Instructor’s Manual
Chapter 3
31
Chapter 3
I
Chapter Outline
3.1
Continuous Random Variables
3.2
The Probability Density Function
3.3
The Cumulative Distribution Function.
 The uniform distribution is introduced.
3.4
The Normal Distribution
 General definition of this distribution.
 Properties of normal distribution
 Presentation of several examples of normal random variables and variables
with non-normal distributions.
3.5
Computing Probabilities for the Normal Distribution
3.6
Sums of Normally Distributed Random Variables
3.7
The Central Limit Theorem
 Examples of applications of this theorem.
 Discussion on the normal approximation to the binomial
3.8
Summary
II
Teaching Tips
1.
The discussion on the normal approximation to the binomial and the central limit
theorem can be enhanced by using the Crystal Ball run option to show in class
how the binomial distribution approaches a bell-shaped curve as the sample size
increases. Other distributions might also be used.
2.
A quick in-class survey of the students height and weight can be used to illustrate
how these two measurements follow normal distributions and how common is
such distribution in nature
III
Answers to Chapter Exercises
3.1
Let X be the site of the traffic incident. X is uniformly distributed between 0 and
30 miles.
(a) P(Travel more than 10 miles) = P(X > 10) = 1 - P(X < 10) = 1 - 10/30 = 2/3.
Manual to accompany Data, Models & Decisions: The Fundamentals of Management Science by Bertsimas and Freund. Copyright
2000, South-Western College Publishing. Prepared by Manuel Nunez, Chapman University.
Instructor’s Manual
Chapter 3
32
(b) P(Travel less than 10 miles with respect to station location) = P(|X - 15| < 10)
= P(5 < X < 25) = P(X < 25) - P(X < 5) = 25/30 - 5/30 = 2/3. Hence, P(Travel
more than 10 miles with respect to station location) = P(|X - 15| > 10) = 1 P(|X - 15| < 10) = 1/3.
3.2
Let T1, T2, and T3 be the running times of Donovan, Frankie, and Ato,
respectively. Each of T1, T2, and T3 is uniformly distributed between 9.75 and
9.95 seconds.
(a) P(Donovan beats record) = P(T1 < 9.86) = (9.86 - 9.75)/(9.95 - 9.75) = 0.55.
(b) Assuming that the running times are independent of each other, P(Winning
time beats record) = P(min{T1, T2, T3} < 9.86) = 1 - P(min{T1, T2, T3} >
9.86) = 1 - (P(T1 > 9.86))3 = 1 - (0.55)3 = 0.83.
(c) It may not be valid because runners lagging behind the leader will try to catch
up and so, their speed will depend on the speed of the leader.
3.3
Let D denote the demand for the kitchen cleanser and let d denote the inventory
level. We want to find d such that P(Stocking out) = P(D > d) = 0.025, or
equivalently, P(D < d) = 0.975. Using the standard normal tables P(Z < z) = 0.975
implies z = 1.96. Therefore, d = 2,550 + (1.96)(415) = 3,363.4 bottles.
3.4
Let X denote the number of miles a pump operates before becoming ineffective.
(a) P(Replace a pump) = P(X < 50,000) = P(Z < (50,000-61,000)/9,000) = P(Z < 1.22) = 0.1112.
(b) P(Pump fails at exactly 50,000) = P(X = 50,000) = 0.
(c) P(Pump fails between 42,000 and 57,000 miles) = P(42,000 < X < 57,000) =
P(-2.11 < Z < -0.44) = 0.3300 – 0.0174 = 0.3126.
(d) We want to find x such that P(X < x) = 0.80. Using the standard normal table
P(Z < z) = 0.80 implies z = 0.845. Therefore, x = 61,000 + (0.845)(9,000) =
68,605 miles.
3.5
Let X denote the time it takes Helen to complete a homework assignment. P(Go
to bed in time) = P(X < 4) = P(Z < (4 - 3.5)/1.2) = P(Z < 0.42) = 0.6628.
