2).e = mw
£)'=]
22 4.dy
aC
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
a
m/2
-
2
ete
5
2
a
;
ae
Aes
ws
|n
3
"2
iia we ~9(n+m)/2
2
n+m
5
(4)
Awe
m/2
B=
Bo
PB
~nt+mw_ 1+(m/n)w
——
(2)""
ay
qantm and
n/2
_(nt+mw)t
eS
eee
See
dt
(m+n)/2
|
(2
(n+m)/2
,
on
n
for 0<w<o
Section 9.5 Estimating Proportions
9-18
(a)
~
P=
(0) =
2070
aangh
7,
69-31 REG)
ih, (EES
205-3999
(c) .676 = .69—.01389<
p<.69+.01389
= .704.
“
Gk pee
eer!
.65 —1.96
a
Aas
Pt Zai24{ pine
Seep. 65 +1.96,|°>.
57.5% < p< oe
@ 193
194
SOLUTIONS TO EXERCISES - CHAPTER 9
;
AD (1—f = Zain} pee
aTRAS = Za/2-0156, which
9-20 p=.57
and 03 =€=Z 92°, /eO—P)
implies that Zy/2 = aren =1.92.
Therefore the level of confidence is 94.52%.
Because of rounding, the newspaper poll likely had a 95% level of confidence.
9-21
We have p =X, where the population X is a Bernoulli trial with probability of
successp. From basic principles E[ p]= ELX]= p and Vari pj=— Ee) = PA
Then
E(pQ.— p)) = Elp)-ELp*]
= p-(Varlp]+ ELpY)
= p- [At pei~ lh
Ee
= p=p)-=*n
= pq|
=
fo
At
q
S =| fl
8|
S
=
l
(1 1g
3
As
o
x1]
FH
See
aaa
Section 9-6 Estimating the Difference Between Means
9-22 (a) (X-Y)-Za/2
en Civ=ity) = (A =F)ae Sadnd a8
a
m
m
n
(14.3-13.8) +1.96-
n
2 (igsue)
fi.g2 1.62
< (14.3-13.8)+1.96- Jaret os
(AC) 59 te, (=
ee
ST end
—.09 seconds
(b)
< (4#y—Hy)
< 1.09 seconds
Since 0 is within the confidence interval we cannot conclude that the estimated
difference in population means is statistically significant at the 95% confidence
level.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
9-23
$2 =
i
(m-1)S? + (n-l)S3
(m+n-2)
eo eI
a
26
ke
= 227,052— 15,0887)
For a 98% confidence interval with (m+n—2)=26
degrees of freedom,
£o1(26) =2.479.
(X-¥) -taya(m+n-2)-Sp,f+
+2 < (uy-uy)
m
n
i
l
< (X-Y)+tg/2(m+n-2)-Sp-,/—+—
on
m
(104 —97) — 2.479 -15.088 ota 2 Une)
oe
9-24
14
(Ti
—1s) ee DS.
Begin by computing the sample statistics.
Aig =10
=
ee
The pooled sample variance is Sp =
i
-,6377
+(9)-.9757
eee =
(3.287—2.468) —1.721-.80 a+
2 inane
9 05950=",00
EOD)
240 < (gp —Lpy ) < 1.398
Section 9.7 Estimating the Sample Size
= [25-233 = 212.07, so sample size is at least 213.
:
@ 195
196 ® SOLUTIONS TO EXERCISES - CHAPTER 9
926 Fe [2]
2
2
zs 2s
= 917.56
be SHOTS.
ce
7)
n= [eee
9.8
= 467.86, so sample at least 468 new professionals.
Chapter 9 Sample Examination
Ve
aca
ee
ee
3
Sy =1.904.
(b)
Since we are using a small sample and the sample standard deviation, we need to
use the Student-t distribution.
Fre)
D
S|<i
= X+taAUG 1)- S
Al
ey
3
204)
Js Shean ade: [12
64 < wy <5.36 hours.
Se,
We are given 112 < wy <120 1s a 95% confidence interval and oy =16.07.
We can use
“=
(4) 21 oe
6
pines
to,
ee
and the margin of erroris ¢ =4.
(hyn {|Ox *Za/2 |-|iE)
z
+
he:
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
(Xna-w? = a
ix)? = SE
we
-
=
2
2
= <7
(nc,
Titus (cet YX)
ky 0B)
i=
(a) X = oo
aie
64.5)? +.
645)?
Cae [(72CER EON
(b) Pte
5]Sine
2
= 5.345,
Field
n
Se
D
58.83 = 64.5-—2.998.
toi(7)
n
ie |Nee c4s.2908/| S388= 07
B
v8
error
(c) The confidence interval for the population standard deviation, oy, is
(n-1)-S?
(n-1)- Se
Le (n-l
Hain? =i)
75.345
mae
aa
Women
<
Oye
X57)
S
5
X95(7)
so that we are 90% confident that 3.77<oay <9.6.
(a) X =21.6 and the margin oferror is € =0.7.
eee
281.75. \,
© n= See) -( 28275)
==) || ee
alt
=\|
—_—_——
=
49.
2
@ 197
198 ® SOLUTIONS TO EXERCISES - CHAPTER 9
yx) -kY = S1(X?2-2.8-X;4¥2)
i
i!
x we <S X; +nX?
=l
i=l
= \\X?l -2X(nX)+nX?
10
nh
LX;
hs
y _ i=l
\e—
ee oe
10
= [lil giel So
710)
2 =
-2
|
n
a
me {Ea
a
=
n-|
=
1,282
—-1-(110)?(110)
282 — 75
9
5.95
:
;
=1).S
ee
= rea= 1575(9) S — < X5259) = 902
=
3.78 =
oe
9S2
eas
<o*
Di
9S-
S s
48
< oO. = 26.67
is a 95% confidence interval for a7.
(2)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
Also,
cae
a
v7(9)
=>
whe Pe[3.3= 350)
Dus
OS
=
<a : Pro? <A =
mee
(9.
OS oe = ao = 21.62 1s a 95% upper confidence interval. (1)
Finally,
es. 90) =
n—|
S?
9
(n-1)S? _
OS
hare
yee Pre
=16.92)
r ete(9)
at = Pnico) <o?a
Thus, 4.26 = ‘oa <a? <@ isa 95% lower confidence interval for a7. (3).
The answer is (1), (2) and (3).
SSeS OE
(E)
ee
J2o
2. yy
eee
Lae <i ; pAy
52
Thus,
It follows that
k=l andr=2
(B)
a
# 199
200 # SOLUTIONS TO EXERCISES - CHAPTER 9
6
y
DX?
Since the means are zero,
DY
=)
~ 77 (6) and
6
Dee
6
a
5104 (oy:
es
Using independence, F(6,9) =-="_——_=1.5-=!_—_ = 1.5W . Then,
95 = PrlW <a] = Prfl.sW <1.5a] = Pr{F(6,9)<1.5a]
(A)
Oars
en
ga OU) ) =
OO
Se
S¥
5S
992
aLX
OF.
A ~ y7(5) and
¥
2
G2
Sy
Sy
—+ ~ y*(9). Thus, F(5,9)= Ox oe
OF
PrBSe Sy|=|
<3=
2k= F(5,9)< 6~.99,
Yo
since F’9;(5,9) = 6.06.
to25(9) =2.262.
Y
(B)
Hence the answer is (A).
to25(8) = 2.306 and the margin of error is f925(8)-
1
i)
S S ox
a
Ws)
The confidence interval is
Let
S=)
4
5+4.61
(D)
Xj ~ 77(8).
i=]
Then, “ws =8 and of =16
>
ws —205
= 8-(2)(4) = 0 (A)
CHAPTER 10: HYPOTHESIS TESTING
Section 10.1 Hypothesis Testing Framework
10-1
(a)
Ay: God does not exist
(b)
A Type I error means rejecting the null hypothesis that God exists even though
God does exist.
