Solutions to Practice Problems for Part II
1. A fund manager is considering investment in the stock of a health care provider. The
manager's assessment of probabilities for rates of return on this stock over the next year
are summarized in the accompanying table. Let A be the event "Rate of return will be
more than 10%" and B the event "Rate of return will be negative."
RATE OF
RETURN
PROBABILITY
a.
Less than - 10%
- 10% to 0%
0% to 10%
10% to 20%
More than 20%
.04
.14
.28
.33
.21
Find the probability of event A.
P A 
b.
Find the probability of event B.
P B 
c.
 P 10% to 20%  More than 20%
 0.33  0.21
 0.54
 P Less than  10%  - 10% to 0%
 0.04  0.14
 0.18
Describe the event that is the complement of A.
A is the event in which "Rate of return will be no more than 10%".
d.
Find the probability of the complement of A.
 
P A
 1  P A 
 1  0.54
 0.46
e.
Describe the event that is the intersection of A and B.
A  B is the event in which the rate of return is both more than 10% and less than zero.
f.
Find the probability of the intersection of A and B.
It is impossible for the rate of return both to be more than 10% and less than zero, so:
P A  B   0
g.
Describe the event that is the union of A and B.
A  B is the event in which the rate of return is either more than 10% or less than zero.
h.
Find the probability of the union of A and B.
P A  B 
i.
 P 10% to 20%  more than 20%  less than - 10%  - 10% to 0%
 0.33  0.21  0.04  0.14
 0.72
Are A and B mutually exclusive?
Yes; they can't both occur at the same time.
j.
Are A and B collectively exhaustive?
No. If, for example, the rate of return is 5%, then neither A nor B has occurred.
2. A manager has available a pool of eight employees who could be assigned to a
project-monitoring task. Four of the employees are women and four are men. Two of
the men are brothers. The manager is to make the assignment at random, so that each of
the eight employees is equally likely to be chosen. Let A be the event "chosen employee
is a man" and B the event "chosen employee is one of the brothers."
a.
Find the probability of A.
P A 

n men
n employees
4
8
 0.5

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Prof. Juran
b.
Find the probability of B.
P B 


c.
2
8
 0.25
Find the probability of the intersection of A and B.
n brothers and men
P A  B  
n employees

d.
n brothers
n employees
2
8
 0.25
Find the probability of the union of A and B.
n
 men or brothers
P A  B 
n employees
4
8
 0.5

3. A department store manager has monitored the numbers of complaints received per
week about poor service. The probabilities for numbers of complaints in a week,
established by this review, are shown in the table. Let A be the event "There will be at
least one complaint in a week," and B the event "There will be less than 10 complaints in
a week."
NUMBER OF
COMPLAINTS
PROBABILITY
a.
1-3
4-6
7-9
10-12
.14
.39
.23
.15
.06
More than
12
.03
Find the probability of A.
P A 
b.
0
 P 1 to 3   4 to 6   7 to 9   10 to 12   more than 12 
 0.39  0.23  0.15  0.06  0.03
 0.86
Find the probability of B.
P B 
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 P 0   1 to 3   4 to 6   7 to 9 
 0.14  0.39  0.23  0.15
 0.91
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c.
Find the probability of the complement of A.
 
P A
 1  P A 
 1  0.86
 0.14
d.
Find the probability of the union of A and B.
P A  B 
e.
Find the probability of the intersection of A and B.
P A  B 
f.
 P 0   1 to 3   4 to 6   7 to 9   10 to 12   more than 12 
 0.14  0.39  0.23  0.15  0.06  0.03
 1.0
 P 1 to 3   4 to 6   7 to 9 
 0.39  0.23  0.15
 0.77
Are A and B mutually exclusive?
No. As we can see in (e), P A  B  0 . Mutually exclusive means they can't both
happen at the same time. If the probability of A intersect B is zero, then A and B are
mutually exclusive. If the probability of A intersect B is more than zero, then A and B
are not mutually exclusive.
g.
Are A and B collectively exhaustive?
Yes. As we can see in (d), P A  B  1.0 . Collectively exhaustive means that at least one
must happen. If the probability of A union B is 100%, then A and B are collectively
exhaustive. If the probability of A union B is anything less than 100%, then A and B are
not collectively exhaustive.
4. A local public-action group solicits donations by telephone. For a particular list of
prospects, it was estimated that for any individual, the probability was .05 of an
immediate donation by credit card, .25 of no immediate donation but a request for
further information through the mail, and .7 of no expression of interest. Mailed
information is sent to all people requesting it, and it is estimated that 20% of these
people will eventually donate. An operator makes a sequence of calls, the outcomes of
which can be assumed to be independent.
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Prof. Juran
a.
What is the probability that no immediate credit card donation will be received
until at least four unsuccessful calls have been made?
Note that there are three events that can result from a phone call:
Event
A
B
C
Definition
Immediate Donation
Request for Info
Not Interested
Probability
0.05
0.25
0.70
Assume that "unsuccessful" means anything other than an immediate donation, an
event symbolized as A .The probability that "no immediate credit card donation will be
received until at least four unsuccessful calls have been made" can be restated as the
probability that A happens on the four phone calls in a row.

P A four times in a row

       
P A P A P A P A
 1  P A  4
 0.95 4
 0.8145
b.
What is the probability that the first call leading to any donation (either
immediately or eventually after a mailing) is preceded by at least four unsuccessful
calls?
Note that there are two possible events that can result from a mailing:
Event
D
E
Definition
DonationMailing
No DonationMailing
Probability
0.20
0.80
Also, the probability of a donation (either immediately or eventually after a mailing) is
P Donation
  P A  D   0.05  0.200.25  0.10

P A  D four times in a row
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
 1  P A  D  4
 0.9 4
 0.6561
Prof. Juran
5. A mail-order firm considers three possible foul-ups in filling an order:
A:
B:
C:
The wrong item is sent.
The item is lost in transit.
The item is damaged in transit.
Assume that event A is independent of both B and C and that events B and C are
mutually exclusive. The individual event probabilities are P(A) = .02, P(B) = .01, and
P(C) = .04. Find the probability that at least one of these foul-ups occurs for a randomly
chosen order.
One way to solve this is to set up a diagram, in which the shaded area represents orders
with no foul-ups:
A
A
(Correct Item)
(Incorrect Item)
BC
(Lost)
0.0098
0.0002
0.0100
BC
(Damaged)
0.0392
0.0008
0.0400
BC
(Not Lost or Damaged)
0.9310
0.0190
0.9500
0.9800
0.0200
Using the diagram, the probability of an order with at least one foul-up is
1  0.931  0.069 . Alternatively, we can use our probability formulae:
P A  B  C

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

 
 
 P A  PB  PC   P A  B  C   P A  B  C  P A  B  C  P A  B  C
 0.0200  0.0100  0.0400  0.0000  0.0002  0.0008  0.0000
 0.0690
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
6. Market research in a particular city indicated that during a week 18% of all adults
watch a television program oriented to business and financial issues, 12% read a
publication oriented to these issues, and 10% do both.
a.
What is the probability that an adult in this city, who watches a television program
oriented to business and financial issues, reads a publication oriented to these
issues?
Let us define the following events:
Event
W
R
W R
Definition
Watches Television Program
Reads Publication
Both Watches and Reads
Probability
0.18
0.12
0.10
Now we can construct a diagram:
R
R
W
W
(Watches) (Doesn't Watch)
(Reads)
0.1
0.02
0.12
(Doesn't Read)
0.08
0.80
0.88
0.18
0.82
P R W
b.

P W  R 
P W 
0.10

0.18
 0.5556

What is the probability that an adult in this city, who reads a publication oriented
to business and financial issues, watches a television program oriented to these
issues?
P W R 
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P R W
P R 
0.10

0.12
 0.8333


Prof. Juran
7. An inspector examines items coming from an assembly line. A review of her record
reveals that she accepts only 8% of all defective items. It was also found that 1% of all
items from the assembly line are both defective and accepted by the inspector. What is
the probability that a randomly chosen item from this assembly line is defective?
Let us define the following events:
Event
AD
Definition
Accepts, given Defective
Probability
0.08
A D
Accepted and Defective
0.01
Bayes' Law: P A D  
P A  D 
P D 
Therefore:
P A D 
0.08
0.08P D 
P D 
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P A  D 
P D 
0.01

P D 
 0.01
0.01

0.08
 0.125

Prof. Juran
8. A bank classifies borrowers as high-risk or low-risk. Only 15% of its loans are made
to those in the high-risk category. Of all its loans, 5% are in default, and 40% of those in
default are to high-risk borrowers. What is the probability that a high-risk borrower will
default?
Let us define the following events:
Event
H
D
H D
Definition
High-Risk
In Default
High-Risk, given In Default
Probability
0.15
0.05
0.40
Using Bayes' Law:
P H D 
0.40
0.400.05 
0.02
P H  D 
P D 
P H  D 

0.05
 P H  D 
 P H  D 

This gives us enough information to construct a diagram:
D
D
H
H
(High-Risk) (Not High-Risk)
(In Default)
0.02
0.03
0.05
(Not In Default)
0.13
0.82
0.95
0.15
0.85
Now apply Bayes' Law again:
P D H
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
737
P D  H
P H 
0.02

0.15
 0.1333


Prof. Juran
9. A quality control manager found that 30% of worker-related problems occurred on
Mondays, and that 20% occurred in the last hour of a day's shift. It was also found that
4% of worker-related problems occurred in the last hour of Monday's shift.
a.
What is the probability that a worker-related problem that occurs on a Monday
does not occur in the last hour of the day's shift?
Let us define the following events:
Event
M
L
L
M L
M L
Definition
Problem on Monday
Problem in Last Hour
Problem Not in Last Hour
Problem on Monday in Last Hour
Problem on Monday Not in Last Hour
Using Bayes' Law:

P LM
b.


