Solutions to Practice Problems for Part III
1. An author receives from a publisher a contract, according to which she is to be
paid a fixed sum of $10,000, plus $1.50 for each copy of her book sold.
Her uncertainty about total sales of the book can be represented by a random
variable with mean 30,000 and standard deviation 8,000. Find the mean and
standard deviation of the total payments she will receive.
Let the random variable X represent total sales of the book, and let the random
variable Y represent the payments to the author.
Y
 $10 ,000  $1.5 X
 $10 ,000  $1.530 ,000 
 $55,000
Y
 $1.5  X
 $1.58 ,000 
 $12,000
2. A charitable organization solicits donations by telephone. Employees are paid
$60 plus 20% of the money their calls generate each week. The amount of money
generated in a week by one employee can be viewed as a random variable with
mean $700 and standard deviation $130. Find the mean and standard deviation
of an employee's total pay in a week.
Let the random variable X represent the amount of money generated, and let the
random variable Y represent the employee's pay.
 $60  $0.2  X
Y
 $60  $0.2700
 $200
Y
 $0.2  X
 $0.2130 
 $26
3. Let the random variable Z follow a standard normal distribution.
(a) Find P(Z< 1.20)
P Z  1.2 
 0.5  0.3849
 0.8849
(b) Find P(Z > 1.33)
P Z  1.33 
 0.5  0.4082
 0.0918
(c) Find P(Z < -1.70)
PZ  1.70
 P Z  1.70 
 0.5  0.4554
 0.0446
(d) Find P(Z > -1.00)
PZ  1.0 
 P Z  1.0 
 0.5  0.3413
 0.8413
(e) Find P(l.20 < Z< 1.33)
P 1.2  Z  1.33 
 PZ  1.33   PZ  1.2 
 0.5  0.4082   0.5  0.3849 
 0.9082  0.8849
 0.0233
(f) Find P(-1.70 < Z < 1.20)
P  1.7  Z  1.2 
 P Z  1.2   P Z  1.7 
 0.5  0.3849   PZ  1.7 
 0.5  0.3849   0.5  0.4554 
 0.8849  0.0446
 0.8403
(g) Find P(-1.70 < Z < -1.00)
P  1.7  Z  1.0 
 PZ  1.0   PZ  1.7 
 P Z  1.0   P Z  1.7 
 0.5  0.3413   0.5  0.4554 
 0.1587  0.0446
 0.1141
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Prof. Juran
4. It is known that amounts of money spent on textbooks in a year by students on
a particular campus follow a normal distribution with mean $380 and standard
deviation $50.
(a) What is the probability that a randomly chosen student will spend less than
$400 on textbooks in a year?
PX  400
400  380 
 P  Z 

50


 P Z  0.4 
 0.5  0.1554
 0.6554
(b) What is the probability that a randomly chosen student will spend more than
$360 on textbooks in a year?
PX  360
360  380 
 P  Z 

50


 PZ  0.4 
 P Z  0.4 
 0.5  0.1554
 0.6554
(c) Draw a graph to illustrate why the answers to parts (a) and (b) are the same.
Graph for Part (a)
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Graph for Part (b)
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Prof. Juran
(d) What is the probability that a randomly chosen student will spend between
$300 and $400 on textbooks in a year?
300  380
400  380 
 P
Z

50
50


P300  X  400
 P Z  0.4   P Z  1.6 
 0.5  0.1554  PZ  1.6 
 0.6554  1  0.5  0.4452 
 0.6554  0.0548
 0.6006
(e) You want to find a range of dollar spending on textbooks in a year that
includes 80% of all students on this campus. Explain why any number of such
ranges could be found, and find the shortest one.
There are an infinite number of pairs of values a and b such that Pa  Z  b   0.8 .
The shape of the bell curve causes the distance between a and b to be minimized
if we center this interval on zero (which means a = -b). Therefore:
0.8
And 0.4
Therefore zb
 Pzb  Z  zb 
 PZ  zb 
 1.28
a , b     1.28 ,   1.28 
 380  1.2850 , 380  1.2850 
 316, 444 
5. The tread life of a particular brand of tire has a normal distribution with mean
35,000 miles and standard deviation 4,000 miles.
(a) What proportion of these tires have tread lives of more than 38,000 miles?
P X  38,000 
38 , 000  35 , 000 

 P Z 

4 , 000


 P Z  0.75 
 1  0.5  0.2734 
 0.2266
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Prof. Juran
(b) What proportion of these tires have tread lives of less than 32,000 miles?
P X  32, 000 
32 , 000  35 , 000 

 P Z 

4 , 000


 P Z  0.75
 P Z  0.75 
 0.2266
(c) What proportion of these tires have tread lives between 32,000 and 38,000
miles?
P 32 , 000  X  38 , 000 
38 , 000  35 , 000 
 32 , 000  35 , 000
P
Z 

4 , 000
4 , 000


 P  0.75  Z  0.75
 P Z  0.75  P Z  0.75
 P Z  0.75  P Z  0.75
 0.5  0.2734  0.5  0.2734
 0.7734  0.2266
 0.5468
(d) Draw a graph of the probability density function of tread lives, illustrating
(i)
Why the answers to (a) and (b) are the same.
(ii)
Why the answers to (a), (b), and (c) sum to one.
(a) and (b) are the same, because these two regions are equal in area. All three of
the regions make up a mutually exclusive and collectively exhaustive set.
Therefore, by our basic rules of probability, they sum to 1.0.
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Prof. Juran
6. An investment portfolio contains stocks of a large number of corporations.
Over the last year the rates of return on these corporate stocks followed a normal
distribution, with mean 12.2%, and standard deviation 7.2%.
(a) For what proportion of these corporations was the rate of return higher than
20%?
P X  0.20 
0.20  0.122 

 P Z 

0.072


 P Z  1.08 
 1.0  0.5  0.3599 
 0.1401
(b) For what proportion of these corporations was the rate of return negative?
P X  0 
0  0.122 