(c)
A Type II error means accepting the null hypothesis that God exists even
though God does not exist.
10-2.
“Answer (A)
10-3.
Y ~ Poisson (Ay =(25)(.1) = 2.5) (see Subsection 4.6.4). Let @ be the significance
level [hen
ee es
10-4
Pres
| 25
yee |e = .758 (D)
~
l
The null hypothesis is that 44 =288.9 with the alternative being
su < 288.9
X =286.9 and
p-value
= PrLX <286.9|
4=288.9] = Ve
286.9
—288.9
3
ma
=,,057
Je
There is only an 8% chance of observing a sample mean of 286.9 or less given that
the population mean is 288.9.
Section 10.2 Hypothesis Testing for Population Means
10-5
¢(24) =
42-45
8/5
="—1.875.
From the table for a two-tail test, we have,
¢49/2(24) =1.711 and ¢95/2(24) = 2.064.
Thus, do not reject at the 5% level, but reject at the 10% level.
10-6
#(20) = SS
= 2.92.
Pr[1(20)|>2.92]=.008461
(D)
Using the function TINV in Excel shows that,
(E)
201
202
10-7
SOLUTIONS TO EXERCISES - CHAPTER 10
Since the surveyor wishes to demonstrate a lower average price we use the left onetail Student-r test:
.
Ho:
=A
21
Aye
4921
From the table, ¢19/2(26—1) = 1.708, so we reject the null hypothesis if f is less than
—1.46, so we do not reject the null
=157038. Then, 4(25)= eee
6.42 / (26
hypothesis at the 5% level of significance. The surveyor’s claim is not substantiated
by the data.
10-8
Since the surveyor wishes to demonstrate a lower average price we use the left onetail Student-t test:
Hg
b—49 71
Hy
up <49.21
From the table, ¢o2/2(24—1) = 2.500, so we reject the null hypothesis if ¢ is less than
—2.500.
Then, (23) = AT37—~ 1-21 _ 4 636, so we do reject the null hypothesis at the 1%
3.42/24
level of significance. The surveyor’s claim is substantiated by the data.
10-9
Since we wish to substantiate the claim of higher starting salaries we use the right
one-tail Student-? test:
Hae
P= 952,000
Heil oa.000
From the table, ¢2/2(9—1) = 2.896, so we reject the null hypothesis if¢is greater
than 2.896.
5
Then 7(8)= PRA
—52
:
Uae = 2.250 , so we do not reject the null hypothesis at the
4000//9
1% level of significance. The salary claim is not substantiated by the data.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
# 203
Section 10.3 Hypothesis Testing for Population Variance
10-10
Ser ae
-1)S?
The test statistic is a Seemv7 (n-1).
Take Hyp :0* =50,
with, H,:07 >50,
so
that we reject the null hypothesis with a sufficiently large value of the test statistic.
This leads to a right one-tail test. Calculate ¥ =13 and S? =119.192.
11971995 =11.92. From the table, 77,(5) =11.07 and
Then, ¥7(5)= be
50
Maps) 1 60. S0R025 <p <,05, (D)
Section 10.4 Hypothesis Testing for Proportions
Mei
MeTH tests tans ieey
ee
ace
Dlowb.
ole)
n
sy.
100
With a@ =.05 we reject the null hypothesis for Z = 2.95 =1.645.
Thus, the rejection region is 20( p-.5)>1.645
If x is the number of “good” then p= saa
n
=>
Te;
100
so reject the null hypothesis for x =100p =58.2.
10-12
p2>.582.
(E)
Presumably, Aaron is claiming he is an at least an 80% shooter, so this is a left one25
tail test and we reject for Z <—zo5
P- Po
—
.694-.8
=—-1.645.
=
p evan 694, and
—1.58, so we cannot reject the null hypothesis
(ee — Po) ap 8: cE 8)
that he rene at least i
10-13.
Since the claim is at least 60% skip, this is a left one-tail test and we reject for
tS,
|
vA
P-P0
p= 2 = 5375,
and,
80
i
SERat
Popo)
67 (1,6)
n
80
ee
114.
so we cannot reject the null hypothesis ofat least 60% skipping class.
204
SOLUTIONS TO EXERCISES - CHAPTER 10
Section 10.5 Hypothesis Testing for Differences in Population Means
10-14,
— 10-44+12-5
S2-= (m-1)S% +(n-l)S? je
P
(m+n—2)
22
Let
5
phere
meer
W=X-Y . This is a right one-tail test and the test statistic is
1(22) =
—27=15___.
2.132,/4
+ +b
= 572. From the table, £/7(22) =1.717, so the p-value
is at least 5% (using TDIST in Excel gives a p-value of 28.6%). (E)
10-15
+(n-1)S7 _
11-4 +11-12 ee
§2 _ (m-N)S¥
ee
.
Chee
20
Let
W=X-Y.
This isa
oo F066.
sien?
left one-tail test with critical value
T= 1 ,,(20) 221725. The test statistics 120) =e ene.
2.966,/4,
+45
11
l
It-T|=|791 + 1.725| = 2.52 (E)
10-16
Assume the female populationX and the male population Y are independent, normal,
and of equal variance. Then we can check,
Ho 2x
— My =5 versus Ho: uty
IC ee ete)
3.84,{4, + a
— uy >5.
Sp=
SE
From the table, ¢)/2(40) =1.684, and ¢(46) <t1/2(40).
Thus,
1.800 = ¢(46) >t (46), and we can reject the null hypothesis.
S2
10-17
Under the null hypothesis f = F(7,8)=
! OR
Sy
9
te
From the table,
F = F'o5(7,8)=3.5. Then |f-F| = |1.778-3.5| = 1.722. (C)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
10-18
Calculate S% =155.6 and S? =28
For L., the test statistic is y*(n-l) =
(n-1)S% _ (5)(155.6)
1556"
BET.
50
This is a right one-tail test with y/;(5) =11.07,
with test statistic
(n-l)S¥ _ (4)(28) =2.24.
Ge
eed
0
so I. is true. II. is a left one-tail test
But, 73,(4)=.71 and therefore we
cannot reject the null hypothesis, and so II. is false. For III., the test statistic is
So
;
:
F(6,4)=
= = ae =5.56.
This is a right one-tail test with Fs (5,4) = 6.26, and
y
SO we cannot reject the null hypothesis.
Thus, III. is false. The answer is (A).
Section 10.6 Chi-Square Tests
10-19
The test statistic is T = y?((r-1)-(c-1)) = 77((4-1)-(4-D) = 779) (A)
10-20
Under the null hypothesis the expected value e for each cell is 24.
There is no parameter, so the degrees of freedom 1s 4.
2 (Aye= (32-24) _ G2- 24)? mh 24)? ple 24)" , 20- 20)
24
24
24
24
24
X25
(4) = 9.49, so the p-value is at least .05S. (E)
10-21
# 205
This is a 2 by 2 table, so the degrees of freedom is 1. The fvalues are:
No Claim
Claim
Group |
40
10
50
Group 2
120
30
150
160
40 |200
The table of e values is identical, so y7(1)=0.
72 (1)- 735(0)| =|0-3.84
=3.84 (D)
yj;(1) =3.84.