P M L
P M 
0.26

0.30
 0.867

Probability
0.30
0.20
0.80
0.04
0.26

Are the events "Problem occurs on Monday" and "Problem occurs in the last hour
of the day's shift" statistically independent?
If these events were statistically independent, then the probability of a Monday problem
being in the last hour (which is 1 - 0.867 = 0.133 in our case) would be the same as the
probability of being in the last hour on all days (which is 0.80). Since these two
probabilities are different, then "Problem occurs on Monday" and "Problem occurs in
the last hour of the day's shift" are not statistically independent.
Stated mathematically:
P L M 
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 P L 
Prof. Juran
10. A lawn care service makes telephone solicitations, seeking customers for the coming
season. A review of the records indicated that 15% of these solicitations produced new
customers, and that, of these new customers, 80% had used some rival service in the
previous year. It was also estimated that, of all solicitation calls made, 60% were to
people who had used a rival service the previous year. What is the probability that a
call to a person who used a rival service the previous year will produce a new customer
for the lawn care service?
Let us define the following events:
Event
N
RN
R
Definition
Solicited Person Becomes New Customer
Customer Used Rival, Given Customer is New
Used Rival
Probability
0.15
0.80
0.60
Using Bayes' Law:
P R N

0.80
0.12
P N  R 
P N 
P N  R 

0.15
 P N  R 

Now to solve the real question:
P N R 
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P N  R 
P R 
0.12

0.60
 0.20

Prof. Juran
11. A survey carried out for a supermarket classified customers according to whether
their visits to the store are frequent or infrequent and to whether they often, sometimes,
or never purchase generic products. The accompanying table gives the proportions of
people surveyed in each of the six joint classifications.
Frequency of Visit
Purchase of Generic Products
OFTEN SOMETIMES
NEVER
.12
.48
.19
.07
.06
.08
Frequent
Infrequent
a.
What is the probability that a customer is both a frequent shopper and often
purchases generic products?
P F  O  0.12
b.
What is the probability that a customer who never buys generic products visits the
store frequently?
P F  N 
P N 
0.19

0.19  0.08
 0.7037
P F N 
c.
Are the events "Never buys generic products" and "Visits the store frequently"
independent?
If they were independent, then P N   P N F  . However:
P N
P N F 

 0.19  0.08
 0.27
P N  F 
P F 
0.19

0.12  0.48  0.19
 0.2405

Therefore, they are not independent.
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Prof. Juran
d.
What is the probability that a customer who infrequently visits the store often buys
generic products?
  P OP FF 
POF 
0.07
0.07  0.06  0.08
 0.3333

e.
Are the events "Often buys generic products" and "Visits the store infrequently"
independent?
 
If they were independent, then P O   P O F . However:
P O 
 
POF
 0.12  0.07
 0.19


P O F
P F
0.07

0.07  0.06  0.08
 0.3333

 
Therefore, they are not independent.
f.
What is the probability that a customer frequently visits the store?
P F   0.12  0.48  0.19  0.79
g.
What is the probability that a customer never buys generic products?
P N   0.19  0.08  0.27
h.
What is the probability that a customer either frequently visits the store or never
buys generic products, or both?
P F  N   0.12  0.48  0.19  0.08  0.87
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Prof. Juran
12. An analyst attempting to predict a corporation's earnings next year believes that the
corporation's business is quite sensitive to the level of interest rates. She believes that if
average rates in the next year are more than 1% higher than this year, the probability of
significant earnings growth is 0.1. If average rates next year are more than 1% lower
than this year, the probability of significant earnings growth is estimated to be 0.8.
Finally, if average interest rates next year are within 1% of this year's rates, the
probability for significant earnings growth is put at 0.5. The analyst estimates that the
probability is 0.25 that rates next year will be more than 1% higher than this year, and
0.15 that they will be more than 1% lower than this year.
a.
What is the estimated probability that both interest rates will be more than 1%
higher and significant earnings growth will result?
Let us define the following events:
Event
H
M
L
G
GH
Definition
Rates more than 1% Higher
Rates within 1%
Rates more than 1% Lower
Significant Earnings Growth
Probability
0.25
0.60
0.15
0.10
GL
0.80
GM
0.50
We can rearrange Bayes' Law as follows:
P H  G 
b.
 P G H P H 
 0.100.25 
 0.025
What is the probability this corporation will experience significant earnings
growth?
P G 
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 P H  G   P M  G   P L  G 
 0.025  P G M P M   P G L P L 
 0.025  0.500.60   0.800.15
 0.025  0.30  0.12
 0.445
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Prof. Juran
c.
If the corporation exhibits significant earnings growth, what is the probability that
interest rates will have been more than 1% lower than in the current year?
P L  G 
P G 
0.12

0.445
 0.2697
P L G 

13. A manufacturer produces boxes of candy, each containing ten pieces. Two machines
are used for this purpose. After a large batch has been produced, it is discovered that
one of the machines, which produces 40% of the total output, has a fault that has led to
the introduction of an impurity into 10% of the pieces of candy it makes. From a single
box of candy, one piece is selected at random and tested. If that piece contains no
impurity, what is the probability that the box from which it came was produced by the
faulty machine?
Let’s define the following events:
Event
I
F
IF
Definition
Piece Contains Impurity
Piece came from Faulty Machine
 
P FI


P F I
P I

Probability
0.40
0.10

 
P F  I   P F  I 
P F P I F 

P F P I F   P F P I F 

P F P I F
0.400.90 
0.400.90   0.601.00 
0.36

0.36  0.60
 0.375

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Prof. Juran
Here’s how to do it with a 2x2 box:
Faulty Machine
0.04
0.36
0.40
Impurity
No Impurity
Not Faulty Machine
0.00
0.60
0.60
0.04
0.96
1. Get the 0.40 and 0.60 at the bottom of the two columns.


2. Put a zero for P I  F because the non-faulty machine produces no impurities.


3. That means P I  F has to be 0.60 (so the right column will add up).
4. 10% of the output from the faulty machine has impurities, so PI  F  must be
0.04.


5. That means P I  F has to be 0.36 (so the left column will add up).
6. The bottom row adds across to a total of 0.96.
7. The question is, if we are in the bottom row (no impurity), what is the chance we
are in the left column (faulty machine)?
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Prof. Juran
14. A student feels that 70% of his college courses have been enjoyable and the
remainder have been boring. He has access to student evaluations of professors and
finds that 60% of his enjoyable courses and 25% of his boring courses have been taught
by professors who had previously received strong positive evaluations from their
students. Next semester the student decides to take three courses, all from professors
who have received strongly positive student evaluations. Assume that his reactions to
the three courses are independent of one another.
a.
What is the probability that he will find all three courses enjoyable?
Let us define the following events:
Event
PV
E
PV E
Definition
Professor got Positive Evaluation
Student will find the Course Enjoyable
Probability
0.70
0.60
0.75
PV E
P E PV 


P E  PV 
P PV 
P PV E P E 

P PV  E   P PV  E
P PV E P E 


 
P PV E P E   P PV E P E
0.600.70 
0.600.70   0.250.30 
0.42

0.42  0.075
 0.8485

Managerial Statistics
745
Prof. Juran
Note that the number of courses the student finds enjoyable (call this number X) is a
binomially distributed random variable, with n = 3 and p = 0.8485.
n 
P X  x    p x 1  p n x
x 
 3
P X  3    p 3 1  p 0
 3


3!
0.8485 3 0.15150
 
 3! 3  3! 
 10.6108 1
 0.6108
b.
What is the probability that he will find at least one of the courses enjoyable?
P at least one is enjoyable  1  P none are enjoyable
 1  P X  0 



3!
0.8485 0 0.1515 3
 1  


0
!
3

0
!


 1  11.0035
 0.9965
15. In a large corporation, 80% of the employees are men and 20% are women. The
highest levels of education obtained by the employees are graduate training for 10% of
the men, undergraduate training for 30% of the men, and high school training for 60%
of the men. The highest levels of education obtained are also graduate training for 15%
of the women, undergraduate training for 40% of the women, and high school training
for 45% of the women.
Let us define the following events:
Event
Definition
Probability
Employee is Female
0.20
F
Employee has Graduate Training
G
Employee has Undergraduate Training
U
Employee has High School Training
H
0.15
G F
U F
0.40
H F
0.45
G F
0.10
U F
0.30
H F
0.60
Managerial Statistics
746
Prof. Juran
a.
What is the probability that a randomly chosen employee will be a man with only a
high school education?
 

 
P F H P H F P F
 0.600.80 
 0.48
b.
What is the probability that a randomly chosen employee will have graduate
training?
P G 

 P F  G   P F  G

  
 P G F P F   P G F P F
 0.150.20   0.100.80 
 0.03  0.08
 0.11
c.
What is the probability that a randomly chosen employee who has graduate
training is a man?

P FG




P F G
P G 

  
PGF P F
0.11
0.100.80 

0.11
 0.7273
d.
Are sex and level of education of employees in this corporation statistically
independent?
No, because:
0.08
P F G

Managerial Statistics

747
 0.088
 P F P G 
 
Prof. Juran
e.
What is the probability that a randomly chosen employee who has not had
graduate training is a woman?
 
P FG


P F G
PG
P F   P F  G 

1  P G 
0.20  P F P G F 

1  0.11
0.20  0.200.15 

0.89
0.17

0.89
 0.191

 
Note that all of these questions can be answered with a diagram:
Female
Male
Managerial Statistics
G
0.030
0.080
0.110
748
U
0.080
0.240
0.320
H
0.090
0.480
0.570
0.200
0.800
Prof. Juran
16. A large corporation organized a ballot for all its workers on a new bonus plan. It
was found that 65% of all night-shift workers favored the plan and that 40% of all
women workers favored the plan. Also, 50% of all employees are night-shift workers,
and 30% of all employees are women. Finally, 20% of the night-shift workers are
women.
Let us define the following events:
Event
Definition
Employee is Female
F
Employee Favors Bonus Plan
B
Employee Works the Night Shift
N
BN
a.
0.40
FN
0.20
What is the probability that a randomly chosen employee is a woman in favor of
the plan?
 P F P B F 
 0.300.40 
 0.12
What is the probability that a randomly chosen employee is either a woman or a
night-shift worker (or both)?
P F  N
c.
0.50
0.65
BF
P F  B 
b.
Probability
0.30

 PF   PN   PF  N 
 0.30  0.50  0.500.20 
 0.70
Is employee sex independent of whether the night-shift is worked?
No because P F   P F N  .
d.
What is the probability that a woman employee is a night-shift worker?
P N F 


P N  F 
P F 
P F N P N

P F 
0.200.50 

0.30
 0.3333
Managerial Statistics
749
Prof. Juran
e.
If 50% of all male employees favor the plan, what is the probability that a randomly
chosen employee both does not work the night-shift and does not favor the plan?