 P Z 

0.072 

 P Z  1.69 
 P Z  1.69 
 1  0.5  0.4545
 0.0455
(c) For what proportion of these corporations was the rate of return between 5%
and 15%?
P 0.05  X  0.15
0.15  0.122 
 0.05  0.122
P
Z 

0.072
0.072


 P  1  Z  0.39 
 P Z  0.39   P Z  1
 0.5  0.1517   P Z  1
 0.6517  1  0.5  0.3413
 0.493
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Prof. Juran
7. A broadcasting executive is reviewing the prospects for a new television series.
According to her judgement, the probability is 0.25 that the show will achieve a
rating higher than 17.8, and the probability is 0.15 that it will achieve a rating
higher than 19.2. If the executive's uncertainty about the rating can be
represented by a normal distribution, what are the mean and variance of that
distribution?
The facts above imply the following:
P X  17.8 
 0.25
P X  19.2 
 0.15
From this, we can see that:
P 17.8  X  19.2 
 0.10
In the standard normal table, we note that an upper-tail probability of 0.15
(found by looking in the middle of the table for a probability near 0.35)
corresponds to about 1.04 standard deviations above the mean, and a probability
of 0.25 corresponds to about 0.67 standard deviations above the mean. Therefore:
17.8    0.67
19.2    1.04
17.8
17.8  0.67
19.2
19.2
1.4
3.78
14.29
   0.67

 17.8  0.67   1.04
 17.8  0.37
 0.37

 2
17.8    0.67
17.8    0.67 3.78 
17.8  2.53  
15.27  
Note: These calculations are subject to rounding error. If you solve the problem
with Excel, you will get more precise answers:
mean = 15.1910594
variance = 14.96156
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Prof. Juran
8. A company can purchase raw material from either of two suppliers and is
concerned about the amounts of impurity the material contains. A review of the
records for each supplier indicates that the percentage impurity levels in
consignments of the raw material follow normal distributions with the means
and standard deviations given in the table. The company is particularly anxious
that the impurity levels in a consignment not exceed 5% and wants to purchase
from the supplier more likely to meet that specification. Which supplier should
be chosen?
MEAN
STANDARD DEVIATION
Supplier A
4.4
.4
Supplier B
4.2
.6
Supplier A:
P X  0.05 
0.05  0.044 

 P Z 

0.004


 P Z  1.5 
 1  0.5  0.4332 
 0.0668
Supplier B:
P X  0.05 
0.05  0.042 

 P Z 

0.006


 P Z  1.33 
 1  0.5  0.4082 
 0.0918
Supplier A is more likely to meet the specification.
9. A company services copiers. A review of its records shows that the time taken
for a service call can be represented by a normal random variable with mean 75
minutes and standard deviation 20 minutes.
(a) What proportion of service calls take less than one hour?
PX  60 
60  75 

 P Z 

20 

 PZ  0.75
 PZ  0.75
 1  0.5  0.2734 
 0.2266
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Prof. Juran
(b) What proportion of service calls take more than 90 minutes?
PX  90 
90  75 

 P Z 

20 

 PZ  0.75
 1  0.5  0.2734 
 0.2266
(c) Sketch a graph to show why the answers to parts (a) and (b) are the same.
This graph would like the one in Problem 5(d) above:
The only difference here is that the area labeled “b” in the graph corresponds to
Part (a) of this question: PZ  0.75  ; and the area labeled “a” corresponds to
Part (b) of this question: PZ  0.75  .
(d) The probability is 0.1 that a service call takes more than how many minutes?
From the normal table we can infer that:
0.10
 P Z  1.28 
Therefore:
X  75
20
1.28

25.6
 X  75
100.6
X
The probability is 0.1 that a service call takes more than 100.6 minutes.
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Prof. Juran
10. It is known that 10% of all the items produced by a particular manufacturing
process are defective. From the very large output of a single day, 400 items are
selected at random.
(a) What is the probability that at least thirty-five of the selected items are
defective?
Note that the number of defective items is binomial with n = 400 and p = 0.10. We
can use the normal approximation to the binomial distribution.
  np
 4000.10 
 40

 n  p 1  p 
 4000.100.90
6
P X  35 
34.5  40 

 P Z 

6


 PZ  0.92 
 0.3212  0.5
 0.8212
(b) What is the probability that between forty and fifty of the selected items are
defective?
(This question is ambiguous; I assume we mean between 40 and 50 inclusive.)
P 39.5  X  50.5
50.5  40 
 39.5  40
P
Z 

6
6


 P Z  1.75  P Z  0.08
 1  0.5  0.4599   1  0.5  0.0319 
 0.4918
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Prof. Juran
(c)
What is the probability that between thirty-four and forty-eight of the
selected items are defective?
(This question is ambiguous; I assume we mean between 34 and 48 inclusive.)
P 33.5  X  48.5
48.5  40 
 33.5  40
P
Z 

6
6


 P Z  1.42   P Z  1.08
 0.5  04222   0.5  0.3599 
(d)
 0.7821
Without doing the calculations, state which of the following ranges of
defectives has the highest probability: 37-39, 39-41, 41-43, 43-45, 45-47.
A bell-shaped distribution is densest in the center, so we expect the probability to
be greatest in the range containing the mean (40). Therefore the range 39-41 has
the greatest probability.
11. A hospital finds that 25% of its bills are at least one month in arrears. A
random sample of forty-five bills was taken.
(a) What is the probability that less than ten bills in the sample were at least one
month in arrears?
Note that the number of bills at least one month in arrears is binomially
distributed with n = 45, p = 0.25, and the following parameters:
  np
 11.25

 np1  p 
 2.905
Using the normal approximation,
PX  10 
9.5  11.25 

 P Z 

2.905 

 PZ  0.60 
 PZ  0.60 
 1  0.5  0.2257 
 0.2743
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Prof. Juran
(b) What is the probability that the number of bills in the sample at least one
month in arrears was between twelve and fifteen (inclusive)?
P12  X  15 
15.5  11.25 
 11.5  11.25
 P
Z