2
206
SOLUTIONS TO EXERCISES - CHAPTER 10
——
a0
[100-50=
30
[2000]
aebard alsa
ea 72 |e 5 waa RE
aC
ees |
BEST otaleey [eet1008 2]eter 0ore eapawanraseny
773) =4.133 (B)
10-23
The given 3 by 4 table of actuals (/):
Number
4
i
x
Z
é
=
=
,
1000
;
a
ya(O)
—
(97ai ace RU ee
Soe
ee
oe
———————————————E
+
1o4
37
3
9
= 11.91,
Then, |v2(6)— 735] = |11.91-12.59] = .68
é
—|])2
yand 75;(6)=1
¢
(E)
Section 10.7 Chapter 10 Sample Examination
Is
The null hypothesis is p=.3,
The test statistic is, Z =
hae sn ee
(.3)(.7)
a
9
with the alternative p<.3.
a
= —2.06.
, and we reject the null hypothesis for
This is equivalent to p <.2056.
100
Then xs np=100-2056=20,56,
(D)
59
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
a
Let
S = X,;+X2+X3
~ N(0,307).
—
# 207
Then,
—_So = 205 =1.645=> 0'=12.16 (C)
(@F
a ee a
[aa
te (2ye
(74-67.03)*
67.03
eo
ar
| =.6703 ~.2681 =.0616
(16-26.81)? F (10-6.16)
26.81
6.16
12
Since the mean of X is zero, we have Oy
oS
67.03
26.81
= 7:49.
(D)
? ~ y7(12). Under the null hypothesis
this means 5DY? ~ y*(12). This is a left one-tail test, so we reject the null
hypothesis if v7 (23272 -.((12)=4.4.
Thus, ae
<4425 0°
¥7 <44
(A)
i=)
We reject the null hypothesis if 77(5) > v6)
(5) =15.09.
(m,-10)° n (m2—-10)10
10
yaad&)
Also,
ny
=
60—m, —(144+8+10+8)=20-—m
i
Ay
eames
10
OO
=
(14—-10)° i (8—10)- is(10-10)- f. (8-10)10
10
10
10
(nm-—l0)* >63.45
(A)
=
eA
m-10
=
> 15.09
10-n,.
Then,
208 # SOLUTIONS TO EXERCISES - CHAPTER 10
6.
This is a two-tail ¢ test in which we reject the null hypothesis if
t(3)
( ) >
15.84-10
¢(3)
= 2.97.
( ) =————
4/f4
025to25(3)
( ) =3.182.
a5
Donot
:
J reject.
(D)
Frequency (f) | 40-Prob (e)
E-. 20 De
>
pia
=
=
a
eed
(11-8)?
(11-8)?
goey £112
—_
(5-4)?
(10-16)?
4 TerrmenTG
=
95
24
Sy
(Oy = oom Db)
8.
This is a two-tail ¢ test with rejection region
(2) 2 ot995(2) t= 4.303
Given that t(2) =4.10, we do not reject the null hypothesis.
9.
(E)
This is a right one-tail chi-squared test in which the test statistic is ican
= ~ y7 (14).
Under the null hypothesis o* =10. Hence reject the null hypothesis for
E
2
P
To Wns (14) = 23,69 eel
A236, ew ()
10.
This 1s a right one-tail test using the normal random variable W. Since the
population variances are both given as 100,
TAA
EAE CaeDe
25
ee eee
ee
Ow
Js
Using the standard normal table, the p-value of this outcome is,
Pr[Z =1.4 = 1—.9192 = .0808. Hence, reject the null hypothesis at the .1 level but
not at the .075 level.
ge
(D)
This is a right one-tail test using the test statistic,
Vee
\ON=
Sian
102 75/10
(9)= OAKS AOS
12
120072 2.79. tyns(O)
=Bea and fyi(9)
=a Rae
Hence, we can reject at the .025 level but not at the .01 level. (B)
CHAPTER
11: THEORY
OF ESTIMATION AND HYPOTHESIS
NN
TESTING
ee
Section 11.1 The Bias of an Estimator
1-1
The CAS Exam 3 tables list the Gamma distribution using @ in place of f. Thus,
ELX] = a6 = 46 and E[Y] = (100)(46) = 4000. Then, e= se =.0025. (A)
11-2.
Let X be exponential with mean @. Then the median m is given by
5 = F(x)
| J)
=1-e"?
=
m=86I1n(2). The density function for Y3-is,
le
= 30[1-—e
7 | |edtel:
6 |) eres
| talile
30)
|= a
+e
0 -—2e 9a
9
|,
allBe
In general, {,xe * dx =? (using the constant fora ["(2,) density). Thus,
6?9
Ae
11-3.
11-4
a.
3
es
1 —6@.
60
55
The bias is
= 9-6 1n(2) = 098 (C)
The distribution function for the minimum random variable is
20
F(y) = Pr(Yoo) < y) = 1-LeA] = 1-e
OA,
Thus,
Then,
B = Y29) 1s exponential with mean
i
The population has exponential distribution with mean E[W]=6 +4.
Therefore, E[W] = 0+4
and so 6 = W —4 isan unbiased estimator for 6.
Section 11.2 Building Estimators
11-5
Let X be the claim size population.
From the distribution type,
pe
Kee
=
ere
_—_—
oes
=
ee
From the data, X= ”.
(x) =1-— (42) for x >10, and
,
ye
=
x10"
=
100/6
=
100/.6=10
—_—
ae
re
Sa
©)
E
209
210 # SOLUTIONS TO EXERCISES - CHAPTER | I
11-6
From the data, n =20 and ¥ = =-[0-141-34--+8-1]=3.1, and,
BU
espe ee
21
24a= 13 oe 1—32)-23.49..
de
at er1+1°--3+---+8
i eaepiee re
Equate.
EIN |=rp=X
=3.1 and Var[N|=rAd
By=f- =3.49.
3.49
Then 1+8 ="
= B=.126 (A)
4
11-7
From the data (in thousands), Y =75 and OG
i=l
=18,458.5. From the Exam 3
20°
ines Gopala-\ ehap ei (a—-1)(a@—2)
Then, #=75 and
©
Sas
eo)
a-l
26?
ee
11-8
= 18,458.5.
To solve, take the square of E| xX] to calculate,
:
37
Oa 2nd
ap 2(a-1)
aie 18,458.5
=
ee helnd(O)
4
From the data, Y =325 and roe =119,/50 se7> =) st a25e 8 tan
i=]
E[X]= af =X =325 and Var[X]=af? =T? =8,125
a= (OB)
=——
=13
ap
$125
11-9
=>
(E
oe
The density function for an exponential distribution with mean £ is
ee
lL -x/B
[te
enwis
0
i
x20
eeu
So the likelihood function is
ib
(P)
l smootthy PScael po 2oy B Se
76
Bs
[
——
Next, find the value of # that maximizes
In{L(B)] = —nIn(P)
(X; + X2 +°°++X,,)
B
Bore
—N
Xp ter tx
Setting the derivative equal to zero, we have 0 = / +=
a
which implies 2 =
Neta Ossie
Hl
te
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
11-10
The maximum likelihood estimate for an exponential distribution is the sample
mean, = ¥ = £50+1123
+67+1257
+435 +1896 = 904.6 hours.
G
11-11
The likelihood function for this discrete distribution is
L(p) = [p-—py* |_p-d- py ]--[p-d— p)* J=p"-d-pystet.
In[L(p)]=n-In(p)
+(x1+.x2+++ xy):In(1=p).
d
inf
gp
LM
See
=
=nm (my +%2=+::'+%n) = 0,
which implies the maximum likelihood estimate is p= Tr
+
11-12
The likelihood function is
L(@)
=
Axo
Oxy"
| ea
ese
-Ox,?7
=
g”
“(X] 23%) o
Soe
Sea
oe
In[L(6)] = n-In(O)
+(0-1): In(xy -x9 ++ Xp).
