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3.
I operate a station that unloads and cleans chemical trucks. Trucks arrive at an average rate of 9 per
hour.
a. What is the probability that no more than 8 arrive in a two-hour period? (1)
b. What is the probability that more than 20 arrive in a two-hour period? (2)
c. What is the probability that between 10 and 22 arrive in a two-hour period? (2)
d. If it takes two hours to unload and clean a truck, how many locations for loading and cleaning do I
need to have at least a 99% chance of providing no-wait service? (2)
e. If trucks arrive at a rate of 9 per hour, what is the probability that no more than 200 arrive in a ten
hour period? (2)
f. If trucks arrive at a rate of 9 per hour, what is the probability that between 100 and 220 arrive in a
ten hour period? (2)
g. If it takes ten hours to unload and clean a truck, how many locations for loading and cleaning do I
need to have at least a 99% chance of providing no-wait service? (2)
Solution: Because this involves a situation in which you are given the average number of events per unit
space or time this would normally be considered to be a Poisson problem.
a) Since the period involved is two hours, the mean is m    29  18 . From the Poisson table for a
b)
c)
mean of 18, Px  8  .00706 .
From the same table Px  20   1  Px  20   1  .73072  .26928 .
From the same table P10  x  22   Px  22   Px  9  .85509  01538  .93971 .
d) On the same table the first number with a cumulative probability above 99% is Px  29   .9906 , so
use 29 locations.
e) Since the period involved is now 10 hours, the mean is now m    109  90 . But we cannot find
this mean on our tables. Thus, since the mean is over 25, we must use the normal distribution with
  90 and   90  9.4868. Using a continuity correction,
f)
(DIAGRAM?)

200 .5  90 
Px  200 .5  P  z 
  Pz  11 .65   Pz  0  P0  z  11 .65   .5000  .5000  1 .
90 

Using the same approximation as in e) PP 100  x  220   PN 99 .5  x  220 .5
 99 .5  90
220 .5  90 
 P
z
  P1.00  z  13 .76   .5  .3413  .1587 . Here PP represents a
90
90 

Poisson probability and PN represents a normal probability.
f)
Again use a normal distribution with   90 and   90  9.4868. From the t-table z .01  2.326 , so
x.01    z .01  90  2.326  90  112 .07 . We would need 113 locations.
1
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4.
Customers who enter my store have a 45% probability of actually buying something.
a. If 14 customers enter my store, what is the probability that at least one will buy something? (2)
b. If 14 customers enter my store, what is the probability that between 3 and 10 will buy something?
(2)
c. On the average, of 14 customers, how many will buy something? (1)
Now, and for the remainder of the problem, assume that customers who enter my store have a 65%
probability of actually buying something.
d. If 14 customers enter my store, what is the probability that at least one will buy something? (2)
e. If 14 customers enter my store, what is the probability that between 3 and 10 will buy something?
(2)
f. If 60 customers enter my store, what is the probability that at least 35 buy something? (2)
g. If 60 customers enter my store, what is the probability that between 40 and 50 buy something? (3)
h. What is the 99th percentile of the distribution in g)? (2)
Solution: Because this involves a constant probability of success, the appropriate distribution is binomial.
a) From the binomial distribution with n  14 and p  .45 . Px  0  Px  1  1  Px  0  1  .00023
 .99977 . Or by direct computation 1  P0  1  C 014 p 0 q14  1  .55 14  1  .0002318  .9997682 .
Recall that Px  C xn p x q n x , and that q  1  p .
b) Using the same part of the binomial table, P3  x  10   Px  10   Px  2
 .98857  .01701  .97156 .
c)
  np  14.45   6.3 . Note that   npq  14 .45 .55   3.465 , and that the expected number of
failures was nq  14 .55   7.7 . Since these are both above 5 , the problems above could have been
done by the normal distribution.
d) We are now using the binomial distribution with n  14 and p  .65 . We have no tables for this, but
we can do the problem in terms of failures. 1 or more successes correspond to 13 or fewer failures with
a probability of failure of .35. From the binomial table with with n  14 and p  .35 ,
Px  13  1.00000 . Or by direct computation with p  .65 ,
e)
f)
1  P0  1  C 014 p 0 q14  1  .35 14  1  .0000004  .999999 .
We are again using the binomial distribution with n  14 and p  .65 . We have no tables for this, but
we can do the problem in terms of failures. 3 successes correspond to 11 failures and 10 sucesses
correspond to 4 failures with a probability of failure of .35. From the binomial table with with n  14
and p  .35 , P4  x  11  Px  11  Px  3  .99997  .27827  .72120 .
We are using the binomial distribution with n  60 and p  .65 . We have no tables for this, but, since
np  60.65   39  5 and nq  60.35   21  5 , we can use a normal approximation. If we let PB
mean a binomial probability and PN mean a normal probability and note that
  npq  60 .65 .35   13 .65  3.6946 , then PB x  35   PN x  34.5