P N B

 1  P N  B 
 1  P N   P B   P N  B 


   
 1  0.50  P B F P F   P B F P F  P B N P N
 1  0.50  0.400.30   0.500.70   0.650.50 
 1  0.50  0.47   0.325
 0.355

17. Subscriptions to American History Illustrated are classified as gift, previous renewal,
direct mail, or subscription service. In January 1979, 8% of expiring subscriptions were
gift; 41%, previous renewal; 6%, direct mail; and 45% subscription service. The
percentages of renewals in these four categories were 81%, 79%, 60%, and 21%,
respectively. In February 1979, 10% of expiring subscriptions were gift; 57%, previous
renewal; 24%, direct mail; and 9% subscription service. The percentages of renewals
were 80%, 76%, 51%, and 14%, respectively.
a.
Find the probability that a randomly chosen subscription expiring in January 1979
was renewed.
For January:
P Re newal 
 PR GPG  PR EPE  PR DPD  PR SPS
 0.810.08   0.790.41  0.600.06   0.210.45 
 0.5192
b.
Find the probability that a randomly chosen subscription expiring in February 1979
was renewed.
For February:
P Re newal 
 PR GPG  PR EPE  PR DPD  PR SPS
 0.800.10   0.760.57   0.510.24   0.140.09 
 0.6482
Managerial Statistics
750
Prof. Juran
c.
Verify that the probability in part (b) is higher than that in part (a). Do you believe
that the editors of American History Illustrated should view the change from January
to February as a positive or negative development?
This is not a positive development, even though the probability of a renewal increased.
The increase is due to a change in the mix of expiring subscriptions. A larger share of
gift and previous renewal subscriptions expired; these types of subscriptions have a
higher probability of being renewed than direct mail or subscription service. Note that,
for each category, the probability of a renewal went down from January to February.
18. The accompanying table shows, for 1,000 forecasts of earnings per share made by
financial analysts, the numbers of forecasts and outcomes in particular categories
(compared with the previous year).
Outcome
Improvement
About the Same
Worse
a.
Forecast
About the Same
82
153
84
Improvement
210
106
75
Worse
66
75
149
Find the probability that if the forecast is for a worse performance in earnings, this
outcome will result.
PW FW 

PW  FW 
PFW 

0.149
0.29
 0.5138
b.
If the forecast is for an improvement in earnings, find the probability that this
outcome fails to result.
 
P I FI


P I  FI
PFI 

0.181
0.391

 0.4629
Managerial Statistics
751
Prof. Juran
19. A corporation produces packages of paper clips. The number of clips per package
varies, as indicated in the accompanying table.
NUMBER OF CLIPS
PROPORTION OF PACKAGES
a.
47
.04
48
.13
49
.21
50
.29
51
.20
52
.10
53
.03
Draw the probability function.
Probability Function
0.350
0.300
Proportion
0.250
0.200
0.150
0.100
0.050
0.000
47
48
49
50
51
52
53
Number of Clips per Box
b.
Calculate and draw the cumulative probability function.
Cumulative Probability Function
1.000
0.900
0.800
Cum. Probability
0.700
0.600
0.500
0.400
0.300
0.200
0.100
0.000
47
48
49
50
51
52
53
Number of Clips per Box
Managerial Statistics
752
Prof. Juran
c.
What is the probability that a randomly chosen package will contain between 49
and 51 clips (inclusive)?
P 49  X  51
d.
 PX  49   PX  50   PX  51
 0.21  0.29  0.20
 0.70
Two packages are chosen at random. What is the probability that at least one of
them contains at least fifty clips?
P at least one  50 
 1  P none  50 
 1  PX  502
 1  0.04  0.13  0.212
 0.8556
20. Refer to the information in Exercise 19.
a.
Find the mean and standard deviation of the number of paper clips per package.
Let’s use Excel:
A
B
C
D
E
F
1 Calculations for Expected Value
2 NUMBER OF CLIPS PROPORTION OF PACKAGES
=A3*B3
3
47
0.04
1.88
4
48
0.13
6.24
5
49
0.21
10.29
6
50
0.29
14.5
7
51
0.2
10.2
8
52
0.1
5.2
9
53
0.03
1.59
=SUM(C3:C9)
10
49.9
11
12
13 Calculations for Variance
Error
Error^2
Prob.
D*E
14
47
49.9
-2.9
8.41
0.04
0.3364
=C15*C15
15
48
49.9
-1.9
3.61
0.13
0.4693
16
49
49.9
-0.9
0.81
0.21
0.1701
17
50
49.9
0.1
0.01
0.29
0.0029
18
51
49.9
1.1
1.21
0.2
0.242
19
52
49.9
2.1
4.41
0.1
0.441
20
53
49.9
3.1
9.61
0.03
0.2883
21
1.95
22
23
1.396424
Managerial Statistics
753
G
H
I
Expected Value
=E17*D17
=SUM(F14:F20)
=SQRT(F21)
Variance
Std. Dev.
Prof. Juran
b.
The cost (in cents) of producing a package of clips is 16 + 2X, where X is the
number of clips in the package. The revenue from selling the package, however
many clips it contains, is $1.50. If profit is defined as the difference between
revenue and cost, find the mean and standard deviation of profit per package.
Recall the Linear Transformation Rule: If X is a random variable, and Y is a random
variable such that in all cases Y = aX + b (for any numbers a and b). Then,
Expected Value
Variance
Standard Deviation
E(Y) = aE(X) + b,
Var(Y) = a2 Var(X),
(Y) = |a|(X).
In this case,
Profit
 $0.02# of clips  $1.50  0.16
EPr ofit 
 0.02Eclips  1.34
 0.0249.9   1.34
 0.342
 profit
  0.02 1.3964 
 0.0279
Managerial Statistics
754
Prof. Juran
21. A college basketball player, who sinks 75% of his free throws, comes to the line to
shoot a "one and one" (if the first shot is successful, he is allowed a second shot, but no
second shot is taken if the first is missed; one point is scored for each successful shot).
Assume that the outcome of the second shot, if any, is independent of that of the first.
Find the expected number of points resulting from the "one and one." Compare this
with the expected number of points from a "two-shot foul," where a second shot is
allowed irrespective of the outcome of the first.
First the “one and one”:
E X 
 0Pmiss 1st   1Pmake 1st, miss 2nd   2Pmake 2 shots in a row 
 00.25  10.750.25  20.750.75
 0  0.1875  1.125
 1.3125
Now, the “two-shot foul”:
EX 
 0Pmiss 2   1Pmake 1st, miss 2nd   1Pmiss 1st, make 2nd   2Pmake 2 
 00.250.25  10.750.25  10.250.75  20.750.75
 0  0.1875  0.1875  1.125
 1.5
Managerial Statistics
755
Prof. Juran
22. A store owner stocks an out-of-town newspaper, which is sometimes requested by a
small number of customers. Each copy of this newspaper costs him 70 cents, and he
sells them for 90 cents each. Any copies left over at the end of the day have no value
and are destroyed. Any requests for copies that cannot be met because stocks have been
exhausted are considered by the store owner as a loss of 5 cents in goodwill. The
probability distribution of the number of requests for the newspaper in a day is shown
in the accompanying table. If the store owner defines total daily profit as total revenue
from newspaper sales, less total cost of newspapers ordered, less goodwill loss from
unsatisfied demand, how many copies per day should he order to maximize expected
profit?
NUMBER OF REQUESTS
PROBABILITY
0
.12
1
.16
2
.18
3
.32
4
.14
5
.08
Let R be the number of requests (a random variable), and S be the number of papers
stocked (a decision variable). Using Excel, we can create the following contingency
table, in which the profit is shown for all possible combinations of S and R.
R=
PROBABILITY
S=0
1
2
3
4
5
0
0.12
$0.00
($0.70)
($1.40)
($2.10)
($2.80)
($3.50)
1
0.16
($0.05)
$0.20
($0.50)
($1.20)
($1.90)
($2.60)
2
0.18
($0.10)
$0.15
$0.40
($0.30)
($1.00)
($1.70)
3
0.32
($0.15)
$0.10
$0.35
$0.60
($0.10)
($0.80)
4
0.14
($0.20)
$0.05
$0.30
$0.55
$0.80
$0.10
5
0.08
($0.25)
$0.00
$0.25
$0.50
$0.75
$1.00
Now, by adding up all of the possible profit levels for each value of S, weighted by their
probabilities, we can determine the expected profit for each option. For example, the
expected profit assuming we order zero papers is:
E X S  0 
 0.120.00   0.16 0.05  0.18 0.10   0.32 0.15  0.14 0.20   0.08 0.25
 0.00  0.008  0.018  0.048  0.028  0.020
 0.122
Performing the same calculations for all values of S, we can see that the best choice is to
order one newspaper:
S
0
1
2
3
4
5
Managerial Statistics
756
E(X)
($0.122)
$0.014
($0.002)
($0.189)
($0.680)
($1.304)
Prof. Juran
23. An investor is considering three strategies for a $1,000 investment. The probable
returns are estimated as follows:
Strategy 1: A profit of $10,000 with probability 0.15 and a loss of $1,000 with probability
0.85.
Strategy 2: A profit of $1,000 with probability 0.50, a profit of $500 with probability 0.30,
and a loss of $500 with probability 0.20.
Strategy 3: A certain profit of $400.
Which strategy has the highest expected profit? Would you necessarily advise the investor
to adopt this strategy?
E X 1 
 10 , 0000.15  1 , 0000.85
E X 2 
 650
 1 , 0000.50   5000.30   5000.20 
E X 3 
 550
 400
Strategy 1 has the highest expected value, but is also the riskiest. Therefore, it might not
necessarily be the “best” strategy for everyone.
Managerial Statistics
757
Prof. Juran
24. The accompanying table shows, for credit card holders with one to three cards, the
joint probabilities for number of cards owned (X) and number of credit purchases made
in a week (Y).
Number of Cards (X)
1
2
3
a.
For a randomly chosen person from this group, what is the probability function for
number of purchases made in the week?
P Y
P Y
P Y
P Y
P Y
b.
Number of Purchases in Week (Y)
0
1
2
3
4
0.08
0.13
0.09
0.06
0.03
0.03
0.08
0.08
0.09
0.07
0.01
0.03
0.06
0.08
0.08
 0
 1
 2
 3
 4
 0.08  0.03  0.01
 0.13  0.08  0.03
 0.09  0.08  0.06
 0.06  0.09  0.08
 0.03  0.07  0.08
 0.12
 0.24
 0.23
 0.23
 0.18
For a person in this group who has three cards, what is the probability function for
number of purchases made in the week?
Note that the probability that a person has 3 cards is:
P X  3 
 0.01  0.03  0.06  0.08  0.08
 0.26
Now, using the conditional probability formula (Bayes' Law):
P Y  0 X  3
P Y  1 X  3
P Y  2 X  3
P Y  3 X  3
P Y  4 X  3
c.
P Y  0  X  3
P X  3 
P Y  1  X  3