Take “ and set equal to zero to find the critical value to be
De
11-13
=n
In(x, -X2°X3°
L(O) =(6
°°"
ee)
+1)" (x; +X_). and In L(A) =nIn(9 +1) + Oln(x)---x,).
Take ise and set equal to zero to find the critical value to be
dé
@ 211
212
11-14
SOLUTIONS TO EXERCISES - CHAPTER | 1
Gye Yorn By end ¥-L-r(n,2).
Se (x) = sees te xt le(’/P)x
for x>0.
(4) @-p:
(b) ~ ETA] = El
= ess fe (x)dx
ie
fon
on
= on _ptle(nlBx by
=
=
Gh
11-15
F(
= :
i
ie
n-1
=
LDL ao
i
nl
(n—2)! =
(=
n
THA = bee
x) == lee —(x/0)
} Bee
eee
n
—
eg een—-1
3 me
toe
a
eo
—
p—(c/8)
=
——
=
¢€
= ini
12
3
We take e=mn( $8. The MLE for 6 =.Y =3.78.
Thus, the MLE for ¢ is inl$3.78) =1,087.
(B)
Note: This is an application ofthe invariance property of MLEs under
transformations.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
11-16
# 213
Let N be negative binomial, the number of failures before the r”” success and
ie B+ltp Then,
para
\
iP
k
B
lank
Pk =Pr| [N=k|/=
| |
(ga) 4
ay:
. The likelihood
function L(r,
8) = po: p3- ps, where po -sa) :
py
SaaNeIne
| )( B )ad
3!
P35
Broly
Resi
5]
su] baa = i hen,
L(y, )= po: ps ps = ps =)
2
2
Mel]
9
bes
3r
Let Y= y,..
= We) — Ml
Themey Cn 8y= ‘5
eve
for O< y;;
for O> y;
n
Prieretoren 1 (0) =
ee
=ere
Gas
0
for
O mint yi,
Va)
for
#>min(j,..-,¥n)
'
‘
ia=min(j,,-..-, y,)
L(@)=0
theta values
As the graph of L(@) shows,
L(@) is an increasing exponential function of @ up to
G=min(Vj,--- y,), and then drops to zero for 8 > min()j,..-, V7 )2 a Dus, the
maximum of L(@) is reached at = MM
eee!
ple)
214
11-18
SOLUTIONS TO EXERCISES - CHAPTER | |
From Exercise 11-13 the MLE is,
.
= ee a ee
a5
Ces
n(x) toe =)
In(.92 -.79-.90-.65 -.86)
11-19
f(x) = F'(x) = (6+1)x°.
“
C=
=
Z
ee
Oe
©)
From Exercise 11-13 the MLE is,
=
Sige
In(x; 20,59) F8 Se UEC Sa
s/s eae
In(.56-.83-.74-.68-.75)
18)
( )
-f Lt
11-20
L(@) =
O"e
xX,
xX}
Xn
=—
=> InL(0)
DOP, 5)
= 10.67
11-21
thousand
(C)
Let p be the probability of winning a bet (success). Each bet can be considered a
Bernoulli trial random variable X with parameter p. The probability function
is fy (x) = p*(l-p)'~* for x =0,1 and 0< p<1.
S = s be the observed number of successes.
Let
be the number of trials and let
Then,
u
i
-
ys
L(p) = [[p*(-p)'™ = pa
i=l
InL(p) = slnp+(n-s)Ini-p)
F
the MLE for pis
A
AY
=>
sl =x)
-0-p)a
= OnE
[In L(p)] oe
-
p
on
:
l-p
a
eat
«
es
.
p=—. Note that this is the same estimate used in standard
n
binomial proportion problems. In the problem at hand there are (20)(3) = 60 trials
e
and 7+(2)(3)=13 successes. Thus, ee
60
n
(A)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 215
11-22
Directly from the data, ¥ =5 and the unbiased sample variance S? =5.5.
Let p be the probability of a claim on an individual policy. From the solution to
Exercise 11-21 above the MLE estimate for pis p = we
= .05.
The annual claim count is binomial with m=100 and probability p.
The ‘variance of the annual claim count is mpg =100 p(1— p).
Therefore, the MLE of the variance in the annual claim count is,
100- p-(1— p) =(100)(.05)(.95) = 4.75.
The difference in the two methods is 5.5-—4.75=0.75
11-23
(D)
ForX exponential both the MLE and the Method of Moments estimator result in X
as the estimate for 9. Hence, 6, 0; = (a (CG)
11-24
a2
11-25
At
From Exercise 11-12 we have
9=————_>»
Leap e9 a
=
In(.25-.50-.40-.80-.65)
= 137
2 ee)
. Thus,
(B)
Let gq be the probability of a$500 loss.
nent
pa giao
2 peel
— gs | >
pss.
Since there are 3 outcomes on each ofthe 3 trials, the probability of observing the
particular outcomes $0, $0, and $1000, using the multinomial formula, is,
3
2
Since p<
peg
p.
Setpie
eel
this probability is maximized for p=
216
SOLUTIONS TO EXERCISES - CHAPTER | |
Section 11.3 Properties of Estimators
11-26
(a)
Foreach x; we have E[x?]=E[X?].
nm
4.2
Then, bs 2 |
= PEL]
n
n
i=l
= E[X?], and so the bias equals zero.
2
2
PT. y2(n-1), and SO vay| |= Var| 72(n-1)| = 2(n-1).
o
o
(b)
On the other hand,
2
rar|
o
2
—l)o*
oy
n
= —Var[T*] = Var{T*] = aioe =ee —> 0,
and hence J? is consistent.
Note: There is a minor complication here since T? is only asymptotically
unbiased.
This is because T* = 415 2 where S? is the unbiased estimator of
o*. Thus, the argument showing that the variance of T” converging to zero
implies consistency requires a slight technical modification.
In f(x) =In3 + InO+2Inx-0x3 => LACES
00
See ge
The CRLB is given by —
oO)
:
:
na]ome
Un f2)] =—
0
= —
: —
=
Ce
is
git
(E)
06?
et
ae
nel
U.
s
1
ee
In Example 11.3-1 we showed that Var[@,] =
n(n+2)
Since this converges to zero as n goes to infinity, the estimator is consistent.
11-29
Since B is binomial, E[B]=np and Var[B] = np(l— p). Let p= ity
n
Ronn
Joes
|
p(l-p)
Then, Z| p|=p and Var[p] = ——-np(l-p) = ———-.
(a
ai
Since this converges to zero as n goes to infinity, the estimator is consistent.
11-30
In f(x) = In A — Ax =
The CRLB
is given by
A
Vise
sul
A)2
hy:
a ete ee
aA2
ng|Slnfe:2) |
=
RA
ary i
)
So
n
(E)
*
97,
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 217
Section 11.4 Hypothesis Testing Theory
11-31
(a)
We use a standard normal z test based on the sample proportion p= in
where X is the number of made shots in 100 tries. The rejection region is
A — 80
z <-1.645, or equivalently, 2H
< -1.645 >
-80-.20
100
X¥<73.42.
We reject the null hypothesis if Grant makes 73 or fewer shots out of 100.
(b)
Under the alternative hypothesis XY is binomial with n =100 and p=.5.
Thus, “x =50 and oy =/100-.8-.2 =4.
Wisk yer = Pr| X >73| p=.5] = Pr)
Z>
11-32
esas
V100-.5-.5
| mee
Use a right one-tailed rejection region of the form X > 4, where,
05 = Pr x2 4|u=10|)= Prizes | >
A=" 1.645.