34 .5  39 
 Pz 
  Pz  1.22   .P 1.22  z  0  Pz  0  .3888  .5  .8888 .
13 .65 

g) Using the same distribution, PB 40  x  50   PN 39.5  x  50.5
 39 .5  39
50 .5  39 
 P
z
.
13 .65 
 13 .65
 P0.14  z  3.11  .P0  z  3.11  P0  z  0.14   .4991  .0557  .4434 .
h) Recall that z .01  2.326 , so that x.01    z.01  39  2.326 3.6946   47.59.
12/20/98 251z9883
2
5.
a.
A department store has the following data on numbers of shoplifters: (Sums in boldface)
Sex
1st time offender 1T 
Repeat offender 1T
60
77
F
Male
 
 
Female F 
44
137
81
125
104
158
If F represents the event that a randomly chosen shoplifter is female and 1T the probability that a
randomly chosen shoplifter is a first time offender, find the following:
125
 .4771
(i)
Solution:
PF  (1)
262
44
 .4231
(ii)
P F 1T (2) Solution:
104
44
 .3520
(iii)
P 1T F (1) Solution:
125
44
 .1679
(iv)
PF  1T  (1) Solution:
262
(v)
Find P1T  , and, using only this probability and some of the probabilities that




you found in (i)-(iv), find PF  1T  . (2) Solution: Since
P 1T  
b.
104
 .3969 ,
262
PF 1T   PF   P1T   PF 1T   .4771  .3969  .1679  .7061
An automatic pitching machine if set up correctly will pitch strikes 70% of the time, if it is
incorrectly set up, it will pitch strikes only 30% of the time. Assume that the prior probability of
setting up the machine correctly is 60%. What is the probability that it will pitch 3 strikes in a row
if it is set up correctly? What if it is set up incorrectly? (3)


3
Solution: P3S C   .7   .3430 , P 3S C  .33  .0270 .
c.
If the machine is tested by having it pitch 3 times and it pitches 3 strikes in a row, what is the
posterior probability that it has been set up correctly.? (4)
P 3S C PC 
Solution: This is a Bayes’ rule problem. P C 3S  
P3S 

P3S C PC 

 
P3S C PC   P 3S C P C

.3460 .6
.2076

.3460 .6  .0270 .4 .2184
 .9505 .
3
262
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6.
Data on teacher earnings is repeated below with years of experience added. (The last three
columns and the symbol column headings have been added.)
Teacher Earnings Experience
x
i
x2
y
xy
y2
1
38
5
190
1444
25
2
41
5
205
1681
25
3
63
8
504
3989
64
4
34
5
170
1156
25
5
47
6
282
2209
36
6
79
7
553
6241
49
7
48
5
240
2304
25
350
41 2144
19004 249
a. Compute and interpret the sample covariance between Earnings and Experience. (3)
 x 350
 xy  2144 ,  x  350 ,  y  41, n  7, x  n  7  50.00,
 y  41  5.857 . So s   xy  nxy  2144  750.00 5.857   15.6667 . The positive
y
Solution:
xy
b.
n
7
n 1
6
sign implies that x and y tend to move together.
Compute and interpret the sample correlation between Earnings and Experience. (3)
Solution:
x
2
 19004 , s x2
s x  250 .6667  15 .832 .
x

2
 nx 2
n 1

y 
2
s y  1.47619  1.21499. So rxy 
249 , s 2y
s xy
19004  750 .00 2
 250 .6667 ,
6
y

2
 ny 2
n 1

249  75.857 2
 1.47619 ,
6
15 .6667
 0.66331  .81444 The sign
250 .6667 1.47619
is explained above. At .66331, the strength is roughly midway between 0 and 1, so is neither weak
or strong.
sx s y


4
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The table below represents the joint distribution of earnings and experience for the district as a whole.
ExperiP y 
y P y  y 2 P y 
ence
Income
c.
y\x
40
50
60
P x 
xPx 
x 2 P x 
5
10
20
.46
.05
.00
.51
.05
.24
.07
.36
.00
.03
.10
.13
2.55 +
3.60 +
2.60 =
8.75
12.75 +
36.00 +
52.00 =
100.75
Compute the population variance for income (2)
yP y   46 .6, E y 2 
Solution: For income,  y 

 2228  46 .62  56 .4397 . For earnings,  x
 
.51
.32
.17
1.00
20.4
16.0
10.2
46.6
816
800
612
2228
   y P y   2228 ,  E y  
  xPx   8.75, E x    x Px   52 .00,
2
2
y
2
2
2
y
2
 x2  E x 2   x2  100 .75  8.75 2  24.1875 .
d.
Compute and interpret the population covariance between Income and Experience. (3)
92 .0  20 
0
Solution: E xy  
 xyPx, y    12.5 
0
120 
42 
30  436 .5 ,
120
 xy  Exy   x  y  436.5  8.7546.6  28.75. The positive sign implies that x and y tend to


move together.  x  4.91808,  y  7.51264
e.
Compute and interpret the population correlation between Income and Experience. (3)
Solution:  xy 
 xy
28 .75
f.