P X  3 
P Y  2  X  3

P X  3 
P Y  3  X  3

P X  3 
P Y  4  X  3

P X  3 






0.01
0.26
0.03
0.26
0.06
0.26
0.08
0.26
0.08
0.26
 0.0385
 0.1154
 0.2308
 0.3077
 0.3077
Are the number of cards owned and number of purchases made statistically
independent?
No, as we can see by comparing the extreme right columns of the two tables shown in
(a) and (b) above. Knowing X has an effect on our predictions about Y.
Managerial Statistics
758
Prof. Juran
25. A market researcher wants to determine whether a new model of a personal
computer, which had been advertised on a late-night talk show, had achieved more
brand name recognition among people who watched the show regularly than among
people who did not. After conducting a survey, it was found that 15% of all people both
watched the show regularly and could correctly identify the product. Also, 16% of all
people regularly watched the show and 45% of all people could correctly identify the
product. Define a pair of random variables as follows:
X
Y
a.
1

0
1

0
if regularly watch the show
otherwise
if product correctly identified
otherwise
Find the joint probability function of X and Y
X
Y
0
1
Total
b.
0
0.54
0.30
0.84
1
0.01
0.15
0.16
Total
0.55
0.45
1.00
Find the conditional probability function of Y, given X = 1.
P Y  0 X  1
P Y  1 X  1
Managerial Statistics
P Y  0  X  1
P X  1
P Y  1  X  1

P X  1

759
0.01
0.16
0.15

0.16

 0.0625
 0.9375
Prof. Juran
26. A production manager knows that 5% of components produced by a particular
manufacturing process have some defect. Six of these components, whose
characteristics can be assumed to be independent of each other, were examined.
a.
What is the probability that none of these components has a defect?
P X  0 
b.
What is the probability that one of these components has a defect?
P X  1
c.
6
  0.05 0 0.95 6
0
 110.7351
 0.7351
6
  0.05 1 0.95 5
 1
 60.050.7738
 0.2321
What is the probability that at least two of these components have a defect?
P X  2 
Managerial Statistics
 1  P X  1
 1  P X  1  P X  0 
 1  0.2321  0.7351
 0.0328
760
Prof. Juran
27. Suppose that the probability is .5 that the value of the U.S. dollar will rise against the
Japanese yen over any given week, and that the outcome in one week is independent of
that in any other week. What is the probability that the value of the U.S. dollar will rise
against the Japanese yen in a majority of weeks over a period of 7 weeks?
Let X be the number of weeks when the dollar rises against the yen. X is binomially
distributed with n = 7 and p = 0.5.
P X  4 
 P X  4   P X  5   P X  6   P X  7 
 7 
  7 
  7 
  7 

  0.5 4 0.5 3    0.5 5 0.5 2    0.56 0.51    0.57 0.50 
 4 
  5 
  6 
  7 

 350.06250.125  210.03130.25  7 0.0156 0.5  10.00781.0 
 0.2734  0.1641  0.0547  0.0078
 0.5
28. A company installs new central heating furnaces, and has found that for 15% of all
installations a return visit is needed to make some modifications. Six installations were
made in a particular week. Assume independence of outcomes for these installations.
a.
What is the probability that a return visit was needed in all of these cases?
P X  6 
b.
What is the probability that a return visit was needed in none of these cases?
P X  0 
c.
6
  0.15 6 0.85 0
6
 10.0000114 1
 0.0000114
6
  0.150 0.856
0
 110.3771
 0.3771
What is the probability that a return visit was needed in more than one of these
cases?
P X  1
 1  P X  0   P X  1


6
 1  0.3771   0.151 0.85 5 
 1


 1  0.3771  6 0.15.4437 
 1  0.3771  0.3993
 0.2236
Managerial Statistics
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Prof. Juran
29. A small commuter airline flies planes that can seat up to eight passengers. The
airline has determined that the probability that a ticketed passenger will not show up
for a flight is 0.2. For each flight, the airline sells tickets to the first ten people placing
orders. The probability distribution for the number of tickets sold per flight is shown in
the accompanying table. For what proportion of the airline's flights does the number of
ticketed passengers showing up exceed the number of available seats? (Assume
independence between number of tickets sold and the probability that a ticketed
passenger will show up.)
NUMBER OF TICKETS
PROBABILITY
6
0.25
7
0.35
8
0.25
9
0.10
10
.05
Let X be the number of reservations, and Y the number who actually show up.
Poverbooking 
 PX  9PY  8 given X  9  PX  10PY  8 given X  10
 9 

 10 

 10 
 0.10 0.89 0.2 0   0.05 0.89 0.2 1   0.810 0.2 0 
 10 
 9 

 9 

 0.1010.1342 1  0.0510 0.1342 0.2   10.1074 1
 0.01342  0.050.2684  0.1074 
 0.0322
30. An automobile dealer mounts a new promotional campaign, in which it is promised
that purchasers of new automobiles may, if dissatisfied for any reason, return them
within two days of purchase and receive a full refund. It is estimated that the cost to the
dealer of such a refund is $250. The dealer estimates that 15% of all purchasers will
indeed return automobiles and obtain refunds. Suppose that fifty automobiles are
purchased during the campaign period.
a.
Find the mean and standard deviation of the number of these automobiles that will
be returned for refunds.
X
X
 np
 500.15 
 7.5
 n  p 1  p 
 500.150.85
 2.525
b.
Find the mean and standard deviation of the total refund costs that will accrue as a
result of these fifty purchases.
Managerial Statistics
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Prof. Juran
$
$
 $2507.5 
 $1 , 875.00
 $250 2.525
 $631.23
31. A company receives a very large shipment of components. A random sample of
sixteen of these components is checked, and the shipment is accepted if fewer than two
of these components are defective. What is the probability of accepting a shipment
containing:
a.
5% defectives?
Note that we accept a shipment if it has either zero or one defective components. The
probability of accepting with 5% defective is:
P accept
b.
P X  0   P X  1
 16 
 16 
  0.050 0.95 16   0.051 0.95 15
0
1
 110.4401  160.050.4633
 0.4401  0.3706
 0.8107

P X  0   P X  1
 16 
 16 
  0.150 0.8516   0.151 0.8515
0
1
 110.0743  160.150.0874 
 0.0743  0.2098
 0.2841

P X  0   P X  1
 16 
 16 
  0.250 0.7516   0.251 0.7515
0
1
 110.0100   160.250.0134 
 0.0100  0.0536
 0.0636
15% defectives?
P accept
c.

25% defectives?
P accept
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Prof. Juran
32. In the past several years, credit card companies have made an aggressive effort to
solicit new accounts from college students. Suppose that a sample of 200 students at
your college indicated the following information as to whether the student possessed a
bank credit card and/or a travel and entertainment credit card:
Travel and Entertainment
Credit Card
Yes
No
60
60
15
65
Bank Credit Card
Yes
No
(a)
Give an example of a simple event.
Let us define these events:
Event
Student has a Bank Credit card
Student does not have a Bank Credit card
Student has a Travel and Entertainment card
Student does not have a Travel and Entertainment card
Symbol
B
B
T
T
There are four simple events:

Student has a Bank Credit card ( B )

Student does not have a Bank Credit card ( B )

Student has a Travel and Entertainment card ( T )

Student does not have a Travel and Entertainment card ( T )
(b)
Give an example of a joint event.
There are four joint events:

Student has a Bank Credit card and has a Travel and Entertainment card ( B  T )

Student has a Bank Credit card and does not have a Travel and Entertainment
card ( B T )

Student does not have a Bank Credit card and has a Travel and Entertainment
card ( B T )

Student does not have a Bank Credit card and does not have a Travel and
Entertainment card ( B T )
Managerial Statistics
764
Prof. Juran
(c)
What is the complement of having a bank credit card?
It is the event of NOT having a bank credit card ( B ).
(d)
Why is "having a bank credit card and having a travel and entertainment credit
card" a joint event?
It is a joint event because it requires two events to both be true.
Here is a probability table, for answering the rest of the questions:
T
T
B
0.300
0.300
0.600
B
0.075
0.325
0.400
0.375
0.625
If a student is selected at random, what is the probability that
(e)
the student has a bank credit card?
P B 
(f)
 0.6
the student has a travel and entertainment credit card?
P T 
(g)
 0.375
the student has a bank credit card and a travel and entertainment card?
P B  T 
(h)
the student has neither a bank credit card nor a travel and entertainment card?

P B T
(i)
 0.3

 0.325
the student has a bank credit card or has a travel and entertainment card?
P B  T 
 PB  PT   PB  T 
 0.6  0.375  0.3
 0.675
(j)
the student does not have a bank credit card or has a travel and entertainment
card?