Vl/n
lin
06 = LF <ajqett] = mz<a
4
Aah
15558
81564571. 050
bhen
Vl/n
Subtract these two equations to get,
fel) eee Scie
a
Pi mee NOE 7
Jl/n
n> (1.645+1.555) =10.24n211 (B)
a
peg
:
Under the null hypothesis 2=1,
Let NV be Poisson. Then, Pr|NV= k] =m a
we have,
(ras)
Reject the null hypothesis if N >k , where k is the smallest integer for which,
Pr[N >k] < .10.
From the table, we have k =3, with
Pr[N >k] = 1-.92 = .08 < .10. Then, the power equals,
1-8 = Pr[N>3|A=2] = 1-Pr[N =0,1,2|/ =2]
50
=
51
e722
52
—
7
= ]—5e-
= 325
(B)
218 ® SOLUTIONS TO EXERCISES - CHAPTER | |
11-34
11-35
The significance level is the probability of rejecting Hy when Ho
false. II. is true, and III. is false. (B)
is true, so I. is
likelihood ratio critical region is of the form,
| and so, the
L(A) =An"eAMte+"+%)
e(nitx E+E)
EX, )
L(1)
LQ)
e(™ +X
|” el
Ftp
Q*etarmetm)
4++-+Xp_)
=< 2° k =e k'
on
<k. Thus,
xX] a X2 =f-se6 Ny
=
Ink’ ,
<
or equivalently,
X <nink'=K. The population X is exponential, meaning that, wy
under the null hypothesis. Using the central limit theorem,
05 =PrLX < K]~Pr
z7< Aa
1/./100
eae G45
1/,/100
=o
=1
= Tiles (B)
Notes:
iN,
since Uy = + , the hypotheses can be restated as:
Ho: uy =1 and Hy: yy =.5, a left one-tail test, so that the form ofthe
rejection region X < K could be deduced without working the NeymanPearson inequality.
2.
The statement of the problem asks for the lower limit of the critical region,
although the solution, 1.1645, is, in fact, an upper limit.
Se
11-36
The fact that Y =1.5 is not used in the solution.
With a significance level of .05, the only possible rejection regions are:
{0}, {0,1}, {2} with @ values of .01, .05, and .05, respectively. The power ofa
rejection region is the probability of the region given 6. Thus, the powers ofthese
regions are .04, .08, and .09, respectively. Therefore, the most powerful region is
{2}. (B)
11-37 The Neyman-Pearson lemma will result in a right one-tailed rejection region of the
form X >A. The calculations are similar to Exercise | 1-32:
05=Pr[¥2 d|q=10] =P z2 ae
Bem
S250 nH
|
95=Pr|X > A] #=1|=Pt]
=
Z>
=
V25/n
3645.
WZOtn
=
“teh
=—1.645.
Subtract these two equations to get,
ANAS
ed
a
Nie 7
n> (25)(2-1.645)? =270.6
oases
=*2*}.645.
2oun
=>n>271
(D)
Then,
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 219
11-38
Let U be the maximum of the 5 observations:
Under the null hypothesis the
population X is uniform on the interval [0,1] and so, Fy (x)=x> for0<x<1. Now,
5|
——
11-39
>
Pr/ Uc]
==
ji—
=1-Fy(c)
Peo)
=1-8
=n
i
eailSiiees a
e-(1
=
=
Sg
=
The calculations are similar to Exercise 11-32 and 11-37:
05 = Pr[¥>C|u=100] = Pe aca so at
V9/n
V9/n
et eel
Os PryZee
ee
ee eG
V9/n
V9/n
;
Subtract these two equations to get,
Ee)
ea eS
rife saan
ee
J9/n
Pies
V9/n
Then, n> (9)(1.645)? =[3(1.645)] (C)
Section 11.5
11-40
More General Likelihood Ratio Tests
025=Pr[X>k|4=L0=15]=Pr|Z>4a!
= AH 1.96 = k=3.94.
B = PrfX<k|w=5,0=15] = Pr] Z<22
2 =-71
Ps
Sea
11-41
2
oe Olle
sel
°
Ex
‘0
09,
a
X|
The likelihood function 10)=[]| 4 +|-() e weg
ort
(CO)
XS
Under
Hp;
T=
8=10
and L(Ho)=100)=(4)
e
l (47
10°? The denominator is the maximum of
L(@) over the parameter space, which by definition, occurs at the MLE for 6. The
MLE for @ is the sample mean,
X= a =o 4.
_
Thus;
L(H,)=L(X) =L(.4) (4) e 94°” and the likelihood ratio is,
CEG) ety
100% ON
44D
e CP8
4)
9.
=
L(A)
=..991
(B)
220 # SOLUTIONS TO EXERCISES - CHAPTER | 1
0
Nowe
ee
=>
(ae
2
ok
Now, under the null hypothesis,
for,
Osx <1
for
x<Oandx>1
>
x<K.
R(x)=0
for x>1.
Thus, .05 = Pr| X <.05 or X >1] (D)
SE
n
I
ECA)
(a, B) Nd fa
prercemerees
a-l
(xj-x2-+
Ese ++ os Xn)
n
]
(x; Xp
Xp, )
e i
The numerator of the likelihood ratio is L(1,1) = ree css)
and the denominator is, Z(2,1) = (x, -x2+ ++: -Xp ye ia
ae
a
LT)
~
h the likelihood
Thus,
ratio is, R(x) — Oe—
1
ee
.
The rejection region takes the form,
xy X2
hp
Ske Six
xg
xy
Si
11-44
Ss Es
k
te
In— s Dinx, >In—-=C (B)
The numerator of the likelihood ratio is f(x;0)=
1
m[l+x-
denominator is the maximum of /(x;@), which is —
| for —0<x<o.
The
(occurring when x=@).
4
Thus, the likelihood ratio R(x) = es eee
+ =>
1x] 2
ssak, fia
form ofthe rejection region. To calculate K, we have,
05 = Pri|X|>K]=\|1- le s
t
=
Then,
tout tan-l(x)
1
x=-K
.05 = 1——tan=+(K)
Wa
=
|-— (tan” (K)—tan7! (-K)]
1
.
= 1-—tan“!(K)
1
K =tan( %..95}=tan(4752)
2
(E)
is toe
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
1145:
eh [le a Sea
n
XG
i=l
it
Xt
Xp,
Bai aceePea
er
# 221
onmeniarer ihe likelihood ratio 1s;
AoX19 "Xp
Dives
ae
Tie
. The denominator of the likelihood ratio is the maximum of
Xp!
L(A), which occurs at the MLE for 2. By Example 11.2-6, the MLE for 2 is X .
Thus, the denominator of the likelihood ratio is, L(Y) =e" +.
Xy!++X,_!
Hens
aaer
rejection takes the form, eae
e7"4o
oe
Mbetn!
es a rari
Cn
Np
Ao
vr nX
_g-ndy
=e
, epnX | 4)
70.
=e
Litts|
eXy+
nX
nX
1S equivalent to Ga
xX
The
= Ae)
n¢
Skene
= Ks (A)
Section 11.6 Bayesian Estimation
11-46
Let 6=1,2,3,4 represent the risk category and let Y =0,1 be the Bernoulli trial
random variable representing whether or not an accident occurs (XY =1 means an
accident occurs). In the solution table below:
Pr[X =1| 6] denotes the conditional probability of an accident for a given 6,
oia(G) denotes the prior distribution of risk categories,
Prion PrX =1) =>
Trew(P)
=
0
Pr xX =—1) 6 )7sq(@), the total of the Prior PrL_X =1] column,
Pr[X =1| O]Zo14 (9) , the posterior distribution of @ using Bayes’
> Pr X =1| A]zo1a
(P)
Formula, Posterior PrLX =1]= > Prf] XA =1\| Ore, (@), the total of the Posterior
‘XY=1] column.