3.6
Let X denote the temperature in Boston and let Y denote the temperature in New
York.
(a) P(Boston’s average temperature above freezing) = P(X > 32) = 1 – P(X < 32)
= 1 – P(Z < (32 – 32.5)/1.59) = 1 - P(Z < -0.31) = 1 - 0.3783 = 0.6217.
(b) Notice that D = Y – X has a normal distribution with mean 35.4 – 32.5 = 2.9
and standard deviation (1.592 + 2.052) = 2.6. Hence, P(Boston’s average
temperature > New York’s) = P(X > Y) = P(D < 0) = P(Z < -2.9/2.6) = P(Z < 1.12) = 0.1314.
(c) It might not be reasonable when we consider that the two cities are close
enough as to be influenced by the same weather systems.
3.7
Let X1, X2, and X3 denote the weights of the three tomatoes in a package. Then
T = X1 + X2 + X3 is the weight of a package.
Manual to accompany Data, Models & Decisions: The Fundamentals of Management Science by Bertsimas and Freund. Copyright
2000, South-Western College Publishing. Prepared by Manuel Nunez, Chapman University.
Instructor’s Manual
Chapter 3
33
(a) E(T) = 3(4.2) = 12.6, VAR(T) = 3(1)2 = 3, and SD(T) = 3 = 1.73.
(b) P(11 < T < 13) = P((11 – 12.6)/1.73 < Z < (13 – 12.6)/1.73) = P(-0.92 < Z <
0.23) = 0.5910 - 0.1788 = 0.4122.
3.8
Let T = 0.3X + 0.7Y denote the total annual return.
(a) E(T) = E (0.3X + 0.7Y) = 0.3(7) + 0.7(13) = 11.2%.
(b) VAR(T) = VAR(0.3X + 0.7Y) = 0.3222 + 0.7282 – 2(0.3)(0.7)(2)(8)(0.4) =
29.03, SD (T) = SD(0.3X + 0.7Y) = (29.03) = 5.4%.
(c) T has a normal distribution with mean  = 11.2% and standard deviation  =
5.4%.
(d) P(10 < T < 15) = P(-0.22 < Z < 0.70) = 0.7580 – 0.4129 = 0.3451.
3.9
(a) P(Dow Jones increases more than 11%) = P(X > 11) = 1 – P(X < 11) = 1 –
P(Z < 0) = 1 – 0.5 = 0.5.
(b) P(Dow Jones decreases more than 11%) = P(X < -11) = P(Z < -1.69) =
0.0455.
(c) P(S&P 500 increase less than 15%) = P(Y < 15) = P(Z < 0.42) = 0.6628.
(d) Let T = 0.3X + 0.7Y. E(T) = (0.3)(11) + (0.7)(10) = 10.3%, VAR(T) = 0.32132
+ 0.72122 + 2(0.3)(0.7)(13)(12)(0.43) = 113.94, SD(T) = (113.94) = 10.7%.
(e) X – Y has a normal distribution with mean  = 11 – 10 = 1%, variance 2 =
132 + 122 - 2(13)(12)(0.43) = 178.84, and standard deviation  = (178.84) =
13.4%. P(X > Y) = P(X – Y > 0) = 1 – P(X – Y < 0) = 1 – P(Z < -0.07) = 1 –
0.4721 = 0.5279.
3.10
Let N denote the number of females in the group.
(a) E(N) = 100(0.522) = 52.2, VAR(N) = 100(0.522)(0.478) = 24.95, and SD(N)
= (24.95) = 5.
(b) P(N < 50)  P(Z < -0.44) = 0.33.
(c) P(45 < N < 55)  P(-1.44 < Z < 0.56) = 0.7123 – 0.0749 = 0.6374.
3.11
Let A and B denote the initial salary and the bonus, respectively.
(a) E(A + B) = $90,000 + $25,000 = $115,000.