P B T
Managerial Statistics



 P B  PT   P B  T
765

Prof. Juran
 0.4  0.375  0.075
 0 .7
33. A company has made available to its employees (without charge) extensive health
club facilities that may be used before work, during the lunch hour, after work, and on
weekends. Records for the last year indicate that of 250 employees, 110 used the
facilities at some time. Of 170 males employed by the company, 65 used the facilities.
(a)
Set up a 2 x 2 table to evaluate the probabilities of using the facilities.
Event
Employee Used Facility
Employee did not Use Facility
Employee is Male
Employee is Not Male
(b)
Symbol
U
U
M
M
Give an example of a simple event.
There are four simple events:
(c)

Employee Used Facility ( U )

Employee did not Use Facility ( U )

Employee is Male ( M )

Employee is Not Male ( M )
Give an example of a joint event.
There are four joint events:
(d)

Employee Used Facility and is Male ( U  M )

Employee Used Facility and is Not Male ( U  M )

Employee did not Use Facility and is Male ( U  M )

Employee did not Use Facility and is Not Male ( U  M )
What is the complement of "used the health club facilities"?
It is the event "did not use the facilities" ( U ).
Here is a probability table for answering the rest of the questions:
Managerial Statistics
766
Prof. Juran
M
M
45/250 = 0.180 110/250 = 0.440
U
65/250 = 0.260
U
105/250 = 0.420 35/250 = 0.140 140/250 = 0.560
170/250 = 0.680 80/250 = 0.320
1.000
What is the probability that an employee chosen at random
(e)
is a male?
P  M  0.68
(f)
has used the health club facilities?
PU  0.44
(g)
is a female and has used the health club facilities?


P M  U  0.18
(h)
is a female and has not used the health club facilities?


P M  U  0.14
(i)
is a female or has used the health club facilities?

  

P M  U  P M  PU   P M  U

 0.32  0.44  0.18
 0.58
(j)
is a male or has not used the health club facilities?


  
P M  U  P M   P U  P M  U

 0.68  0.56  0.42
 0.82
(k)
has used the health club facilities or has not used the health club facilities?

  

P U  U  PU   P U  P U  U

 0.44  0.56  0.00
 1.00
Managerial Statistics
767
Prof. Juran
34. Each year, ratings are compiled concerning the performance of new cars during the
first 90 days of use. Suppose that the cars have been categorized according to two
attributes, whether or not the car needs warranty-related repair (yes or no) and the
country in which the company manufacturing the car is based (United States, not
United States). Based on the data collected, the probability that the new car needs a
warranty repair is .04, the probability that the car is manufactured by an Americanbased company is .60, and the probability that the new car needs a warranty repair and
was manufactured by an American-based company is .025.
(a)
Set up a 2 x 2 table to evaluate the probabilities of a warranty-related repair.
Event
Car Needs Warranty-related Repair
Car Does Not Need Warranty-related Repair
USA Manufacturer
Non-USA Manufacturer
W
(b)
Symbol
W
W
U
U
U
0.025
W
0.575
U
0.015
0.385
0.4
0.04
0.96
1
0.6
Give an example of a simple event.
There are four simple events:




(c)
Car Needs Warranty-related Repair ( W )
Car Does Not Need Warranty-related Repair ( W )
USA Manufacturer ( U )
Non-USA Manufacturer ( U )
Give an example of a joint event.
There are four joint events:




(d)
Car Needs Warranty-related Repair and USA Manufacturer ( W  U )
Car Does Not Need Warranty-related Repair and USA Manufacturer
( W U )
Car Needs Warranty-related Repair and Non-USA Manufacturer ( W  U )
Car Does Not Need Warranty-related Repair and Non-USA Manufacturer
( W U )
What is the complement of "manufactured by an American-based company"?
It is the event that there is a Non-USA Manufacturer ( U )
Managerial Statistics
768
Prof. Juran
What is the probability that a new car selected at random
(e)
needs a warranty-related repair?
P W  0.04
(f)
is not manufactured by an American-based company?
 
P U  0.4
(g)
needs a warranty repair and is manufactured by a company based in the United
States?
PW  U  0.025
(h)
does not need a warranty repair and is not manufactured by a company based in
the United States?


P W  U  0.385
(i)
needs a warranty repair or was manufactured by an American-based company?
PW  U  PW   PU   PW  U 
 0.04  0.6  0.025
 0.615
(j)
needs a warranty repair or was not manufactured by an American-based
company?


  
P W  U  PW   P U  P W  U

 0.04  0.4  0.015
 0.425
(k)
needs a warranty repair or does not need a warranty repair?

  

P W  W  PW   P W  P W  W

 0.04  0.96  0.000
 1.000
Managerial Statistics
769
Prof. Juran
35. Recall the following data from a sample of 200 students in Problem 32 above.
Travel and Entertainment
Credit Card
Yes
No
60
60
15
65
Bank Credit Card
Yes
No
(a)
Assume we know that the student has a bank credit card. What is the
probability, then, that he or she has a travel and entertainment card?
Probabilities:
T
T
B
0.300
0.300
0.600
B
0.075
0.325
0.400
0.375
0.625
Bayes' Law:
PT B
(b)
Assume that we know that the student does not have a travel and entertainment
card. What, then, is the probability that he or she has a bank credit card?
 
P BT
(c)
PT  B
P  B
0 .3

0 .6
 0.5


P BT
PT
0 .3

0.625
 0.48



Are the two events, having a bank credit card and having a travel and
entertainment card, statistically independent? Explain.
If these two events were independent, then knowledge about one of them would not
affect our expectation about the other.
For example, if we are ignorant about whether or not someone has a bank credit card,
we expect that there is a 0.375 probability that they have a travel and entertainment
card. If the two events were independent, then the probability that a person has a travel
and entertainment card, given that we know the person has a bank credit card, should
also be 0.375.
Managerial Statistics
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Prof. Juran
In this case, the answer to (a) above (in which we found that the probability that a
person has a travel and entertainment card, given that we know the person has a bank
credit card, is really 0.5) reveals that the two events are not independent. Specifically,
people with bank credit cards are more likely to have travel and entertainment cards
than other people.
36. Use the data from Problem 33 above (in which a company has made health club
facilities available to its employees) to answer the following:
(a)
Suppose that we select a female employee of the company. What, then, is the
probability that she has used the health club facilities?
Probabilities:
M
U
M
0.260 0.180 0.440
U
0.420 0.140 0.560
0.680 0.320
Bayes' Law:

PUM
Managerial Statistics

771

PUM
PM
0.18

0.32
 0.5625

 

Prof. Juran
(b)
Suppose that we select a male employee of the company. What, then, is the
probability that he has not used the health club facilities?

PUM
(c)


PUM
P M 
0.42

0.68
 0.618


Are the gender of the individual and the use of the health club facilities
statistically independent? Explain.


 
No. We know that P U M  0.618  P U  0.56 . If the two events were independent,
these probabilities would be the same. It turns out that men are less likely than women
to use the health club.
37. Use the new car performance ratings data from Problem 34 above to answer the
following:
(a)
Suppose we know that the car was manufactured by a company based in the
United States. What. then, is the probability that the car needs a warranty repair?
Probabilities:
W
U
0.025
W
0.575
U
0.015
0.385
0.4
0.04
0.96
1
0.6
Bayes' Law:
PW U 
Managerial Statistics
772
PW  U 
PU 
0.025

0 .6
 0.042

Prof. Juran
(b)
Suppose we know that the car was not manufactured by a company based in the
United States. What, then, is the probability that the car needs a warranty repair?

PWU
(c)


P W U
PU
0.015

0 .4
 0.038

 

Are need for a warranty repair and location of the company manufacturing the
car statistically independent?
No. American companies' cars are more likely to need a warranty repair.
38. A standard deck of cards is being used to play a game. There are four suits (hearts,
diamonds, clubs, and spades), each having 13 faces (ace. 2, 3, 4, 5, 6, 7, 8, 9, 10, jack,
queen, and king), making a total of 52 cards. This complete deck is thoroughly mixed,
and you will receive the first two cards from the deck without replacement.
(a)
What is the probability that both cards are queens?
Let Q1 be the event that you get a queen on the first card, and Q2 be the event that you
get a queen on the second card, given Q1. On the first card, you have four chances to get
a queen out of 52 cards. On the second card (assuming you get a queen on the first card)
there are three remaining chances out of 51 cards.
PQ1  Q2 
 PQ1  PQ2 Q1

4
3
*
52 51
 0.004525
Managerial Statistics
773
Prof. Juran
(b)
What is the probability that the first card is a 10 and the second card is a 5 or 6?
On the first card, you have four chances to get a 10 out of 52 cards. On the second card
(assuming you get a 10 on the first card) there are eight remaining chances to get a 5 or
6 out of 51 cards.
P10 first   5 or 6 second 10 first 

4
8

52 51
 0.012066
(c)
If we were sampling with replacement, what would be the answer in (a)?
Now, the second draw would consist of 52 cards, four of which are queens so
4
PQ1  PQ 2  
.
52
PQ1  Q2 
 PQ1  PQ2 Q1
 PQ1  PQ 2 
 4 
 
 52 
2
 0.005917
(Drawing two queens is more likely with replacement than without replacement.)
(d)
In the game of Blackjack, the picture cards (jack, queen, king) count as 10 points
and the ace counts as either 1 or 11 points. All other cards are counted at their
face value. Blackjack is achieved if your two cards total 21 points. What is the
probability of getting blackjack with two cards?
Let's examine the possibilities for getting Blackjack in two cards:

It is impossible to get Blackjack in two cards if the first card drawn is neither a 10, a
Picture, or an Ace (32 out of 52).

If the first card drawn is either a 10 or a Picture (16 out of 52), then you can get
Blackjack if the second card is an Ace (4 out of 51).

