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Selected Solutions - 19

This document provides solutions to selected problems from Chapter 19. It includes the needed or assumed information to calculate an expected value, solutions to probability distribution and expected value problems, and a description of the steps to solve a probability problem and compare the results to the given probabilities. It also demonstrates how to calculate sample variance and range from a probability distribution, and how to use a spreadsheet random number generator to simulate cash flows from a normal distribution for a capital budgeting problem.

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30 views4 pages

Selected Solutions - 19

This document provides solutions to selected problems from Chapter 19. It includes the needed or assumed information to calculate an expected value, solutions to probability distribution and expected value problems, and a description of the steps to solve a probability problem and compare the results to the given probabilities. It also demonstrates how to calculate sample variance and range from a probability distribution, and how to use a spreadsheet random number generator to simulate cash flows from a normal distribution for a capital budgeting problem.

Uploaded by

moonisq
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOC, PDF, TXT or read online on Scribd
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SOLUTIONS TO SELECTED PROBLEMS

Student: You should work the problem completely before referring to the solution.

CHAPTER 19
Solutions included for problems: 2, 5, 8, 11, 14, 17, and 20
19.2 Needed or assumed information to be able to calculate an expected value:
1. Treat output as discrete or continuous variable.
2. If discrete, center points on cells, e.g., 800, 1500, and 2200 units per week.
3. Probability estimates for < 1000 and /or > 2000 units per week.
19.5

(a)

P(N) = (0.5)N
N
P(N)
F(N)

N = 1,2,3,... is discrete
1
0.5
0.5

2
0.25
0.75

3
0.125
0.875

4
0.0625
0.9375

5
etc.
0.03125
0.96875

P(L) is a triangular distribution with the mode at 5.


f(mode) = f(M) = 2 = 2
5-2 3
F(mode) = F(M) = 5-2 = 1
5-2

19.8

(b)

P(N = 1, 2 or 3) = F(N 3) = 0.875

(a)

Xi
F(Xi)

(b)

P(6 X 10) = F(10) F(3) = 1.0 0.6 = 0.4


P(X = 4, 5 or 6) = F(6) F(3) = 0.7 0.6 = 0.1

(c)

P(X = 7 or 8) = F(8) F(6) = 0.7 0.7 = 0.0

1
0.2

2
0.4

3
0.6

6
0.7

9
0.9

10
1.0

No sample values in the 50 have X = 7 or 8. A larger sample is needed to


observe all values of X.

Chapter 19

19.11 Use the steps in Section 19.3. As an illustration, assume the probabilities that are
assigned by a student are:

P(G = g) =

0.30
0.40
0.20
0.10
0.00
0.00

G=A
G=B
G=C
G=D
G=F
G=I

Steps 1 and 2: The F(G) and RN assignment are:


RNs
0.30 G=A 00-29
0.70 G=B 30-69
F(G = g) =
0.90 G=C 70-89
1.00 G=D 90-99
1.00 G=F
-1.00 G=I
-Steps 3 and 4: Develop a scheme for selecting the RNs from Table 19-2. Assume
you want 25 values. For example, if RN1 = 39, the value of G is B. Repeat for
sample of 25 grades.
Step 5: Count the number of grades A through D, calculate the probability of each
as count/25, and plot the probability distribution for grades A through I. Compare
these probabilities with P(G = g) above.
19.14 (a) Convert P(X) data to frequency values to determine s.
X
1
2
3
6
9
10

P(X)
.2
.2
.2
.1
.2
.1

XP(X)
.2
.4
.6
.6
1.8
1.0
4.6

f
10
10
10
5
10
5

X2
1
4
9
36
81
100

Sample average: Xbar = 4.6


Sample variance: s2 = 1630 50 (4.6)2 = 11.67
49
49
s = 3.42
19.14 (cont) (b) Xbar 1s is 4.6 3.42 = 1.18 and 8.02
25 values, or 50%, are in this range.
Chapter 19

fX2
10
40
90
180
810
500
1630

Xbar 2s is 4.6 6.84 = 2.24 and 11.44


All 50 values, or 100%, are in this range.
19.17 P(N) = (0.5)N
E(N) = 1(.5) + 2(.25) + 3(.125) + 4(0.625) + 5(.03125) + 6(.015625) +
7(.0078125) + 8(.003906) + 9(.001953) + 10(.0009766) + ..
= 1.99+
The limit to the series N(0.5)N is 2.0, the correct answer.
19.20 Use the spreadsheet Random Number Generator (RNG) on the tools toolbar to
generate CFAT values in column D from a normal distribution with = $2040 and
= $500. The RNG screen image is shown below. (This tool may not be available
on all spreadsheets.)

19.20 (cont)

Chapter 19

The decision to accept the plan uses the logic:


Conclusion: For certainty, accept the plan if PW > $0 at MARR of 7% per year.
For risk, the result depends on the preponderance of positive PW values
from the simulation, and the distribution of PW obtained.

Chapter 19

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