(b) VAR(A + B) = 200002 + 50002 = 425,000,000 and SD(A + B) = $20,616.
(c) E(1.2A + B) = (1.2)($90,000) + $25,000 = $133,000.
(d) VAR(1.2A + B) = (1.2)2200002 + 50002 = 601,000,000 and SD(1.2A + B) =
$24,515.
(e) P(1.2A + B > $140,000) = P(Z > 0.29) = 1 – P(Z < 0.29) = 1 – 0.6141 =
0.3859.
3.12
Let X denote the exchange rate.
(a) P(X > 2.08) = P(Z > 0.63) = 1 – P(Z < 0.63) = 1 – 0.7357 = 0.2643.
(b) P(X < 1.85) = P(Z < -2.25) = 0.0122.
(c) P(2 < X < 2.2) = P(-0.38 < Z < 2.13) = 0.9834 – 0.3520 = 0.6314.
Manual to accompany Data, Models & Decisions: The Fundamentals of Management Science by Bertsimas and Freund. Copyright
2000, South-Western College Publishing. Prepared by Manuel Nunez, Chapman University.
Instructor’s Manual
Chapter 3
34
3.13
Let A, B, and C denote the annual rate of return of each company, respectively.
Let EX and IN denote the annual rate of return of the two portfolios, respectively.
Notice that EX = 0.7A + 0.15B + 0.15C and IN = 0.34A + 0.33B + 0.33C. EX has
a normal distribution with mean  = (0.7)(8) + (0.15)(11) + (0.15)(17) = 9.8%,
variance 2 = (0.7)2(0.5)2 + (0.15)2(6)2 + (0.15)2(20)2 = 9.93, and standard
deviation  = (9.93) = 3.15%. IN has a normal distribution with mean  =
(0.34)(8) + (0.33)(11) + (0.33)(17) = 12%, variance 2 = (0.34)2(0.5)2 +
(0.33)2(6)2 + (0.33)2(20)2 = 47.51, and standard deviation  = (47.51) = 6.9%.
P(EX > 0) = P(Z > -3.11) = 1 – P(Z < -3.11) = 1. P(IN > 0) = P(Z > -1.74) = 1 –
P(Z < -1.74) = 1 – 0.0409 = 0.9591. Therefore, the Explorer Fund has a greater
probability of not losing money.
3.14
Let A and B denote the score of each player, respectively.
(a) Notice that log(A) – log(B) has a normal distribution with mean  = 0.15 and
standard deviation  = (2)(0.205) = 0.29. Therefore, P(A > B) = P(log(A) >
log(B)) = P(log(A) – log(B) > 0) = P(Z > -(0.15)/0.29)) = P(Z > -0.52) = 1 –
P(Z < -0.52) = 1 – 0.3015 = 0.6985  0.7.
(b) Let N be the number of times that A beats B in a series of 5 games. N has a
binomial distribution with parameters n = 5 and p = 0.7 (from part (a)). P(N
>= 3) = P(N = 3) + P(N = 4) + P(N = 5) = 0.309 + 0.36 + 0.168 = 0.837.
3.15
(a) Let N be the number of packages returned from January’s sales. N has a
binomial distribution with mean  = 2500(0.10) = 250, variance 2 =
2500(0.10)(0.90) = 225, and standard deviation  = (225) = 15.
(b) Let R denote the revenue from January sales after returns. R = (2500)(500) –
500N = 1,250,000 – 500N. E(R) = 1,250,000 – 500(250) = $1,125,000,
VAR(R) = (500)2(225) = 56,250,000, and SD(R) = $7,500.
(c) P(R > $1,300,000)  P(Z > 23.33) = 0.
3.16
(a) If p = 0.5, then we must have n >= 5/0.5 = 10. If p = 0.1, then we must have n
>= 5/0.1 = 50. If p = 0.01, then we must have n >= 5/0.01 = 500. If p = 0.001,
then we must have n >= 5/0.001 = 5000.
