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1.
2.
Answer Key for Homework 3 (Due date: February 14, 2003)
Notice that you are working alone for this homework. If I were you, I would start doing it ASAP
even though you have lots of time. It will save you from last minute frustrations.
Make sure to write your name and section number on each page.
Staple everything together. Place it on the table with your section before class starts. There is a
penalty of 5 pt. if you do not staple them together.
Each question worth different with the total 100 pt. for the homework.
Exercise 3.10
Let A be the number of pumps in use at six-pump station and B be the number of pumps in use at fourpump station. A can take the values 0 to 6 (integers) and B can take the values 0 to 4 (integers).
(a) T=0,1,2,3,4,5,6,7,8,9,10
(b) X=-4,-3,-2,-1,0,1,2,3,4,5,6
(c) U=0,1,2,3,4,5,6
(d) Z=0,1,2
Exercise 3.14. (a)-(c)
(a)
5
5
y 1
y 1
 ky  k  y  k (1  2  3  4  5)  15k  1
then k=1/15
3
y
1 3
1
6

y  (1  2  3) 


15 y 1
15
15
y 1 15
4
y
1
9
(c) P(2X4)= 
 (2  3  4) 
15
15
y  2 15
(b) P(X3)=
y2
1 5 2
1 2
11

y 
(1  2 2  3 2  4 2  5 2 ) 
 1 which is not possible for


50 y 1
50
10
y 1 50
5
(d)
pmf.
3.
Exercise 3.16.
X: number among the four who have earthquake insurance=0,1,2,3,4
p(Ai)=P(ith homeowner have earthquake insurance)=0.30, i=1,2,3,4
(a)
P(X=0)=P(none of the homeowners have earthquake insurance)= P(A 1) P(A2) P(A3)
P(A4)=(0.7)4=0.2401
P(X=1)=P(exactly 1 out of 4 homeowners will have earthquake insurance)=4 P(A 1) P(A2) P(A3)
P(A4)=4(0.3)(0.7) 3=0.4116
P(X=2)=P(exactly 2 out of 4 homeowners will have earthquake insurance)=6 P(A 1) P(A2) P(A3)
P(A4)=6(0.3) 2 (0.7) 2=0.2646
P(X=3)=P(exactly 3 out of 4 homeowners will have earthquake insurance)=4 P(A1) P(A2) P(A3)
P(A4)=4(0.3) 3 (0.7) 1=0.0756
P(X=4)=P(all of the homeowners have earthquake insurance)= P(A1) P(A2) P(A3)
P(A4)=(0.3)4=0.0081
The probability distribution of X is
x
0
1
2
3
4
otherwise
p(x)
0.2401 0.4116 0.2646 0.0756 0.0081 0
(d) P(X2)=1-P(X1)=1-[P(X=0)+P(X=1)]=0.3483
4.
The number of residential homes that a fire company can serve depends on the distance (in city
blocks) that a fire engine can cover in a specified (fixed) period of time. Let X be the number of blocks that
fire engine can move in the specified time interval. The following is the cdf for x
x  21
0,
0.05,
21  x  22