Prior
Pr[X ==1 |0]
7 old (A)
Pr[X =1]
Posterior
ao
Pr[X = 1]
=0095
(a)
Privat
X all = Aew(1). 21584
(b)- > Prior Pri
X =1.= .0303
(cy 29 Posterior Pr
X = 1] =".0352
222
SOLUTIONS TO EXERCISES - CHAPTER | 1
Notes:
11-47
(1)
Since there was an accident, the posterior probability of an accident (in the
next period) increases.
(2)
Since X is a Bernoulli trial, Pr_X =1] = p = E[X]. Thus, the answers to (b)
and (c) can also be interpreted as the prior and posterior expected values of X.
The joint distribution of @ and - is discrete in 0 ot continuous in x. The prior
distribution of 6 is Pr[@=1]
= —5 and Pr[@ =3]= —. The conditional density for
Ais f(eie=1)= aoa . The joint distribution evaluated at x =5 and @=1 is,
x+
FGM (OOS Piet = [
(541)?
fA@e)) =
: "
(5+3)*
Pr[@=1| X =5]=
17
Ie) = = Stiritaclyt
5) = ee
J\.2
Next, the posterior probabilities for 6 are,
1/72
=
ask
3721
ne
Bie =4 be
eS
3/128
= 6279.
a3798
The conditional CDF for X given @ (using the nea distribution) is,
F(x|6)=1-—
Then, Pr X>8|0=1]=4+ and PrL¥>8|0=3]=>.
Using the posterior probabilities for 6,
PrLX >8]
11-48
PrLX >8]6=1]Pr[@ =1]+PrLX >8|6 =3]Pr[0 =3]
S$(5}orn+( 4 |e279 = 2126.
Let X; and X> represent the number of claims in year one and year two,
respectively. Then, the conditional probability f(2,.2|q)
X; =land X2 =0 is f(1,0|g)=q(1-q).
cs
evaluated at
probability function evaluated at
X, =land
X> =0 is, f(1,0,q)=q(1-q)a(q) =——
~—(a" —q°) for. 6<q<.8.
The marginal probability for X, and X> ne: at X,; =l and X> =0
,
f (1,0)
=
Pr[ X) slot
os=
ap FU; 0, q) dq
i
=
oe is(q* i
“OF
Then, the posterior density for g is given by,
4
f0,0¢) _ 974-7)
Tr
|
ant
2
5
=
.
Then, Pr{q>.7] = J
8
(at=4°) dg = 156
“
ame
014 (q* =2>
7 <.6
)10l.0<
is given by,
2) ig=
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
11-49
# 223
Let @ be the class of policyholders.
Then the prior distribution is given by, Pr[@=/] = 2/3 and Pr[@=J/] = 1/3.
Now, E[X|@=1] = 12,875, and E| X|0=1]=6,675.
Given X = 250, the posterior probabilities for @ are given by,
Pr/@=1|.¥
r[9=1|
BPR
r[9=1|
=250]
=
(.5)(2/3)
eal
X=250] (52/3) 00/3) > 17° 28%
eA er
(.7)(1/3)
X =250] Gio
senda a ye
Posterior ELY] = 12, 875(72)+6 6754] = 10,322.
11-50
Let @ be the class of policyholders. Then the prior distribution is given by,
Pr[9 =1] = 1/2, Pr[9=2] = 1/3 and Pr[6=3] = 1/6. Let Xbe the number of
claims. Fromithe table sh((O=1)
= 1, E[X|\e@=2| = 2,and Aix
10 =3)= 3.
Given X =1, the posterior probabilities for 6 are given by,
iy ESL
Pr[9@
eel
ag) esee
6)(1/3)+(0)1/6)
(1/3)(1/2)+(1/
3
(1/6)(1/3)
=1] =
X Ms
=2|
ame
ch
Mat2)
ea
nOnediayeOliey
(iam
he
Oey
(1/3)(1/2)+(1/6)(1/3) +(0)(1/6)
Posterior ELX] = (1)(3/4) +(2)(1/ 4) +(3)(0) = 1.25.
11-51
The joint density is feima=(4 }
0<x,x%2
<O=>O> max(x,x2).
Then, with Y; =400 and >
500
Q?
4’
* i
4° toe
=U eeLhe,
5000+. Note that
Let f(x,,x2) be the marginal density for 11, X2.
=600, (400,600)= [a 5000+d6 =
Thus, the posterior density for @ is given by,
f (400,600, 0) =(3
£(400,600)
500
(3)(600)°
(3)(600)
39-4.
cpr
Since
X is uniform on [0,9], we have ELX |6]= £ , and the Posterior
x
ELX] = (2
(3)(600)30-+ dé = (3)(600)'
63d0 = 450.
p)
600
600 ey
Note: In some ofthe tabular calculations that follow we have rounded numbers,
typically to 4 decimal places. This may result in some columns not summing exactly
to the number in the totals row. Students attempting to replicate the calculations
should avoid rounding numbers in intermediate calculations, either by using exact
fractional arithmetic, or by using storage cells on a calculator or spreadsheet.
224
11-52
SOLUTIONS TO EXERCISES - CHAPTER | |
Let @ be the class of coin, with 6 = A corresponding to coins 1-4,
9=B
corresponding to coin 5, and
9=C
corresponding to coin 6. Since a coin is chosen
at random the prior probabilities of A, B, and C are 2/3, 1/6, and 1/6, respectively.
The given sequence of outcomes is HHTH
and Pr| HHT.fee P| = p(1-p).
Prior
Moa (O) |Pr{H
|0] |PrLHHTH
|@]| Pr[HATH] | tye (A)
(4)4
1/2
A | 22
.0625)(2/3) |
|4
~o613
(.0625)(2/3)
ae
=.0417
~ 6809
Posterior
Pr[H]|
(1/ 2)(.6809)
= 3404
(1/.4)3(3/4) | (.0117)(1/6) | R222 | (1/4)(.0319)
=.0117
= .0020
~ 0319
(3/4)°(1/4)
(1055)(176) ere.
=,1055
= .0176
.0176
FTRSETEY ESCs [Citic Gs IC
= .0080
|GL4\2872)
/
= 2154
ets
Cok
Uy Me
The probability of heads on the 5" toss is .5638.
11-53
The prior distribution of @ is @ =1,2,3 with equal probabilities of 1/3 each. Let Xbe
the claim amount. The conditional density f(x|@)= per
F(x|q@) =1- (
|
a
. Given the observation
probability function /(20,q@) = A aloe
x+10)@+!
With CDF given by
X = 20, anda
value of @, the joint
\¢2 The problem requires calculating
P| X > 30) using the posterior distribution of @, given Y =20.
We will use
P[X >30|a] =1- F(30|a) = (<2+10 i a (5)
Joint
@
1
Posterior
| Xoa(P) | f[20|a] | f(20,a) | tney(P) | Pr[X >30|a@] | Pr[X>30]
he
AUS
LO.
302
0037
= 5000
ay
0025
5)
3
}
;
= 1250
PN
eS
1/3
200
0074
0625
(.0625)(.3333)
}
se
=3333
vail
= 0208
1/3
3000
SS 000R es
e002
0156
(.0156)(.1667)
= 00124
= 16en
= 0026
0074
1.000
1484
ie
Totals}
Foe
(.25)(.5000)
5
.0074
3
1.000
30
3 «| 0074
The revised (posterior) probability that the next claim exceeds 30 is .1484.
A
PROBABILITY AND STATISTICS WITH APPLICATIONS:
11-54
The prior density for 6 is given by (0) =2;1/2<6<1
A PROBLEM SOLVING TEXT
® 225
and the conditional
probability function for
X given @ is f(x|6)= i;
Jore-GO) 20h =0,1,2. The
joint probability function evaluated at Y =1
is
f0,A) = 200-6) 2(@) = 40(1-0); 1/2<@<1.