2.905
2.905 

15.5  11.25 
 11.5  11.25
 P
Z

2.905
2.905 

 P0.09  Z  1.46 
 PZ  1.46   PZ  0.09 
 0.5  0.4279   0.5  0.0359 
 0.392
12. The tread life of a brand of tire can be represented (as in Exercise 5) by a
normal distribution with mean 35,000 miles and standard deviation 4,000 miles.
A sample of 100 of these tires is taken. What is the probability that more than 25
of them have tread lives of more than 38,000 miles?
Recall from Exercise 5 that the probability that a single tire will have a tread life
of longer than 38,000 miles is 0.2266. Therefore, the number of tires in a random
sample of 100 that have a tread life of longer than 38,000 miles has a binomial
distribution with n = 100, p = 0.2266, and the following parameters:
  np
 22.66

 np1  p 
 4.186
Using the normal approximation,
PX  25 
25.5  22.66 

 P Z 

4.186 

 PZ  0.68
 1  0.5  0.2517 
 0.2483
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Prof. Juran
13. An investor puts $2,000 into a deposit account with a fixed rate of return of
10% per annum. A second sum of $1,000 is invested in a fund with expected rate
of return of 16% and standard deviation of 8% per annum.
(a) Find the expected value of the total amount of money this investor will have
after a year.
Let's define some variables:
$X
= Principal amount in deposit account
 $2000
$Y
= Principal amount in fund
 $1000
M
a
X
b
= Total amount after one year
= Proportion of portfolio in fund
 0.667
 0.10
 0.333
Y
= Rate of return on fund
Random:  Y  0.16 ,
= Proportion of portfolio in deposit account
= Rate of return on deposit account
M
 Y  0.08
 $ X  $Y   1  EaX  bY 

 $3,000  1  aX  bY

 $3,000  1  0.667  0.10   0.333  0.16 
 $3,000  1.12
 $3,360
(b) Find the standard deviation of the total amount after a year.
  aX bY 
 $3000  a 2  X2  b 2  Y2
 $3000 
4
1
0   0.0064 
9
9
 $3000  0.02667
 $80
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Prof. Juran
14. It has been found that times taken by people to complete a particular tax form
follow a normal distribution with mean 100 minutes and standard deviation 30
minutes.
(a) What is the probability that a randomly chosen person takes less than 85
minutes to complete this form?
PX  85 
85  100 

 P Z 

30 

 PZ  0.5 
 PZ  0.5
 1  0.5  0.1915 
 0.3085
(b) What is the probability that a randomly chosen person takes between 70 and
130 minutes to complete this form?
P70  X  130 
130  100 
 70  100
 P
Z

30
 30

 P  1  Z  1
 2  P0  Z  1
 20.3413 
 0.6826
(c) Five percent of all people take more than how many minutes to complete this
form?
0.05
 PZ  X 
 PZ  1.645
Note: 1.645 was found by looking in the standard normal table for a probability
near 0.5 - 0.05 = 0.45. This value is between 1.64 and 1.65 standard deviations
above the mean; we interpolate to 1.645.
X
 1.645 standard deviations above the mean
   1.645
 100  1.64530 
 149.35
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Prof. Juran
(d) Two people are chosen at random. What is the probability that at least one of
them takes more than an hour to complete this form?
First, figure out the probability that one person takes more than an hour:
 60  100

 P
 Z
30


P60  X 
 P 1.33  Z 
 PZ  1.33
 0.5  0.4082
 0.9082
Pat least one 
 1  Pzero 
2
 1   0.9082 0 0.0918 2 
0
 1  110.00843
 0.99157
(e) Four people are chosen at random. What is the probability that exactly two of
them take longer than an hour to complete this form?
Observe that the number of people out of four who take longer than one hour is
binomially distributed with n = 4 and p = 0.9082.
P X  2 
 4
  0.9082 2 0.0918 2
2
4!
0.82480.0084 
2!2!
 60.82480.0084 

 0.0416
(f) For a randomly chosen person, state in which of the following ranges
(expressed in minutes) time to complete the form is most likely to lie.
70-90
Managerial Statistics
90-110
110-130
847
130-150
Prof. Juran
(g) For a randomly chosen person, state in which of the following ranges
(expressed in minutes) time to complete the form is least likely to lie.
70-90
90-110
110-130
130-150
Here we have divided up the distribution into "bins", as in a histogram. From the
shape of the normal distribution, we expect that observations will most often fall
in the bin that contains the mean, and are least likely to fall in a bin that is far
away from the mean. Therefore, in (f) the answer is the bin containing the mean,
or 90-110. Similarly, in (g) the answer is the bin farthest from the mean, or 130150.
15. A market research organization has found that 40% of all supermarket
shoppers refuse to cooperate when questioned by its pollsters. If 1,000 shoppers
are approached, what is the probability that fewer than 500 will refuse to
cooperate?
The number of shoppers refusing to cooperate will be binomially distributed
with n = 1000 and p = 0.4. Using the binomial transformation, we assume that the
number of shoppers refusing to cooperate will be approximately normally
distributed with the following parameters:
  np
 10000.4 
 400

 np1  p 
 10000.40.6
 15.49
Therefore, using the binomial transformation with the continuity correction:
PX  500 
499.5  400 

 P Z 

15.49 

 PZ  6.42 
 1.0
Managerial Statistics
848
Prof. Juran
16. A Chicago radio station believes that 40% of its listeners are younger than 25
years of age. Six hundred listeners are chosen at random.
(a) If the station's belief is correct, what is the probability that more than 260 of
these listeners are younger than 25?
The number of listeners younger than 25 will be binomially distributed with n =
600 and p = 0.4 (if the station's belief is correct). Using the binomial
transformation, we assume that the number of listeners younger than 25 will be
approximately normally distributed with the following parameters:
  np
 6000.4 
 240

 np1  p 
 6000.40.6
 12
Therefore, using the binomial transformation with the continuity correction:
P260  X 
 260.5  240