If the first card drawn is an Ace (4 out of 52), then you can get Blackjack if the
second card is either a 10 or a Picture (16 out of 51).
Managerial Statistics
774
Prof. Juran
Events:
Event
Ace on First Card
Ace on Second Card
10 or Picture on First Card
10 or Picture on Second Card
PBlackjack in two cards
Symbol
A1
A2
P1
P2
 PP1  A2  P A1  P2
4   4
16 
 16
           
 52   51   52   51 
 0.0241  0.0241
 0.0483
39. Suppose that the probability that a person has a certain disease is 0.03. Medical
diagnostic tests are available to determine whether the person actually has the disease.
If the disease is actually present, the probability that the medical diagnostic test will
give a positive result (indicating that the disease is present) is 0.90. If the disease is not
actually present, the probability of a positive test result (indicating that the disease is
present) is 0.02. Given this information, we would like to know the following:
(a)
If the medical diagnostic test yields a positive result (indicating that the disease is
present), what is the probability that the disease is actually present?
Events
Disease
No Disease
Test Positive
Test Not Positive
Test Positive, Given Disease
Test Positive, Given No Disease
Managerial Statistics
775
Symbol
D
D
T
T
TD
TD
Probability Given
0.03
0.90
0.02
Prof. Juran
Probability Table:
T
T
D
D
0.0270 0.0194 0.0464
0.0030 0.9506 0.9536
0.0300 0.9700
Note: Here is how these probabilities can be calculated in an Excel spreadsheet:
1
A
B
Events
Probability Given
0.030
0.900
0.020
Disease
Test Positive, Given Disease
4 Test Positive, Given No Disease
2
3
C
=C11*B3
5
6
7
D
E
F
G
=D11*B4
=C11-C9
=SUM(C9:D9)
D
T 0.0270
Not T 0.0030
0.0300
8
9
10
11
12
13
14
15
16
Not D
0.0194 0.0464
0.9506 0.9536
0.9700
=1-E9
=1-C11
=B2
=D11-D9
The question for Part (a) is "If the medical diagnostic test yields a positive result, what is
the probability that the disease is actually present?" In other words, what is the
probability of D given T?
PD T 

PD  T 
PT 

0.027
0.0464
 0.582
(b)
If the medical diagnostic test has given a negative result (indicating the disease is
not present), what is the probability that the disease is not present?
 
P DT


P DT
PT

0.9506
0.9536


 0.997
Managerial Statistics
776
Prof. Juran
40. The Olive Construction Company is determining whether it should submit a bid for
a new shopping center. In the past, Olive's main competitor, Base Construction
Company, has submitted bids 70% of the time. If Base Construction Company does not
bid on a job, the probability that the Olive Construction Company will get the job is .50.
If Base Construction Company does bid on a job, the probability that the Olive
Construction Company will get the job is .25.
(a)
If the Olive Construction Company gets the job, what is the probability that the
Base Construction Company did not bid?
Events
Base Bids
Base Does Not Bid
Olive Wins Job
Olive Does Not Win Job
Olive Wins, Given Base Does Not Bid
Symbol
B
B
O
O
OB
Probability Given
0.70
OB
0.25
Olive Wins, Given Base Bids
0.50
Probability Table:
O
O
B
0.175
0.525
0.700
 
P BO
(b)
B
0.150
0.150
0.300

0.325
0.675
1
P BO
PO 
0.15

0.325
 0.4615


What is the probability that the Olive Construction Company will get the job?
PO 


 PO  B  P O  B
 
 PO BPB  P O B P B
 0.250.70   0.500.3
 0.175  0.15
 0.325
Managerial Statistics
777
Prof. Juran
41. A municipal bond service has three rating categories (A, B, and C). Suppose that in
the past year, of the municipal bonds issued throughout the United States, 70% were
rated A, 20% were rated B, and 10% were rated C. Of the municipal bonds rated A, 50%
were issued by cities, 40% by suburbs, and 10% by rural areas. Of the municipal bonds
rated B, 60% were issued by cities, 20% by suburbs, and 20% by rural areas. Of the
municipal bonds rated C, 90% were issued by cities, 5% by suburbs, and 5% by rural
areas.
(a)
If a new municipal bond is to be issued by a city, what is the probability it will
receive an A rating?
First, organize what we know:
Event
City Issuer
Suburban Issuer
Rural Issuer
A Rating
B Rating
C Rating
U given A
S given A
R given A
U given B
S given B
R given B
U given C
S given C
R given C
Symbol
U
S
R
A
B
C
UA
SA
RA
UB
SB
Probability Given
0.70
0.20
0.10
0.50
0.40
0.10
0.60
0.20
0.20
0.90
0.05
0.05
RB
UC
SC
RC
Using Bayes' Law, we can fill in the following table:
Rating
Issuer
A
0.50*0.70 =
0.35
B
0.60*0.20 =
0.12
C
0.90*0.10 =
0.09
0.35 + 0.12 + 0.09 =
0.56
S
0.40*0.70 =
0.28
0.20*0.20 =
0.04
0.05*0.10 =
0.005
0.28 + 0.04 + 0.005 =
0.325
R
0.10*0.70 =
0.07
0.20*0.20 =
0.04
0.05*0.10 =
0.005
0.07 + 0.04 + 0.005 =
0.115
0.70
0.20
0.10
U
Managerial Statistics
778
Prof. Juran
Now, if a new municipal bond is to be issued by a city, what is the probability it will
receive an A rating?
This is asking for the probability of A , given U .
PA U 

P A  U 
PU 

0.35
0.56
 0.625
(b)
What proportion of municipal bonds are issued by cities?
PU 
(c)
 0.56
What proportion of municipal bonds are issued by suburbs?
PS 
 0.325
42. Using the company records for the past 500 working days, the manager of Torrisi
Motors, a suburban automobile dealership, has summarized the number of cars sold
per day into the following table:
Number of Cars
Sold per Day
0
1
2
3
4
5
6
7
8
9
10
11
Total
Managerial Statistics
779
Frequency
of Occurrence
40
100
142
66
36
30
26
20
16
14
8
2
500
Prof. Juran
(a)
Form the empirical probability distribution (i.e., relative frequency distribution)
for the discrete random variable X, the number of cars sold per day.
Here's how to set this up in Excel:
A
B
1 Number of Cars Frequency
2
Sold per Day of Occurrence
3
0
40
4
1
100
5
2
142
6
3
66
7
4
36
8
5
30
9
6
26
10
7
20
11
8
16
12
9
14
13
10
8
14
11
2
15
Total
500
(b)
C
0.080
0.200
0.284
0.132
0.072
0.060
0.052
0.040
0.032
0.028
0.016
0.004
1.000
D
E
=B3/$B$15
Compute the mean or expected number of cars sold per day.
A
B
1 Number of Cars
Frequency
2
Sold per Day of Occurrence
3
0
40
4
1
100
5
2
142
6
3
66
7
4
36
8
5
30
9
6
26
10
7
20
11
8
16
12
9
14
13
10
8
14
11
2
15
Total
500
Managerial Statistics
C
D
E
F
=A3*C3
0.080
0.200
0.284
0.132
0.072
0.060
0.052
0.040
0.032
0.028
0.016
0.004
1.000
0.000
0.200
0.568
0.396
0.288
0.300
0.312
0.280
0.256
0.252
0.160
0.044
3.056
780
=SUM(D3:D14)
Prof. Juran
(c)
Compute the standard deviation.
A
B
1 Number of Cars Frequency
2
Sold per Day of Occurrence
3
0
40
4
1
100
5
2
142
6
3
66
7
4
36
8
5
30
9
6
26
10
7
20
11
8
16
12
9
14
13
10
8
14
11
2
15
Total
500
16
(d)
C
Prob.
0.080
0.200
0.284
0.132
0.072
0.060
0.052
0.040
0.032
0.028
0.016
0.004
1.000
D
E
F
G
H
I
=A3-$D$15
Weighted
Weighted =E3^2
Prob.
Error
Error^2
by Prob.
0.000
-3.056
9.339
0.747
0.200
-2.056
4.227
0.845
=F3*C3
0.568
-1.056
1.115
0.317
0.396
-0.056
0.003
0.000
0.288
0.944
0.891
0.064
0.300
1.944
3.779
0.227
0.312
2.944
8.667
0.451
0.280
3.944
15.555
0.622
0.256
4.944
24.443
0.782
0.252
5.944
35.331
0.989
0.160
6.944
48.219
0.772
0.044
7.944
63.107
0.252
=SUM(G3:G14)
3.056
Variance
6.069
=SQRT(G15)
Std. Dev.
2.464
What is the probability that on any given day
(i)
fewer than four cars will be sold?
P X  4 
 P X  0   P X  1  P X  2   P X  3 
 0.080  0.200  0.284  0.132
 0.696
(ii)
at most four cars will be sold?
P X  4 
 P  X  0   P X  1   P  X  2   P  X  3   P X  4 
 0.080  0.200  0.284  0.132  0.072
 0.768
(iii)
at least four cars will be sold?
P X  4 
 PX  4   PX  5  ...  PX  11
 0.072  0.060  0.052  0.040  0.032  0.028  0.016  0.004
 0.304
Managerial Statistics
781
Prof. Juran
Alternatively,
P X  4 
 1  P X  4 
 1  0.696
(from Part d(i) above)
 0.304
(iv)
exactly four cars will be sold?
P X  4 
(v)
 0.072
more than four cars will be sold?
P X  4 
 PX  5  PX  6   ...  PX  11
 0.060  0.052  0.040  0.032  0.028  0.016  0.004
 0.232
Alternatively,
P X  4 
 1  P X  4 
 1  0.768
(from Part d(ii) above)
 0.232
43. In the carnival game Under-or-over-Seven, a pair of fair dice are rolled once, and the
resulting sum determines whether or not the player wins or loses his or her bet. For
example, the player can bet $1.00 that the sum will be under 7 — that is, 2, 3, 4, 5, or 6.
For such a bet the player will lose $1.00 if the outcome equals or exceeds 7 or will win
$1.00 if the result is under 7. Similarly, the player can bet $1.00 that the sum will be over
7 — that is, 8, 9, 10, II, or 12. Here the player wins $1.00 if the result is over 7 but loses
$l.00 if the result is 7 or under. A third method of play is to bet $1.00 on the outcome 7.
For this bet the player will win $4.00 if the result of the roll is 7 and lose $1.00 otherwise.
Managerial Statistics
782
Prof. Juran
(a)
Form the probability distribution function representing the different outcomes
that are possible for a $1.00 bet on being under 7.
Here's a way to set this up in Excel:
A
B
C
1
2
1st Die
2nd Die
Sum
3
1
1
2
4
1
2
3
5
1
3
4
6
1
4
5
7
1
5
6
8
1
6
7
9
2
1
3
10
2
2
4
11
2
3
5
12
2
4
6
13
2
5
7
14
2
6
8
15
3
1
4
16
3
2
5
17
3
3
6
18
3
4
7
19
3
5
8
20
3
6
9
21
4
1
5
22
4
2
6
23
4
3
7
24
4
4
8
25
4
5
9
26
4
6
10
27
5
1
6
28
5
2
7
29
5
3
8
30
5
4
9
31
5
5
10
32
5
6
11
33
6
1
7
34
6
2
8
35
6
3
9
36
6
4
10
37
6
5
11
38
6
6
12
39 Number of Winning Outcomes
40 Probability of Winning
41
D
E
F
Results
Bet Under 7 Bet Over 7 Bet = 7
Win
Lose
Lose
Win
Lose
Lose
Win
Lose
Lose
Win
Lose
Lose
Win
Lose
Lose
Lose
Lose
Win
Win
Lose
Lose
Win
Lose
Lose
Win
Lose
Lose
Win
Lose
Lose
Lose
Lose
Win
Lose
Win
Lose
Win
Lose
Lose
Win
Lose
Lose
Win
Lose
Lose
Lose
Lose
Win
Lose
Win
Lose
Lose
Win
Lose
Win
Lose
Lose
Win
Lose
Lose
Lose
Lose
Win
Lose
Win
Lose
Lose
Win
Lose
Lose
Win
Lose
Win
Lose
Lose
Lose
Lose
Win
Lose
Win
Lose
Lose
Win
Lose
Lose
Win
Lose
Lose
Win
Lose
Lose
Lose
Win
Lose
Win
Lose
Lose
Win
Lose
Lose
Win
Lose
Lose
Win
Lose
Lose
Win
Lose
15
15
6
0.417
0.417
0.167
G
H
I
=IF(C38=7,"Win","Lose")
=COUNTIF(F3:F38,"Win")
=F39/36
So, for a bet on "Under 7", here are the possibilities:
Outcome
Win
Lose
Managerial Statistics
Payoff
$1
-$1
783
Probability
0.417
0.583
Prof. Juran
(b)
Form the probability distribution function representing the different outcomes
that are possible for a $1.00 bet on being over 7.
Same deal:
Outcome
Win
Lose
(c)
Probability
0.417
0.583
Form the probability distribution function representing the different outcomes
that are possible for a $1.00 bet on 7.
Outcome
Win
Lose
(d)
Payoff
$1
-$1
Payoff
$4
-$1
Probability
0.167
0.833
Show that the expected long-run profit (or loss) to the player is the same — no
matter which method of play is used.
Bet
Under 7
Over 7
=7
EX 
EX 
EX 
Expected Value Calculation
 1 * 0.417    1 * 0.583  $0.167
 1 * 0.417    1 * 0.583  $0.167
 4 * 0.167    1 * 0.833  $0.167
44. Suppose that warranty records show the probability that a new car needs a warranty
repair in the first 90 days is 0.05. If a sample of three new cars is selected,
(a)
what is the probability that
(i)
none needs a warranty repair?
We have a binomial distribution with n = 3 and p = 0.05.
P X  0 