The marginal for X evaluated at
1
X=lis4 f(0,0)= |,
400-0) do = + rsa (Gye! ;ue =120(1-6);1/2<@<1.
11
The Bayes’ estimate for squared error loss is E[@]= [,1207-6) ater 16.
11-55 E[,] = 5000- a ; # 5000 =
6,, is biased. I. is false.
Since @, doesn’t depend on any observations of X, Var[ 6, | =(, and therefore
6, iS
consistent. II. is true. The mean square error is given by,
5000) Jean
|_
E|(Ala
- 0) | = Eso 10
= _ 5000}- £|(2~
612
so III. is true. (D)
11-56
The mean square error of an estimate c of X is given by E |(X-c)? |. As shown in
the text the mean square error is minimized when c = E[X]=a@@=300.
11-57
A
f)
2
(D)
9
The mean square error is given by E|(6-0)? |= i (V0) ie Je.where
values of @ are denoted by y in the integral. Using tabular integration by parts, we
have,
10. ( =
Bia OO
y?
YY
10 \|
|
2
29)
EL8 at iy ede
cs oe
pal (9-0) 5-20-98) Tony oT a
st | pees
- (10 fee29"
“X90 J 4410-11-12
For @=100, the mean square error is
1007 _
66
\=2
226
11-58
SOLUTIONS TO EXERCISES - CHAPTER | |
The bias of a point estimator 1s Bias = E [6] — 0.
alG - E{6}) +(z14]- ay
MSE(6) = E[(6-6)]
II
£|(6~ £[6]) + ae
AG : £6) |i 2E|Bias8
= Var[6]+2-
£{6))|+ E[Bias?]
Bias -(0)+ E[Bias?] = Var[@]+ Bias?
Section 11.7 Simple Linear Regression: Basic Formulas
11-59
Wehave
N=9,X =2,Y =10,Syy =164,Syy =100. From formula (5),
S2, Ey aya aes SPAT
S2,
ee wa
EAT IIOG
enamel
y = T164)(100) aa?
=> Syy =—112.695 (negative since the correlation is negative).
be a Pond 0 AY ORY =1137 oY Sao
Then,
tt
oor,
XX
For X¥=6,Y
11-60
=d+BX =11.37-.69-6=7.251.
Using the given data for_X and Y, we have,
N =5,X =3,Y =13,Syy =10,Syy =34,Syy =17. Then,
R? ee
eee(or)5. (D)
Sixx Sy
Note:
Since the values of Y, are also given, one could calculate ESS and use
formula (8): ESS = Syy -(1—-R?).
11-61
Syy =10,Sxy =195=> berSxy =) 719.5, and
Sixx
@=¥ — BX =45-(19.5)3) =-13.5 > Y =@+BX =-13.5419.5-1.2=9.9.
(B)
11-62
N=d,X
s45.Y =1.8286- Sp =700,Syy
0804s Sap = nS
p= 2X = 0336 and @ =Y—-BX =.3179
> ¥ =@+B-75 =2.8357. (A)
11-63
N=12,X
=12,¥ =145.1667,S yy =572, Syy =59,793.6667,S xy =5792=>
ie
=10.13 @=Y-BX =23.66 = Y =23.66+10.13.X
(E)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
227
11-64 We have InY =Ina+
BX , and,
Ce
Rea ee
ree
ase NS|B 9119.01 7.030" |Om 49A 221,098)
Prowisp is] [3523 |55|248.28|106.23
N=5,X =3,In¥ = 7.0463,Sxv=10,S(inyynv) =-0294,Sxiny =-5314 =>
B =23+4 =.0531 and In@ = 7.0463-.0531-3 = 6.8869 =>
InY = 6.8869+.0531-6 = 7.2057 =>. Y = e797 = 13475. (D)
39,
OT
11-65
11-66
=g
_ 23 ee
7ee
_ BF OT RC
ers s
N=5,X =7.1,Y =1.06,Syy =.2,Syy =.332,Sxy = 18>
(a)
p= Sa =.9 and @=Y - BX =-5.33
(b)
Y=-5.33+.9-7=.97
(c)
2_
Ra
(d)
ESS =Syy -(I-R?)=.17
(ec)
TSS = ESS+RSS
Sey
ae
_= 488
=>
RSS = TSS—ESS
= .332-—.17 = 162
11-67
wehave
Y = a@+ PX , so that the regression line passes through
Since
@=Y —BX
Dah
Minit eH G@+2f. Also, 3.3 = a+3B.
Eliminating @ between these two equations gives,
B=3.3-Y.
228 @ SOLUTIONS TO EXERCISES - CHAPTER | |
Next, note that Y = fo"
=
"w = 5Y —12° Then,
Sxy = (0)(5)+()G.5)
+(2)3) +3w+t(4)(5)- S)(2)Y
=129
5 3(57 -12)_-10Y = 5Y —65,
Now, £= 22158 fete OO Combine this with the previous expression,
Sixx
10
i= ieee
aver
aS
282
Sy
ey 19
ae
eee,
This implies Y = 30 and i Sef
11-68
Weused Excel to generate the following chart.
Actuary Income
§ 60000
”
o 50000
E[e)
S 40000
y = 2884.9x + 43759
R?ae= 0.9119
Se
#® 30000
0
1
ie
3
4
)
Number of Examinations Passed
(a)
Notice that the regression equation y = 2884.9x + 43,759 should be interpreted
as follows: The intercept of 43,759 should be thought of as a base salary.
(b)
The slope 2,884.9 implies for each examination passed, companies compensate
their employees an additional $2885. So if Hilary had passed five
examinations while in school, we would predict that her starting salary would
be near y = 2884.9(5)+ 43759 = $58,184.
Note: The accuracy of the regression equation is measured by R-. However, ifthe
data is not chosen randomly or disobeys our assumptions, then our work and
associated statistics are meaningless.
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT # 229
Procedure to Use Excel®
11-69
li
Enter the data in two adjacent columns, x first, then y.
2.
Highlight the data and make an XY scatter plot. You should label the axes and
the chart.
3.
Left-click on one of the diamonds then right-click to add a trend-line. Choose
a linear trend-line for simple linear regression. If a different model fits the data
more closely, then another model may be preferable. Such a discussion is
beyond the scope of this text.
4.
Click on the trend-line to format it. Under options, add the R-squared value
and the equation.
(a)
Salary and Weight
400
350
y = 9.7297x - 189.05
R? = 0.997
Ena
|
|
2
|
2g
|
|
=
55
Salary
(b)
The correlation is ./.997
cenit
=.998 (positive, since the slope is positive).
_39 99.
¥ =200 then ¥ = 220+187:09
Oey
5
230
SOLUTIONS TO EXERCISES - CHAPTER | |
11-70 (a)
Children and Income
y = -2.9173x + 24.898
20 4
R? = 0.8329
10 |
(1000)
Income
Disposable
af
>
0)
0
2
4
6
8
10
Number of Children
(b)
The correlation is —/.8329 =-—.9126 (negative, since the slope is negative).
OMT at 4hen
9 0173
04 RoR
FI29F
ee
11-71
ACT and Grade in Stat
400
.
2
2 90 8
S
a
*
y = -2.5691x + 140.58
R2=0.9723
80
&
®
PRS
6
|
£ 60
[tS
|
a
50 |
40
15
20
25
ACT Score
Correlation is —4/.9723 =—.986.
30
35
40
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
Section 11.8 Simple Linear Regression: Estimation of Parameters
9)
11-72
Given Var[o7]=4.