 P
 Z
12


 P1.71  Z 
 1  PZ  1.71
 1  0.5  0.4564 
 0.0436
Managerial Statistics
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Prof. Juran
(b) If the station's belief is correct, the probability is 0.6 that more than how many
of these 600 listeners are younger than 25?
If there is a 0.6 probability that more than X out of 600 are younger than 25, then
X must lie somewhere below the mean of 240. Specifically,
0.6
0 .1
 PZ  X 
 P X  Z  0 
 PZ  0.25
Note: 0.25 was found by looking in the standard normal table for a probability
near 0.10. This value is between 0.25 and 0.26 standard deviations above the
mean; we choose 0.25 because it is the closest.
X
 0.25 standard deviations below the mean
   0.25
 240  0.2512 
 237
In fact, the normal approximation doesn’t help answer this question exactly, as
shown in this Excel spreadsheet:
A
B
C
D
E
F
G
H
I
J
K
n
600
1
p
0.4
2
3
=1-BINOMDIST(A4,$B$1,$B$2,1)
4
230
78.54%
5
231
76.02%
6
232
73.34%
7
233
70.53%
8
234
67.58%
9
235
64.53%
236
61.37%
10
237
58.15% <--- The given probability of 60% can only be approximated in this discrete example.
11
12
238
54.86%
13
239
51.55%
14
240
48.23%
15
16 Using the Continuity Correction
17
230.5
78.54%
18
231.5
76.02%
19
232.5
73.34%
20
233.5
70.53%
21
234.5
67.58%
22
235.5
64.53%
236.5
61.37%
23
237.5
58.15% <--- Apparently the Excel BINOMDIST function rounds non-integer values down to the nearest integer.
24
25
238.5
54.86%
26
239.5
51.55%
Managerial Statistics
850
L
Prof. Juran
17. Consider the following two stocks:
Expected Return Std Dev Return
Stock A
24%
20%
Stock B
18%
5%
(a) Assuming the returns on both stocks are normally distributed, which is more
likely to lose money?
Z transformations:
Stock A:
z

0  0.24
0.20
 1.20
Stock B:
z

0  0.18
0.05
 3.60
Pz  1.2 
 0.1151
Use either the normal table ( 0.5  0.3849  0.1151 ) or the Excel function
=NORMSDIST(-1.2).
Pz  3.6 
 0.0002
If your normal table doesn't go to -3.6 standard deviations, use the Excel function
=NORMSDIST(-3.6).
Managerial Statistics
851
Prof. Juran
(b) What is the expected value and standard deviation of the return on a
portfolio consisting of 70% Stock A and 30% Stock B, assuming that their
returns have a correlation of 0.00?
EaX  bY 
 aX  bY
 .7 .24   .3.18 
 0.2220
 22.2%
  aX  bY 
 a 2 X2  b 2 Y2  2abCov XY 
 .7 2 .20 2  .3 2 .05 2  2.7 .30.20.05
 0.019825
 0.1408
 14.08%
(c) What is the expected value and standard deviation of the return on a
portfolio consisting of 80% Stock A and 20% Stock B, assuming that their
returns have a correlation of -0.40?
EaX  bY 
 aX  bY
 .8.24   .2.18 
 0.2280
 22.8%
  aX  bY 
 a 2 X2  b 2 Y2  2 abCov XY 
 .8 2 .20 2  .2 2 .05 2  2.8.2  .40 .20 .05
 0.02442
 0.1563
 15.63%
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852
Prof. Juran
18. Consider the following two stocks:
Expected Return Std Dev Return
Stock A
15%
4%
Stock B
5%
2%
(a) Assuming the returns on both stocks are normally distributed, which is more
likely to lose money?
Z transformations:
Stock A

z
0  0.15
0.04
 3.75
Stock B

z
0  0.05
0.02
 2.50
Pz  3.75
0.000088
Use the Excel function =NORMSDIST(-3.75).
Pz  2.5
0.0062
From the normal table, 0.5 - 0.4938 = 0.0062, or 0.62%.
(b) What is the expected value and standard deviation of the return on a
portfolio consisting of 40% Stock A and 60% Stock B, assuming that their
returns have a correlation of 0.00?
E aX  bY 
 a X  bY
 .4.15  .6.05
 0.09
 9.0%
 aX bY 
 a 2 X2  b 2 Y2  2abCov XY 
 .4 2.04 2  .62.02 2  2.4.60.04.02
 0.0004
 0.02
 2.00%
Managerial Statistics
853
Prof. Juran
(c) What is the expected value and standard deviation of the return on a
portfolio consisting of 30% Stock A and 70% Stock B, assuming that their
returns have a correlation of 0.60?
E aX  bY 
 a X  bY
 .3.15  .7.05
 0.0800
 8.00%
 aX bY 
 a 2 X2  b 2 Y2  2abCov XY 
 .32.04 2  .7 2.02 2  2.3.70.6.04.02
 0.0005416
 0.0233
 2.33%
19. Consider a Dow Jones Industrial Average index fund and a growth stock,
with the following returns per $1000 invested:
(a)
Probability
Economic Scenario
DJIA Fund
Growth Stock
0.2
Recession
-$100
-$200
0.5
Stable Economy
+$100
+$50
0.3
Expanding Economy
+$250
+$350
Calculate the expected value and standard deviation for the dollar return
per $1000 invested for each of the two investments.
Dow Jones (X)
Expected Value
X
n
  P X  x i x i 
i 1
 0.2  100   0.5 100   0.3 250 
 $105
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854
Prof. Juran
Standard Deviation
X
 PX  x x
n

i
i 1
i
  X 2

0.2  100  1052  0.5100  1052  0.3250  1052

 14 ,725
 $121.35
Growth Stock (Y)
Expected Value
Y
n
  P Y  y i y i 
i 1
 0.2  200   0.5 50   0.3 350 
 $90
Standard Deviation
Y

 PY  y y
n
i 1

i
i
 Y 2

0.2  200  902  0.550  902  0.3350  902
 37 ,900
 $194.68
(b)
Calculate the covariance and correlation coefficient between the two
investments.
COVXY
 EXY    X Y
 0.2  100 * 200   0.5 100 * 50   0.3 250 * 350   105 * 90 
 4,000  2,500  26,250  9,450
 23,300
 XY