 x
n!
 p 1  p nx
 
 x! n  x ! 


3!
0.05 0 1  0.05 30
 
 0! 3  0 ! 
 110.95  3
 0.857
Managerial Statistics
784
Prof. Juran
(ii)
at least one needs a warranty repair?
P X  1
 P X  0 
 1  P X  0 
 1  0.857
 0.143
(iii)
more than one needs a warranty repair?
P X  1
 1  PX  0   PX  1


 

3!
0.05 1 1  0.05 31  
 1  0.857  

 1! 3  1! 
 


 1  0.857  30.050.95 2

 1  0.857  0.135 
 1  0.993
 0.007
Here's how to do the previous questions in Excel:
A
1
2
3
4 (i) probability that none needs a warranty repair?
5 (ii) probability that at least one needs a warranty repair?.
6 (iii) probability that more than one needs a warranty repair?
7
8
(b)
B
n
p
C
D
E
F
G
3
0.05
=BINOMDIST(0,$C$1,$C$2,0)
0.857
0.143
0.007
=1-BINOMDIST(0,$C$1,$C$2,0)
=1-BINOMDIST(1,$C$1,$C$2,1)
What assumptions are necessary in (a)?

For each of the three cars, there are only two possible outcomes: Either it
needs a warranty repair or it doesn't. (These two outcomes are mutually
exclusive and collectively exhaustive.)

The probability of any one car needing warranty repair is the same for all
cars, and is independent of what happens with any other car.
Managerial Statistics
785
Prof. Juran
(c)
What are the mean and the standard deviation of the probability distribution in
(a)?
E(X)
(X)
 np
 30.05 
 0.15
 np( 1  p )
 30.05( 0.95 )
 0.143
 0.377
(d)
What would be your answers to (a)-(c) if the probability of needing a warranty
repair was 0.10?
P X  0 
 0.729
P X  1
 0.271
P X  1
 0.028
E(X)
 0.30
(X)
 0.520
45. Suppose that the likelihood that someone who logs onto a particular e-commerce
site will purchase an item is 0.20. If the site has 10 people accessing it in the next minute,
what is the probability that
(a)
none of the individuals will purchase an item?
P X  0 

 x
n!
 p 1  p nx
 


x
!
n

x
!



 0
10!
0.2 1  0.2 100
 


0
!
10

0
!


 110.8 10
 0.107
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Prof. Juran
(b)
exactly 2 individuals will purchase an item?

 x
n!
 p 1  p nx
 


x
!
n

x
!


P X  2 

 2
10!
0.2 1  0.2 10 2
 


2
!
10

2
!


 3 ,628 ,800  2
0.2 0.8 8
 


2
40
,
320


 450.040 0.168
 0.302
(c)
at least 2 individuals will purchase an item?
P X  2 
 1  PX  0   PX  1


 
 1
10!
0.2 1  0.2  101  
 1  0.107  

 1! 10  1! 
 
 1  0.107  10 0.2 0.134 
 1  0.107  0.268 
 1  0.376
 0.624
(d)
at most 2 individuals will purchase an item?
P X  2 
 PX  0   PX  1  PX  2 
 0.107  0.268  0.302
 0.678
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Prof. Juran
(e)
(f)
If 20 people accessed the site in the next minute, what would be your answers to
(a)-(d)?
P X  0 
 0.012
P X  2 
 0.137
P X  2 
 0.931
P X  2 
 0.206
If the probability of purchasing an item was only 0.10, what would be your
answers to (a)-(d)?
Managerial Statistics
P X  0 
 0.349
P X  2 
 0.194
P X  2 
 0.264
P X  2 
 0.930
788
Prof. Juran
46. An important part of the customer service responsibilities of a telephone company
relate to the speed with which troubles in residential service can be repaired. Suppose
past data indicate that the likelihood is 0.70 that troubles in residential service can be
repaired on the same day.
(a)
For the first five troubles reported on a given day, what is the probability that
(i)
all five will be repaired on the same day?
P X  5 

 x
n!
 p 1  p nx
 
 x! n  x ! 

 5
5!
0.7 1  0.7  5  5
 
 5! 5  5! 
 10.7 5 0.30
 10.168 1
 0.168
(ii)
at least three will be repaired on the same day?
P X  3 
 P X  3   P  X  4   P  X  5 

 

 3
 4
5!
5!
0.7 1  0.7  5 3   
0.7 1  0.7  5 4   0.168
 
 3! 5  3! 
  4! 5  4 ! 

 120  3
  120  4

0.7 0.32   
0.7 0.31   0.168
 
 62  
  241 

 10 0.3430.09   50.240 0.3  0.168
 0.309  0.360  0.168
 0.837
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Prof. Juran
(iii)
P X  2 
fewer than two will be repaired on the same day?
 PX  0   PX  1
 5!  0
 5!  1
0.7 0.3 5  
0.7 0.3 4
 
0
!

5

!
1
!

4

!




 120 
 120 
10.0081  
0.70.0024 
 
 1120  
 124  
 0.0024  0.0284
 0.031
(b)
(c)
What assumptions are necessary in (a)?

For each of the five problems reported, there are only two possible outcomes:
Either they get repaired on the same day or they don't. (These two outcomes
are mutually exclusive and collectively exhaustive.)

The probability of any one problem getting repaired on the same day (0.70) is
independent of what happens with any other reported problem.
What are the mean and the standard deviation of the probability distribution in
(a)?
E(X)
(X)
 np
 50.7 
 3.5
 np( 1  p )
 50.7 (0.3)
 1.05
 1.025
Managerial Statistics
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Prof. Juran
Here is how to answer these questions with Excel:
1
2
3
4
5
6
7
8
9
A
n
p
B
5
0.7
C
D
a(i)
E
0.168
F
G
H
I
a(ii)
0.837
=1-BINOMDIST(2,$B$1,$B$2,1)
a(iii)
0.031
=BINOMDIST(1,$B$1,$B$2,1)
c
3.500
1.025
=BINOMDIST(5,$B$1,$B$2,0)
mean =B1*B2
=SQRT((B1*B2*(1-B2)))
(d)
What would be your answers in (a) and (c) if the probability is 0.80 that troubles
in residential service can be repaired on the same day?
(e)
Compare the results of (a) and (d).
Probability
Event
p = 0.7
p = 0.8
All five will be repaired on the same day
0.168
0.328
At least three will be repaired on the same day
0.837
0.942
Fewer than two will be repaired on the same day
0.031
0.007
3.5
4.0
1.025
0.894
Expected value
Std Dev
47. Suppose that a student is taking a multiple-choice exam in which each question has
four choices. Assuming that she has no knowledge of the correct answers to any of the
questions, she has decided on a strategy in which she will place four balls (marked A, B,
C, and D) into a box. She randomly selects one ball for each question and replaces the
ball in the box. The marking on the ball will determine her answer to the question.
(a)
If there are five multiple-choice questions on the exam, what is the probability
that she will get
The number of correct answers out of five is binomially distributed with n = 5 and p =
0.25.
Managerial Statistics
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Prof. Juran
(i)
five questions correct?
P X  5 

 x
n!
 p 1  p nx
 


x
!
n

x
!




5!
0.25 5 1  0.25 5  5
 
5
!

5

5

!


 10.25 5 0.750
 10.0010 1
 0.0010
(ii)
at least four questions correct?
P X  4 
 P X  4   P X  5 



5!
0.25 4 1  0.255 4   0.0010
 
 4! 5  4! 