28.75 2

24.1875 56.4397 
0.60547  .778125 At
24 .1875 56 .44
.60547, the strength is roughly midway between 0 and 1, so is neither weak or strong.
 x y

Next year income will rise by 9% and experience by 1, using the results of c-e and appropriate
formulas, find population variance for income, population covariance for experience with income
and the population correlation of experience with income. (3)
Solution: For next year x will become x  1 and y will become 1.09 y . So let w  ax  b  1x  1 ,
so that a  1 and b  1 , and let v  cy  d  1.09 y  0 , so that c  1.09 and d  0 .
We have the formulas Varcy  d   c 2Var  y   1.09 2 56 .4397   67.048 ,
Covax  b, cy  d   acCovx, y   11.09 28.75   31.34 ,
Corrax  b, cy  d   SignacCorrx, y   Sign11.09 .778125   .778125 .
5
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7.
a. I am a party planner for McDonalds and have two birthday parties today.
The first party is for 13 5-year olds. The average 5-year old drinks 10 oz. of soda with a standard
deviation of 3 oz.
The second party is for 11 10-year olds. The average 10-yeqr old drinks 12 oz. of soda with a
standard deviation of 5 oz.
Assume that the above distributions are normal and independent of one another, and that I have 140 oz
of soda on hand for each party. What is the probability that.
(i)
The 5-year old party will run out of soda? (3)
Solution: The distribution is Normal. So x ~ N 10,3 . Recall that if x ~ N  ,   ,
x ~ N  ,  x  , where  x 
(ii)
(iii)
(iv)
(v)

. So if  x 

3
 0.8320 , (DIAGRAM?)
13
140 
10 .794  10 


P
x  140  P x 
  Px  10 .794   P z 
  Pz  0.92 
13 
0.8320 


 .5  .3212  .1788 . Call this P A .
The 10-year old party will run out of soda? (1)

5

 1.5076 ,
Solution: So y ~ N 12,5 .  y 
(DIAGRAM?)
n
11
140 
12 .727  12 


P
y  140  P y 
  P y  12 .727   P z 
  Pz  0.48 
11
1.5076 







n

n
 .5  .1844  .3156 . Call this PB  .
Both parties will run out of soda? (1)
Solution: If A and B are independent, P A  B   P A PB   .1788 .3156   .0564 .
At least one party will run out of soda? (2)
Solution: P A  B   P A  PB   P A  B   .1788  .3156  .0564  .4340
The 10-year old party will run out of soda, given that the 5-year old party runs out of
soda? (1)
Solution: If A and B are independent, P B A  PB  .3156.
 
b. If a user of Vanceril Double Strength Inhalation Aerosol (an asthma treatment) has a 22% chance of
reporting a headache in a 4-week trial, what is the chance that in a group of 10 none will report
headaches? (2)
Solution: Because we are asking for the probability of a number of successes in a number of trials
when the probability of success is constant, this is the binomial distribution.
If p  .22 and n  10 , P0  C 010 p 0 q10  .78 10  .0834 .
c. If I interview the individuals in the trial in b) in sequence, what is the chance that the 4 th person that
I interview is the first to complain of a headache. (2)
Solution: Because we are asking for the probability of the first success occurring on the 4 th trial in a
number of trials when the probability of success is constant, this is the geometric distribution.
If p  .22 , P4  q 3 p  .78 3 .22   .1044 .
6
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d. If I have a group of 100 individuals, half of whom are using Vanceril inhaler and half using Vanceril
Double Strength Inhalation Aerosol, what is the probability that if I randomly select a group of 8
people to interview in depth, at least one will be using the inhaler? (2)
Solution: Because we are asking for the probability of a picking a number of individuals of one type
from a finite population which is divided into two types of individuals, this is the Hypergeometric
distribution.
50  49  48  47  46  45  44  43
C 050 C850
50  49  48  47  46  45  44  43
8  7  6  5  4  3  2 1
1  P0  1 
 1
 1
100
100

99

98

97

96

95

94

93
100
 99  98  97  96  95  94  93
C8
8  7  6  5  4  3  2 1
 1  .001317  .99868 .
IV. Extra Credit
1.
The lifetime of a new brand of light bulb follows an exponential distribution with a mean of 2100
hours. Find the probability that a randomly selected light bulb lasts less than 2600 hours. (3)
1
1
Solution: For the Exponential distribution, F x 1  e cx where c  . So c 
and
2100


2.
1
2600
2100
Px  2600   F 2600   1  e
 1  e 1.23809  1  .2899  .7101
The shelf life of a loaf of bread is approximated by a exponential distribution with a mean of 3.0
days. What fraction of loaves stocked today will still be fresh after 4.5 days? (3)
1
1
Solution: For the Exponential distribution, F x 1  e cx where c  . So c  and
3

1

 4.5  
  e 1.5  .22313 .
Px  4.5  1  Px  4.5  1  F 4.5  1  1  e 3




See also 251y9882
7