COVXY

23 ,300
121.35 * 194.68
 X Y
 0.9863
Managerial Statistics
855
Prof. Juran
(c)
What is the expected return and standard deviation on a portfolio
consisting of 50% DJIA index and 50% Growth Stock?
EaX  bY 
 a X  bY
 0.5105   0.590 
 $97.50
 aX bY 
 a 2 X2  b 2 Y2  2abCovXY 
 0.5 2 14,725  0.5 2 37 ,900  20.50.523,300
 24 ,806.25
 $157.50
(d)
Recalculate the portfolio expected return and the portfolio risk if 30% are
invested in the Dow Jones index fund and 70% in the growth stock.
E P 
 $94.50
0.3 2 14 ,725  0.7  2 37 ,900   20.30.7 23 ,300 
P
 $172.29
(e)
Recalculate the portfolio expected return and the portfolio risk if 70% are
invested in the Dow Jones index fund and 30% in the growth stock.
E P 
P

 $100.50
0.7  2 14 ,725  0.3 2 37 ,900   20.7 0.323 ,300 
 $142.87
Managerial Statistics
856
Prof. Juran
(f)
Which of the three investment strategies (30%, 50%, or 70% in the Dow
Jones index stock) would you recommend? Why?
30% Dow
50% Dow
70% Dow
Expected Return
94.5
97.50
100.50
Standard Deviation
172.29
157.50
142.87
The 70% portfolio seems to dominate the other two; it has the least risk while
also having the highest expected return.
Risk vs. Return
(DJIA Fund and Growth Stock)
$106
$104
$102
Expected Return
$100
90% DJIA
$98
$96
70% DJIA
$94
50% DJIA
$92
30% DJIA
$90
10% DJIA
$88
$100
$110
$120
$130
$140
$150
$160
$170
$180
$190
$200
Risk (StDev)
Managerial Statistics
857
Prof. Juran
20. You are trying to develop a strategy for investing in two different stocks. The
anticipated annual return for a $1,000 investment in each stock has the following
probability distribution:
Returns
Probability
Stock X
Stock Y
0.1
-$100
$50
0.3
0
100
0.3
80
-20
0.3
150
100
Compute the
(a)
expected return for stock X.
(b)
expected return for stock Y
(c)
standard deviation for stock X.
(d)
standard deviation for stock Y
For Stock X:
Probability
Stock X
E(X)
Error
Error^2 Weighted
0.1
$(100.00)
$ 59.00
$(159.00)
25281
2528.1
0.3
$-
$ 59.00
$ (59.00)
3481
1044.3
0.3
$ 80.00
$ 59.00
$ 21.00
441
132.3
0.3
$ 150.00
$ 59.00
$ 91.00
8281
2484.3
Expected
$ 59.00
6189
Variance
$ 78.67
StDev
For Stock Y:
(e)
Probability
Stock Y
E(Y)
Error
Error^2 Weighted
0.1
$ 50.00
$ 59.00
$ (9.00)
81
8.1
0.3
$ 100.00
$ 59.00
$ 41.00
1681
504.3
0.3
$ (20.00)
$ 59.00
$ (79.00)
6241
1872.3
0.3
$ 100.00
$ 59.00
$ 41.00
1681
504.3
Expected
$ 59.00
2889
Variance
$ 53.75
StDev
Covariance of stock X and stock Y
COVXY
 EXY    X Y
 0.1 100 * 50   0.3 0 * 100   0.3 80 * 20   0.3 150 * 100   59 * 59 
  500  0  480  4 ,500   3,481
 39
Managerial Statistics
858
Prof. Juran
(f)
Do you think you will invest in stock X or stock Y? Explain.
Stock Y appears to be a superior investment to Stock X; it has the same expected
return at a lower risk.
(g)
Suppose you wanted to create a portfolio that consists of stock X and stock
Y. Compute the portfolio expected return and portfolio risk for each of the
following proportions invested in stock X.
(i)
.10
(ii)
.30
(iii)
.50
(iv)
.70
(v)
.90
Here is how you can solve this in Excel:
A
1
2
3
4
5
6
7
8
9
10
11
(h)
Expected Return
StDev
Covariance
Proportion in X
Expected Return
StDev
B
C
D
E
Stock X
Stock Y
$ 59.00 $ 59.00
$ 78.67 $ 53.75
39
=($B$2*B6)+($C$2*(1-B6))
$
$
0.1
59.00
49.08
$
$
0.3
59.00
44.60
$
$
0.5
59.00
47.84
0.7
$ 59.00
$ 57.52
F
0.9
$ 59.00
$ 71.06
=SQRT(((C6^2)*($B$3^2))+(((1-C6)^2)*($C$3^2))+(2*(C6)*(1-C6)*$B$4))
On the basis of the results of (g), which portfolio would you recommend?
Explain.
It looks like the best portfolio consists of 30% Stock X and 70% Stock Y.
Managerial Statistics
859
Prof. Juran
21. You are trying to set up a portfolio that consists of a corporate bond fund and
a common stock fund. The following information about the annual return (per
$1,000) of each of these investments under different economic conditions is
available along with the probability that each of these economic conditions will
occur.
Probability
State of the Economy
Corporate Bonds
Common Stocks
0.10
Recession
-$30
-$150
0.15
Stagnation
50
-20
0.35
Slow growth
90
120
0.30
Moderate growth
100
160
0.10
High growth
110
250
Compute the
(a) expected return for corporate bonds.
(b) expected return for common stocks.
(c) standard deviation for corporate bonds.
(d) standard deviation for common stocks.
For Bonds:
Probability Corporate Bonds Expected
Error
Error^2 Weighted
0.1
$ (30.00)
$ 77.00
$(107.00)
11449
1144.9
0.15
$ 50.00
$ 77.00
$ (27.00)
729
109.35
0.35
$ 90.00
$ 77.00
$ 13.00
169
59.15
0.3
$ 100.00
$ 77.00
$ 23.00
529
158.7
0.1
$ 110.00
$ 77.00
$ 33.00
1089
108.9
$ 77.00
1581
Variance
$ 39.76
StDev
For Stocks:
Probability Common Stocks Expected
Error
Error^2 Weighted
0.1
$ (150.00)
$ 97.00
$(247.00)
61009
6100.9
0.15
$ (20.00)
$ 97.00
$(117.00)
13689
2053.35
0.35
$ 120.00
$ 97.00
$ 23.00
529
185.15
0.3
$ 160.00
$ 97.00
$ 63.00
3969
1190.7
0.1
$ 250.00
$ 97.00
$ 153.00
23409
2340.9
$ 97.00
Managerial Statistics
860
11871
Variance
$108.95
StDev
Prof. Juran
(e) Covariance of corporate bonds and common stocks.
28
29
30
31
32
33
34
35
H
I
J
Probability Corporate Bonds Common Stocks
0.1
$
(30.00) $
(150.00)
0.15
$
50.00 $
(20.00)
0.35
$
90.00 $
120.00
0.3
$
100.00 $
160.00
0.1
$
110.00 $
250.00
K
L
M
N
=H29*I29*J29
450
-150
3780
4800
2750
7469
4161
=E(Bonds)*E(Stocks)
=SUM(K29:K33)-K34
(f) Do you think you should invest in corporate bonds or common stocks?
Explain.
There is no right or wrong recommendation here; there is a clear trade-off
between risk and return, which would have to be resolved differently for each
individual investor.
(g) Suppose you wanted to create a portfolio that consists of corporate bonds and
common stocks. Compute the portfolio expected return and portfolio risk for
each of the following proportions invested in corporate bonds.
(i)
0.10
(ii)
0.30
(iii)
0.50
(iv)
0.70
(v)
0.90
Proportion in Bonds
0.1
0.3
0.5
0.7
0.9
Expected Return
$95.00
$91.00
$87.00
$83.00
$79.00
StDev
$101.88
$87.79
$73.78
$59.92
$46.35
(h) On the basis of the results of (g), which portfolio would you recommend?
Explain.
Once again, there is a trade-off between risk and return.
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861
Prof. Juran
22. Given a standardized normal distribution (with a mean of 0 and a standard
deviation of 1), answer the following:
(a)
What is the probability that
(i)
Z is less than 1.57?
From the table, the probability of Z being between 0 and 1.57 is
0.4418. We add the probability that Z is less than zero to get the
answer to the question: 0.5 + 0.4418 = 0.9418.
(ii)
Z exceeds 1.84?
0.5 – 0.4671 = 0.0329.
(iii)
Z is between 1.57 and 1.84?
– 0.0329) - 0.9418 = 0.9671 – 0.9418 = 0.0253
(iv)
Z is less than 1.57 or greater than 1.84?
0.9418 + 0.0329 = 0.9747
(v)
Z is between -1.57 and 1.84?
0.4418 + 0.4671 = 0.9089
(vi)
Z is less than –1.57 or greater than 1.84?
1.0 – 0.9089 = 0.0911
(b)
What is the value of Z if 50.0% of all possible Z values are larger?
Remember that the normal distribution is symmetric around the mean.
50% of the values lie above the mean, which is where Z = 0.0.
(c)
What is the value of Z if only 2.5% of all possible Z values are larger?
From the Z table, we see that 0.5 – 0.025 = 0.475 corresponds to a Z-value
of 1.96.
(d)
Between what two values of Z (symmetrically distributed around the
mean) will 68.26% of all possible Z-values be contained?
We look in the table for 0.6826/2 = 0.3413, which corresponds to a Z-value
of 1.0.
Managerial Statistics
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Prof. Juran
Here is how to answer all of these using Excel:
1
2
3
4
5
6
7
8
9
10
11
12
13
A
a
1
2
3
4
5
6
B
C
D
E
F
G
less than
greater than
between
less than
between
less than
b
0.5
0
c
0.025
1.96
H
I
J
K
=NORMSDIST(C2)
1.57
1.84
1.57
1.57
-1.57
-1.57
and
or greater than
and
or greater than
1.84
1.84
1.84
1.84
0.9418
0.0329
0.0253
0.9747
0.9089
0.0911
=1-NORMSDIST(C3)
=NORMSDIST(E4)-NORMSDIST(C4)
=NORMSDIST(C5)+(1-NORMSDIST(E5))
=NORMSDIST(E6)-NORMSDIST(C6)
=NORMSDIST(C7)+(1-NORMSDIST(E7))
d
0.6826
=NORMSINV(B9)
=NORMSINV(1-B11)
=NORMSINV(0.5+(B13/2))
1.000
23. Given a standardized normal distribution (with a mean of 0 and a standard
deviation of 1), determine the following probabilities.
(a)
P(Z > +1.34)
(b)
P(Z < +1.17)
(c)
P(0 < Z < +1.17)
(d)
P(Z < -1.17)
(e)
P(-1.17 < Z < +1.34)
(f)
P(-1.17 < Z < -0.50)
1
2
3
4
5
6
A
a
b
c
d
e
f
B
greater than
less than
between
less than
between
between
C
D
1.34
1.17
0 and
-1.17
-1.17 and
-1.17 and
E
F
=1-NORMSDIST(C1)
=NORMSDIST(C2)
1.17 =NORMSDIST(E3)-NORMSDIST(C3)
=NORMSDIST(C4)
1.34 =NORMSDIST(E5)-NORMSDIST(C5)
-0.5 =NORMSDIST(E6)-NORMSDIST(C6)
G
0.0901
0.8790
0.3790
0.1210
0.7889
0.1875
24. Given a normal distribution with  = 100 and  = 10,
(a)
what is the probability that
(i)
X > 75?
z


X

75  100
10
 2.5
Now we look up this value for z in the z-table, and find that it corresponds to a
probability of 0.4938. We add this to 0.5 and get 0.9938
Managerial Statistics
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Prof. Juran
(ii)
X < 70?
z


X

70  100
10
 3
Now we look up this value for z in the z-table, and find that it corresponds to a
probability of 0.4987. We subtract this from 0.5 and get 0.0013.
(iii)
75 < X < 85?
z