 120 

 
0.00390.75  0.0010
 24 

 0.0146  0.0010
 0.0156
(iii)
no questions correct?
P X  0 


5!
0.25 0 1  0.25 5  0
 
 0! 5  0 ! 
 120 

10.2373
 120 
 0.2373
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Prof. Juran
(iv)
no more than two questions correct?
P X  2 
 PX  0   PX  1  PX  2 



5! 
5!
0.25 1 1  0.25 5  1  
0.25 2 1  0.25 5  2
 0.2373  
1
!

5

1

!
2
!

5

2

!




 120 
 120 
4
3
1
2
 0.2373  
0.25 0.75  
0.25 0.75
 24 
 12 
 0.2373  50.250.3164   10 0.0625 0.4219 
 0.2373  0.3955  0.2637
 0.8965
(b)
(c)
What assumptions are necessary in (a)?

For each of the five questions, there are only two possible outcomes: Either
she gets it right or she doesn't. (These two outcomes are mutually exclusive
and collectively exhaustive.)

The probability of getting any one question correct is the same for all
questions (0.25), and is independent of what happens with any other
question.
What are the average and the standard deviation of the number of questions that
she will get correct in (a)?
E(X)
(X)
 np
 50.25 
 1.25
 np( 1  p )
 50.25(0.75)
 0.9375
 0.9682
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Prof. Juran
(d)
Suppose that the exam has 50 multiple-choice questions and 30 or more correct
answers is a passing score. What is the probability that she will pass the exam by
following her strategy? (Use Microsoft Excel to compute this probability.)
The Excel function =1-BINOMDIST(29,50,0.25,1) returns the probability: 0.00000016 (not
bloody likely).
48. Suppose that a survey has been undertaken to determine if there is a relationship
between place of residence and ownership of a foreign-made automobile. A random
sample of 200 car owners from large cities, 150 from suburbs, and 150 from rural areas
was selected with the following results.
Car Ownership
Own foreign car
Do not own foreign car
Total
(a)
Type of Area
Large City Suburb
90
60
110
90
200
150
Rural
25
125
150
Total
175
325
500
If a car owner is selected at random, what is the probability that he or she
(i)
owns a foreign car?
Events
Own foreign car
Do not own foreign car
Large City
Suburb
Rural
P F 

Symbols
F
F
C
S
R
175
500
 0.35
(ii)
lives in a suburb?
P S 

150
500
 0.3
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Prof. Juran
(iii)
owns a foreign car or lives in a large city?
PF  C 
 PF   PC   PF  C 

175 200 90


500 500 500
 0.35  0.4  0.18
 0.57
(iv)
lives in a large city or a suburb?
PC  S 
 PC   PS   PC  S 
200 150
0


500 500 500

 0.4  0.3  0
 0.7
(v)
lives in a large city and owns a foreign car?
PC  F 

90
500
 0.18
(vi)
lives in a rural area or does not own a foreign car?

P RF

 
 P R   P F  P R  F


150 325 125


500 500 500
 0.3  0.65  0.25
 0.70
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795
Prof. Juran
(b)
Assume we know that the person selected lives in a suburb. What is the
probability that he or she owns a foreign car?
PF S 

PF  S 
PS 

0.12
0.30
 0.4
(c)
Is area of residence statistically independent of whether the person owns a
foreign car? Explain.
If these two variables were independent, then knowing about a person's area of
residence would not affect our expectations about the type of car they own.
Mathematically:
PF C   PF S   PF R  PF 
In fact, the above probabilities are not equal, and therefore the two events are not
independent.
49. The finance society at a college of business at a large state university would like to
determine whether there is a relationship between a student's interest in finance and his
or her ability in mathematics. A random sample of 200 students is selected and they are
asked whether their interest in finance and ability in mathematics are low, average, or
high. The results are as follows:
Interest in Finance
Low
Average
High
Total
(a)
Ability in Mathematics
Low
Average
High
60
15
15
15
45
10
5
10
25
80
70
50
Total
90
70
40
200
Give an example of a simple event.
A student shows a low interest in finance.
(b)
Give an example of a joint event.
A student shows a low interest in finance and an average ability in mathematics.
Managerial Statistics
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Prof. Juran
(c)
Why are high interest in finance and high ability in mathematics a joint event?
This is a joint event because two separate conditions must both be true.
Event
Low interest in finance
Average interest in finance
High interest in finance
Low ability in mathematics
Average ability in mathematics
High ability in mathematics
(d)
Symbol
LF
AF
HF
LM
AM
HM
If a student is chosen at random, what is the probability that he or she
(1)
has a high ability in mathematics?
P  HM 

50
200
 0.25
(2)
has an average interest in finance?
P AM 

70
200
 0.35
(3)
has a low ability in mathematics?
P LM 

80
200
 0.4
(4)
has a high interest in finance?
P HF 

40
200
 0.2
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Prof. Juran
(5)
has a low interest in finance and a low ability in mathematics?
PLF  LM 

60
200
 0.3
(6)
has a high interest in finance and a high ability in mathematics?
PHF  HM 

25
200
 0.125
(7)
has a low interest in finance or a low ability in mathematics?
PLF  LM 
 PLF   PLM   PLF  LM 

90
80
60


200 200 200
 0.45  0.4  0.3
 0.55
(8)
has a high interest in finance or a high ability in mathematics?
PHF  HM 
 PHF   PHM   PHF  HM 

40
50
25


200 200 200
 0.2  0.25  0.125
 0.325
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Prof. Juran
(9)
has a low ability in mathematics or an average ability in mathematics or a
high ability in mathematics? Are these events mutually exclusive? Are
they collectively exhaustive? Explain.
Event
Low ability in mathematics
Average ability in mathematics
High ability in mathematics
Symbol
LM
AM
HM
Probability
0.4
0.35
0.25
These three probabilities add up to 1.0. They are mutually exclusive because only one of
them can occur at a time (a student can't have average ability and have low ability at the
same time). They are collectively exhaustive because one of them must occur (every
student must fall into one of the three categories).
(e)
Assume we know that the person selected has a high ability in mathematics.
What is the probability that the person has a high interest in finance?
PHF HM 

PHF  HM 
PHM 

0.125
0.25
 0.5
(f)
Assume we know that the person selected has a high interest in finance. What is
the probability that the person has a high ability in mathematics?
PHM HF 

PHF  HM 
PHF 

0.125
0.2
 0.625
(g)
Explain the difference in your answers to (e) and (f).
Note that the numerators are the same (representing the probability that a student has
both a high ability in mathematics and a high interest in finance. The denominators are
different.
Managerial Statistics
799
Prof. Juran
In (e) we are interested in the subset of students who have a high ability in
mathematics, and would like to know how likely they are to have a high interest in
finance. In (f) we are interested in the subset of students who have a high interest in
finance, and would like to know how likely they are to have a high ability in
mathematics.
(h)
Are interest in finance and ability in mathematics statistically independent?
Explain.
No they are not. On the whole, 25% of students have a high ability in mathematics.
Knowing that a student has a high interest in finance shouldn't affect this probability if
the two events are independent. In question (f) above, we see that this bit of information
about a student increases the probability that he/she has a high ability in mathematics
from 25% to 62.5%.
50. On the basis of past experience, 15% of the bills of a large mail-order book company
are incorrect. A random sample of three current bills is selected.
We assume the number of incorrect bills is binomially distributed with n = 3 and p =
0.15.
(a)
What is the probability that
(1)
exactly two bills are incorrect?
P X  2 

 x
n!
 p 1  p nx
 


x
!
n

x
!




3!
0.15 2 1  0.15 3 2
 


2
!
3

2
!


 3 0.0225 0.85 
 0.0574
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Prof. Juran
(2)
no more than two bills are incorrect?
P X  2 
 1  P X  3 


3!
0.15 3 1  0.150
 1  


0
!
3

0
!


 1  10.0034 1
0.9966
(3)
at least two bills are incorrect?
P X  2 
 P X  2   P X  3 
 0.0574  0.0034
 0.0608
(b)
What assumptions about the probability distribution are necessary to solve this
problem?
We assume that a bill must be either correct or incorrect, that the probability of any bill
being incorrect is 15%, and that the result with any one bill is independent of whether
any other bill is incorrect.
(c)
What would be your answers to (a) if the percentage of incorrect bills was 10%?
(1)
0.0270
(2)
0.9990
(3)
0.0280
Managerial Statistics
801
Prof. Juran
51. Suppose that on a very long mathematics test, the probability is that Lauren would
get 70% of the items right.
The number of correct items is binomially distributed with n = 10 and p = 0.7. We'll
solve this one using Excel.
(a)
For a 10-item quiz, calculate the probability that Lauren will get
(1)
at least 7 items right.
(2)
fewer than 6 items right (and therefore fail the quiz).
(3)
9 or 10 items right (and get an A on the quiz).
(b)
What is the expected number of items that Lauren will get right? What
proportion of the time will she get that number right?
(c)
What is the standard deviation of the number of items that Lauren will get right?
(d)
What would be your answers to (a)-(c) if she typically got 80% correct?
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
B
n
10
a
(1) at least 7 items right.
(2) fewer than 6 items right (and therefore fail the quiz).
(3) 9 or 10 items right (and get an A on the quiz).
0.650
0.150
0.149
(b) What is the expected number of items that Lauren will get right?
What proportion of the time will she get that number right?
7
0.2668
(c) What is the standard deviation of the number of items that Lauren will get right?
1.4491
d
(1) at least 7 items right.
(2) fewer than 6 items right (and therefore fail the quiz).
(3) 9 or 10 items right (and get an A on the quiz).
0.879
0.033
0.376
Managerial Statistics
C
p
0.7
D
E
F
=1-BINOMDIST(6,$B$2,$C$2,1)
=BINOMDIST(5,$B$2,$C$2,1)
=1-BINOMDIST(8,$B$2,$C$2,1)
=B2*C2
=BINOMDIST(B9,B2,C2,0)
=SQRT(B2*C2*(1-C2))
=1-BINOMDIST(6,$B$2,$C$19,1)
=BINOMDIST(5,$B$2,$C$19,1)
=1-BINOMDIST(8,$B$2,$C$19,1)
0.8
802
Prof. Juran