From formula (16), Var[ B= ca . From the data,
DOG
Seat co Var p\== ==.
GE
CE
(n-2)
B
oi
gee
S yy
11-74
a0
Sie;
Syyv
ara
as
i
ge fe
ae
ar
Ve GsYe A |
Sixx
Formula (26) gives,
Pe
eeSS
Te (N=D)RSS= (N= 2)RSS Ts
,
ESS /(N—2)
ESS
TSS.
ESS”
From formula (15), TSS = RSS+ESS. Also, R? = = $0
ESS
TSS : 1 Th
es RSS ESS ype 5 ESS!
TSS’ TSS
ame ns Paks
iy
Le
He
oo
S89)
TSS
ESS
2
1— R?
From the data, N =20,¥ =5,Y =6,Syy =1600,Syy = 6400, Syy =1920=>
peek,
SAC
Saeed 7
asm
100086400
(18)(.36)
Mares CV=2 RSS Ber SS BB)
i
S516
0 105.
® 231
232
SOLUTIONS TO EXERCISES - CHAPTER | |
a elit
i
el
at de et
Ce
;
1-75
pe Ory ee 1okU
=
=
i Sxx
1600
eS
no
eleee
=1.2 and 6 = Y-BX = 6-1.2-5 =
Then, Y =1.2X. For X¥ =10,Y =12.
) 6 = S?= ESS. z ee ee
=409
= (1-.36)(6400
ESS = (1-R?)JTSS
tos (18) = 1.734.
(a)
Using formula (23), Y% + ts/2(N—-2)
;
;
rv is, 12 ++1.734
interval
|2, the confidence
1
(10-5)
2
pee
1600 Jexr56)= (5.3,18.7).
(b) Using formula (24), Y + ts;2(N-2)
interval is, 12+1.734
11-76
i eH
N
Sixx
fe
1
+—+ eB) S* , the confidence
OSE |ae SO)
We use the relationship, (1, N—2)= ies
— R?
,
= (—15,39),
with
Nos and ho 67 36 hen es) -=
at 212.942 From the F table
ae)
Fos (1,23) =4.28 , and /'¢,
(1, 23) = 7.88. rit reject in each instance.
11-77
ESS =7SS —RSS =43—37 =6.. Then;
F(,N-2)
=
RSS.
—lg~ = aitSA Al oes aay Rl
41.83 =11.91.
(n-2)
Section 11.9 Chapter 11 Sample Examination
ie
Maximum Likelihood:
L(@) = O°(x-+-x5)?!
eGae.
=>
InZ(@) = 5In@+(O-1)In(x:--xs)
ae rae
=> {InL(O)) = F+ini%s)=0
Ale
>
Method of Moments:
ELX] =
Gives §=5
=1.07, Then, R—S
=
Ox® dx =
R
i
osae
in(21-43-56-.67.72)
aa and X
= .27.
(A)
=13 Nn
>|
=.518. Equating these
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT @ 233
Ay: fo(x) = 1/10 for 0<x<10,
and0 elsewhere,
Ay: fi(x) = 1/5
and 0elsewhere.
for 5<x<10,
undefined
for
x<0
00
ie)
Msp
sea
Ai See
fOvaees
LO)
undefined
for
The likelihood ratio is {1/19
x>10
This means the rejection region C must be a subset of [5,10].
_ length of C
level is .05, we have .05 = Pr[C]
=> C can be any subset of[5,10] of
10
total length equal to .5. Then,
length of C
5
Pu etersSel
Ve
Reny =ial)
+ ky
» [hen
ELY]
=
Since the significance
ee
AK E(Y )i + kxE|Y> |
ey
=
lS
hak,
since
Y,¥,,¥2 are all unbiased estimates and therefore have the same expectation. Then,
Var[Y] = k?Var[Y%,]+k3Var[¥o] = (447 +k? War[Y]
= (4k? + (1-1)?
)Var[ Ya|.
To minimize this expression, differentiate with respect to A, and set equal to zero:
(Var{Y]) = [8k -20-m)|Var[y] = 0 > h=02 (B)
au ies
~ F(12,11).
Under the null hypothesis,
S7 ony
wh
edUe
SO,
S050,
e on _ F(12.11).
KS,
From the alternative hypothesis, this is a right one-tail rejection region.
The null hypothesis is rejected if F(12,11) > Fos(12,11)=2.79.
05 = Pr
i2
-
==
ave
i
Se
o
FQ
IN) 2 a
IRB O.5o)
Oe PL
=
OK SZ,
22.79
(E)
Then,
234 @ SOLUTIONS TO EXERCISES - CHAPTER | |
5:
Let X ~ N(20,27) be the side of the square.
£
Z
Then, Y ~ N [20,22 ;
The estimate for the area is X? and E[X?] = Var[X]+E[XP?
=
This will overestimate the true area by .8 meter. (D)
N
=7,W =1.86,Y =6,Sww =18.6,Syy =138,Swy =50
p= ae =2.65 and @ = Y-BW
=108
=>
=> Y = 1.08+265/X.
AX
ab
Let_Xbe lognormal, with XY =e”, where W ~ N(,07).
From the properties of the
lognormal distribution in Section 6.3.4 we have, E[X]=e4*(/)>"
ELX?]=e2#+2"
Then, InELX] = io
o? = InE[X*]—2InE[X].
8.
and
and InELX?] = 24+207.
Thus,
The method of moments estimator for o? is
In 4.4817 —21n1.8682 =.25 , and so the estimator for
o =./.25 =.5.
Since X is exponential, E[X]=6 and Var[X]=6?.
Thus, XY = y(o2|
Pr[Type II error] = Pr[_¥ <11,000|6
\||
d
(B)
12,500]
‘
oe
sl _ 11,000 12,500 |
Pridé <~-1.2] = 1-@(1.2)
=.115
(8)
PROBABILITY AND STATISTICS WITH APPLICATIONS: A PROBLEM SOLVING TEXT
Cy
@ 235
Reject the null hypothesis if
Sa
mene
S¥
4
=
F(n-1,n-1)
=
—
iNas:
=
3,077
=
Fos(n—1,n-1).
From the F table F'(9,9) =3.1789 and F(10,10) = 2.9782, so the minimum sample
size to reject.occurs fom —1 = 10 =S"e7=11, (EB)
10.
Reject the null hypothesis for X >C, C sufficiently large. Equivalently, reject for
a
ee
ee
se
C= 81942 (C)
n-1|
ie
The density function for Y, is f(y)=n x
ee
ana So; 212, |=
n
n
ar!
Then the bias is | Z 1a = eee
n+l
n+]
Thus, we need —_T
+]
<a)
a= a
20
eat
ol
So AW
es
eae
Therefore, 1 must be at least 20. (B)
D
x= 2 Sohne = a Bs
625) ee) and
G30
NOS KIS) 193.
Fess
tocar 78-11032 = 50.4,
(504541103
50
aa yee
Thus, 2009 is the first year during which premiums exceed 150 and 2010 is the first
year premiums exceed 150 for the whole year.
Either (C) or (D) is correct.
)
Sakae
eine 7 oT
eS
a
——
oer
Pa:
7
*
d
-‘ ; 7
a
~
et
it
aed ahs
.
mee
ao
a:
eee
ae
*
of
<2°s
7
:
:
am ae
So
4
oe
eer a
a
*
he— f wan
ro
es
eat
7
ia
<
s
Pa:
¢
a
iy
|
bree : or
ms
‘.
-
val
—
e
7
a
.
‘the
a
an
ISBN: 978-1-56698
il
it81566 987226
Probability and Statistics with An=
| TU
7
cf] Ws-AAV-