X

75  100
10
 2.5
z


X

85  100
10
 1.5
In the z-table, we see that the probabilities are 0.4938 (for 2.5 standard deviations
from the mean) and 0.4332 (for 1.5 standard deviations). We subtract 0.4938 –
0.4332 = 0.0606
(b)
10% of the values are less than what X value?
This is the same as asking for a value that has 40% probability between it and the
mean. We look up in the middle of the z-table for a probability near 0.4000 and
find that when z = 1.28, the corresponding probability is 0.3997. In our case, we
want z = -1.28 (because the value we want will be below the mean).
Now, we plug in this value of z and solve for X:

- 1.28

- 12.8
 X  100
87.2
Managerial Statistics
X
-1.28

X  100
10
X
864
Prof. Juran
(c)
80% of the values are between what two X values (symmetrically
distributed around the mean)?
In Part (b) we figured out that 40% of the distribution is between the mean and
1.28 standard deviations. Here, we will construct an interval that extends 1.28
standard deviations on either side of the mean; each side will contain 40%
probability and the total probability content of the interval will be 80%
 100  1.2810 
  1.28
 100  12.8
or 87.2, 112.8
(d)
70% of the values will be above what K value?
If 70% of the values are above K, then 30% of the values will be below K and 20%
of the values will be between K and the mean. We look in the body of the z-table
for a probability near 0.2000 and find that 0.1985 corresponds to 0.52 standard
deviations from the mean.
   0.52
K
 100  0.5210 
 100  5.2
 94.8
Here’s how to do the whole thing in Excel:
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
mean
stdev
(a) what is the probability that
(1) X>
(2) X <
(3)
(b) 10% of the values are less than what X value?
B
100
10
C
D
E
75
0.9938
70
0.0013
75 <X< 85 0.0606
10%
87.2
(c) 80% of the values are between what two X values?
lower 10%
upper 90%
87.2
112.8
(d) 70% of the values will be above what K value?
Managerial Statistics
70%
94.8
865
F
G
H
I
J
K
L
=1-NORMDIST(B5,$B$1,$B$2,1)
=NORMDIST(B6,$B$1,$B$2,1)
=NORMDIST(D7,$B$1,$B$2,1)-NORMDIST(B7,$B$1,$B$2,1)
=NORMINV(B9,$B$1,$B$2)
=NORMINV(B12,$B$1,$B$2)
=NORMINV(B13,$B$1,$B$2)
=NORMINV(1-B15,$B$1,$B$2)
Prof. Juran
25. Toby's Trucking Company determined that on an annual basis, the distance
traveled per truck is normally distributed with a mean of 50.0 thousand miles
and a standard deviation of 12.0 thousand miles.
(a)
What proportion of trucks can be expected to travel between 34.0 and 50.0
thousand miles in the year?
z

X

34 ,000  50 ,000

12 ,000
 1.33
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Prof. Juran
z

X

50 ,000  50 ,000

12 ,000
 0.00
In the z-table, we see that the probabilities are 0.4082 (for 1.33 standard
deviations from the mean) and 0.0000 (for 0.00 standard deviations). We subtract
0.4082 – 0.0000 = 0.4082.
(b)
What is the probability that a randomly selected truck travels between
34.0 and 38.0 thousand miles in the year?
z

X

34 ,000  50 ,000

12 ,000
 1.33
z

X

38,000  50 ,000

12 ,000
 1.00
In the z-table, we see that the probabilities are 0.4082 (for 1.33 standard
deviations from the mean) and 0.3413 (for 1.00 standard deviations). We subtract
0.4082 – 0.3413 = 0.0669. (Using Excel, we would get a more precise 0.06744.)
(c)
What percentage of trucks can be expected to travel either below 30.0 or
above 60.0 thousand miles in the year?
z

X

30 ,000  50 ,000

12 ,000
 1.67
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z

X

60 ,000  50 ,000

12 ,000
 0.83
In the z-table, we see that the probabilities are 0.4525 (for 1.67 standard
deviations from the mean) and 0.2967 (for 0.83 standard deviations). We add
(0.5000 - 0.4525) + (0.5000 - 0.2967) = 0.0475 + 0.2033 = 0.2508.
(d)
How many of the 1,000 trucks in the fleet are expected to travel between
30.0 and 60.0 thousand miles in the year?
From Part (c) we have the two z-values of –1.67 and 0.83. This time we are
looking for the probability between these two values.
(0.4525) + (0.2967) = 0.7492
Out of 1,000 trucks, we expect 749 will travel between 30,000 and 60,000 miles.
(e)
How many miles will be traveled by at least 80% of the trucks?
We could restate this question as “What value has only 20% of the values below
it?” The value will lie below the mean, and correspond to a probability between
it and the mean of 30%. In the middle of the z-table, we look for a probability
near 30%, and find that 0.2995 corresponds to a z-value of -0.84.
 0.84
 0.84

X

X  50 ,000

12 ,000
 0.84 * 12 ,000   50 ,000  X
39,920
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X
Prof. Juran
(f)
What will your answers be to (a)-(e) if the standard deviation is 10.0
thousand miles?
One of the nice things about Excel is that you can answer a question like this one
simply by changing the cell that has the standard deviation. Here is the
spreadsheet set up with the original value (note that some of the answers are
slightly different because of rounding error):
1
2
3
4
5
6
7
8
9
10
11
A
mean
stdev
B
50,000
12,000
C
D
E
F
G
H
I
J
K
L
=NORMSDIST(G4)-NORMSDIST(F4)
=(C4-$B$1)/$B$2
=NORMSDIST(G5)-NORMSDIST(F5)
a
b
c
d
between
between
below
between
e
greater than
34,000
34,000
30,000
30,000
and
and
or above
and
out of
50,000
38,000
60,000
60,000
1,000
80%
-1.33
-1.33
-1.67
-1.67
0.00
-1.00
0.83
0.83
-0.84
0.4088
0.0674
0.2501
0.7499
750
39,901
=(1-NORMSDIST(G6))+NORMSDIST(F6)
=NORMSDIST(G7)-NORMSDIST(F7)
=H7*E8
=NORMSINV(1-C9)
=B1+(F9*B2)
Cell B2 has the standard deviation. We change that to 10,000 and Excel does all
the rest:
1
2
3
4
5
6
7
8
9
10
11
A
mean
stdev
B
50,000
10,000
C
D
E
F
G
H
I
J
K
L
=NORMSDIST(G4)-NORMSDIST(F4)
=(C4-$B$1)/$B$2
=NORMSDIST(G5)-NORMSDIST(F5)
a
b
c
d
e
between
between
below
between
greater than
34,000
34,000
30,000
30,000
and
and
or above
and
out of
50,000
38,000
60,000
60,000
1,000
80%
-1.60
-1.60
-2.00
-2.00
0.00
-1.20
1.00
1.00
-0.84
0.4452
0.0603
0.1814
0.8186
819
41,584
=(1-NORMSDIST(G6))+NORMSDIST(F6)
=NORMSINV(1-C9)
=NORMSDIST(G7)-NORMSDIST(F7)
=H7*E8
=B1+(F9*B2)
26. Plastic bags used for packaging produce are manufactured so that the
breaking strength of the bag is normally distributed with a mean of 5 pounds per
square inch and a standard deviation of 1.5 pounds per square inch.
(a)
What proportion of the bags produced have a breaking strength of
(i)
between 5 and 5.5 pounds per square inch?
P5.0  X  5.5 
5.5   
 5.0  
 P
Z