0.25,
22  x  23

F ( x)  0.55,
23  x  24
0.80,
24  x  25

25  x  26
0.95,
1,
x  26

a. What is the pmf of X?
x
21
22
p(x)
0.05
0.20
23
0.30
24
0.25
25
0.15
26
0.05
otherwise
0
b. What is the expected number of blocks that a fire engine can move in the specified time interval?
=E(X)=0.05(21)+….+0.05(26)=23.4
c. What is the variance of the number of blocks that a fire engine can move in the specified time interval?
2=Var(X)=E(X2)- 2=[212(0.05)+….+ 262(0.05)]- 23.42=549.1-23.42=1.54
d. Calculate E[X(X-1)]=E(X2)-E(X)?
E[X(X-1)]=E(X2)-E(X)=549.1-23.4=525.70
  2  ?
P(   2    X    2   )  P(20.92  X  25.88)=0.95
e. What is the probability of
f. Suppose they developed a new random variable, Y=2X-1. What is the pmf of Y?
y
41
43
45
47
49
51
otherwise
p(y)
0.05
0.20
0.30
0.25
0.15
0.05
0
g. Suppose they developed a new random variable, Y=2X-1. . What is the expected value of Y?
E(Y)=2E(X)-1=45.8 or E(Y)=
 y  p( y) =45.8
y
h. Suppose they developed a new random variable, Y=2X-1. What is the standard deviation of Y?
Var(Y)=4Var(X)=4(1.54)=6.16 then the standard deviation is 2.48
or Var(Y)= E(Y2)- E(Y)2 =6.16 can be used to compute the standard deviation
i. What is the probability of at most 23 blocks fire engine can move in the specified time interval?
P(X23)=F(23)=0.55
j. What is the probability of between 24 and 26 blocks fire engine can move in the specified time interval?
(inclusive)
P(24  X  26)=F(26)-F(23)=1-0.55=0.45
5.
x
p(x)
Exercise 3.37
1
1/6
2
1/6
3
1/6
4
1/6
5
1/6
6
1/6
1 1 1 1 1 1 1 
1 3 1
E (h( X ))  E      p ( x)          0.4083
6 1 2 3 4 5 6 
 X  x 0 x
Since 0.4083 > 1/3.5=0.2857, we would gamble.
6.
Exercise 3.47(a)(b)
X: number of seconds (goblets with flaws) ~ Binomial(n=6,p=0.10)
(a) P(X=1)=
 6
 (0.1)1 (0.9) 5  0.3543
1 
(b) P(X2) = 1-P(X<2)=1-P(X1)=1-P(X=0)-P(X=1)
=
7.
8.
 6
 6
1   (0.1) 0 (0.9) 6   (0.1)1 (0.9) 5  1  0.5314  0.3543  0.1143
 0
1 
Exercise 3.54
h(X)=1(X)+2.5(25-X) =62.5-1.5(X)
E(h(X))=62.5-1.5E(X)=62.5-1.5(15)=$40
where X: the number of passenger cars ~ Binomial(25,0.6) then E(X)=np=25(0.60)=15
Exercise 3.65
10  10 
 

x 15  x 

(a) X~hypergeometric with N=10+10=20, M=10, n=15 then P ( X  x) 
,x=5,..,10
 20 
 
15 
(b) P(all
specimens
of
one
of
the
two
types
selected)=P(all
are
granite)=P(all
are
10 10  10 10 
     
10  5    5 10  =0.0325
basaltic)=P(X=10)+P(X=5)=
 20 
 20 
 
 
15 
15 
10 10  10 10 
     
 7  8    8  7 
(c) P ( | X -  |  1 ) = P(6.51  X  8.49)=P(X=7)+P(X=8)=
 20 
 20 
 
 
15 
15 
=0.3483+0.3483=0.6966
M
=15(10/20)=7.5
N
N n
M  M  20  15
10  10 
 n   1   
 15  1   =0.9868 then =0.9934
2=Var(X) =
N 1
N 
N  20  1
20  20 
where =E(X) = n 
9.
Exercise 3.79
e 8 8 5
(a) P(X=5)=
=0.0916
5!
P(X5)=1-0.1=0.9
P(X10)=1-0.717=0.283
(b) in 90 min. period, =8(1.5)=12
standard deviation of X= 12 =3.4641
(c) in 2.5 hour period, =8(2.5)=20
P(X20)=1-0.47=0.53
P(X10)=0.011
10.
Exercise 3.83
=1/mean time between failures=1/0.5 year =2 per year
(a)  = 2(2)=4 during 2 year period
(b) X: number of loads in 2 years ~Poisson( =4)
P(X>5)=1-P(X5)=1-0.785=0.215
(c) P(X=0) =
e   0
 e   0.1 then -  ln(0.1) and   -ln(0.1)=2.3
0!
Since  =(the expected rate per year)(number of years)=2(number of years), at least 2.3/2=1.15
years.