(b) Suppose that N is a random variable and has a binomial distribution with
parameters p = 0.01 and n = 20. Hence, N has mean  = 0.2 and standard
deviation  = 0.44. P(N <= 0) = P(N = 0) = (0.01)20 = 10-40. Using the normal
approximation, P(N <= 0)  P(Z < -0.2/0.44) = P(Z < -0.45) = 0.3264.
Therefore, the approximation is off by 0.3264.
3.17
Let D denote the distance traveled.
(a) Since the mean of D is 150 feet and the net is 30 feet long, he should place the
nearest edge of the net at 150 – 15 = 135 feet from the cannon. Therefore, the
net will cover a landing between 135 and 165 feet from the cannon, which is
the interval where the given normal distribution has the most area, and hence
the most probability of landing on the net.
Manual to accompany Data, Models & Decisions: The Fundamentals of Management Science by Bertsimas and Freund. Copyright
2000, South-Western College Publishing. Prepared by Manuel Nunez, Chapman University.
Instructor’s Manual
Chapter 3
35
(b) P(135 < D < 165) = P(-1.5 < Z < 1.5) = 0.9332 - 0.0668 = 0.8664.
3.18
(a) There are two assumptions: whether a car has a faulty electrical wiring is
independent of another car having or not a faulty electrical wiring, and the
probability that a given car has a faulty electrical wiring is the same for each
car in the sample. The assumptions appear to be reasonable and, unless there
is a consistent error in the production process of this type of car, the
experiment is binomial.
(b) Let N be the number of cars having the problem in the sample. N has mean 
= 2000(0.30) = 600 and standard deviation  = (2000x0.30x0.70) = 20.5.
P(N <= 5)  P(Z < (5 – 600)/20.5) = P(Z < -29) = 0.
(c) According to (b), it is not likely that the true proportion of cars having the
problem is 0.30.
3.19
Let A and B denote the annual returns of the two stocks, respectively.
(a) If we invest a fraction p in A, then the annual return of the resulting portfolio
is T = pA + (1 – p)B = B + p(A – B). The expected return of T is E(T) = E(B)
+ pE(A – B) = 20 - 10p. Since we want E(T) = 13%, then we need 20 – 10p =
13, that is, p = 0.70. Therefore, we should invest 70% in A and 30% in B.
(b) VAR(T) = (0.70)2(4)2 + (0.30)2(10)2 – 2 (0.70)(0.30)(4)(10)(0.20) = 13.5 and
SD(T) = (13.5) = 3.7%.
(c) Let S denote the resulting portfolio. E(S) = 50(0.02)(20) = 20%, VAR(S) =
50(0.02)2(20)2 = 8, and SD(S) = 8 = 2.8%.
(d) P(18 < S < 24) = P(-0.71 < Z < 1.43) = 0.9236 – 0.2389 = 0.6847.
3.20
(a) P(Car returned) = 1 - P(Car is not defective) = 1 - (0.80)(0.90) = 0.28.
(b) Let N denote the number of cars returned for rework. E(N) = 1000(0.28) =
280 and SD(N) = (1000x0.28x0.72) = 14.2.
(c) P(N <= 200)  P(Z < -5.63) = 0.
(d) X has a binomial distribution with parameters n = 1000 and p = 0.20. Y has a
binomial distribution with parameters n = 1000 and p = 0.10.
(e) X + Y can be approximated by a normal random variable with parameters  =
1000(0.20) + 1000(0.10) = 300 and standard deviation  = (1000(0.2)(0.8) +
1000(0.1)(0.9)) = 15.8. P(X + Y <= 300)  P(Z < 0) = 0.5.
Manual to accompany Data, Models & Decisions: The Fundamentals of Management Science by Bertsimas and Freund. Copyright
2000, South-Western College Publishing. Prepared by Manuel Nunez, Chapman University.