 
 
5.5  5.0 
 5.0  5.0
 P
Z

1.5 
 1.5
 P0.00  Z  0.33 
 0.1293  0.0000
 0.1293
Managerial Statistics
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Prof. Juran
(ii)
between 3.2 and 4.2 pounds per square inch?
P3.2  X  4.2 
4.2   
 3.2  
 P
Z

 
 
4.2  5.0 
 3.2  5.0
 P
Z

1.5 
 1.5
 P 1.20  Z  0.53 
 0.3849  0.2019
 0.1830
(iii)
at least 3.6 pounds per square inch?
P  3 .6  X 
 3. 6  

 P
 Z



 3.6  5.0

 P
 Z
 1.5

 P 0.93  Z 
 0.5000  0.3238
 0.8238
(iv)
less than 3.17 pounds per square inch?
PX  3.17 
3.17   

 P Z 




3.17  5.0 

 P Z 

1.5


 PZ  1.22 
 0.5000  0.3888
 0.1112
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Prof. Juran
(b)
Between what two values symmetrically distributed around the mean will
95% of the breaking strengths fall?
A 95% interval around the mean contains 47.5% probability on either side of the
mean, a probability that corresponds to 1.96 standard deviations. Therefore, the
lower and upper limits on this interval will be:
 5  1.961.5 
  1.96
or
 5  2.94
2.06, 7.94 
Here is a spreadsheet version of Parts (a) and (b):
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
C
D
E
F
G
H
I
5
1.5
between
between
at least
less than
5 and
3.2 and
3.6
3.17
5.5
4.2
0.00
-1.20
-0.93
-1.22
J
K
L
=NORMSDIST(G5)-NORMSDIST(F5)
=(C5-$C$1)/$C$2
a
1
2
3
4
=NORMSDIST(G6)-NORMSDIST(F6)
0.33 0.1306
-0.53 0.1818
0.8247
0.1112
=1-NORMSDIST(F7)
=NORMSDIST(F8)
=NORMSINV(0.5+(C11/2))
=F11*C2
b confidence interval
(c)
95%
1.96 plus/minus
lower limit
2.06
upper limit
7.94
2.94
=C1-H11
=C1+H11
What will your answers be to (a) and (b) if the standard deviation is 1.0
pound per square inch?
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
B
mean
standard deviation
B
mean
standard deviation
C
D
E
F
G
H
I
5
1
between
between
at least
less than
5 and
3.2 and
3.6
3.17
5.5
4.2
0.00
-1.80
-1.40
-1.83
K
L
=NORMSDIST(G5)-NORMSDIST(F5)
=(C5-$C$1)/$C$2
a
1
2
3
4
J
0.50 0.1915
-0.80 0.1759
0.9192
0.0336
=NORMSDIST(G6)-NORMSDIST(F6)
=1-NORMSDIST(F7)
=NORMSDIST(F8)
=NORMSINV(0.5+(C11/2))
=F11*C2
b confidence interval
Managerial Statistics
95%
1.96 plus/minus
lower limit
3.04
upper limit
6.96
871
1.96
=C1-H11
=C1+H11
Prof. Juran
27. A statistical analysis of 1,000 long-distance telephone calls made from the
headquarters of Johnson & Shurgot Corporation indicates that the length of these
calls is normally distributed with  = 240 seconds and  = 40 seconds.
Here is an Excel spreadsheet to answer this question, using the NORMDIST and
NORMINV functions. They are nice for this sort of situation, because they don’t
require any z calculations.
A
1
2
3
4
5
6
7
8
9
10
(a)
(a)
(b)
(c)
(d)
(e)
B
mean
stdev
C
240
40
D
E
F
G
H
I
J
K
L
M
=NORMDIST(C4,$C$1,$C$2,1)
less than 180
0.0668
=NORMDIST(E5,$C$1,$C$2,1)-NORMDIST(C5,$C$1,$C$2,1)
between 180 and
300
0.8664
less than 180 or more than 300
0.1336 =NORMDIST(C6,$C$1,$C$2,1)+(1-(NORMDIST(E6,$C$1,$C$2,1)))
between 110 and
180
0.0662
1%
What
of all
is calls
the length
are shorter
of a particular call
146.95
if only 1% of =NORMDIST(E7,$C$1,$C$2,1)-NORMDIST(C7,$C$1,$C$2,1)
all calls are shorter?
=NORMINV(B8,C1,C2)
What percentage of these calls lasted less than 180 seconds?
6.68%
(b)
What is the probability that a particular call lasted between 180 and 300
seconds?
86.64%
(c)
How many calls lasted less than 180 seconds or more than 300 seconds?
13.36% of 1,000 is about 134 calls.
(d)
What percentage of the calls lasted between 110 and 180 seconds?
6.62%
(e)
What is the length of a particular call if only 1% of all calls are shorter?
146.95 seconds.
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Prof. Juran