SOLUCIONARIO

Chapter 19

More on Variation and Decision Making Under Risk

Solutions to Problems

19.1 (a) Continuous (assumed) and uncertain – no chance statements made.

(b) Discrete and risk – plot units vs. chance as a continuous straight line between 50 and 55 units.

(c) 2 variables: first is discrete and certain at $400; second is continuous for  $400, but uncertain (at this point). More data needed to assign any probabilities.

(d) Discrete variable with risk; rain at 20%, snow at 30%, other at 50%.

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.

19.3 (a) N is discrete since only specific values are mentioned; i is continuous from 0 to 12.

(b) The P(N), F(N), P(i) and F(i) are calculated below.

N 0 1 2 3 4

P(N) .12 .56 .26 .03 .03

F(N) .12 .68 .94 .97 1.00

i 0-2 2-4 4-6 6-8 8-10 10-12

P(i) .22 .10 .12 .42 .08 .06

F(i) .22 .32 .44 .86 .94 1.00

(c) P(N = 1 or 2) = P(N = 1) + P(N = 2)

= 0.56 + 0.26 = 0.82

or

F(N  2) – F(N  0) = 0.94 – 0.12 = 0.82

P(N  3) = P(N = 3) + P(N  4) = 0.06

(d) P(7%  i  11%) = P(6.01  i  12)

= 0.42 + 0.08 + 0.06 = 0.56

or

F(i  12%) – F(i  6%) = 1.00 – 0.44

= 0.56

19.4 (a) $ 0 2 5 10 100

F($) .91 .955 .98 .993 1.000

The variable $ is discrete, so plot $ versus F($).

(b) E($) = $P($) = 0.91(0) + ... + 0.007(100)

= 0 + 0.09 + 0.125 + 0.13 + 0.7

= $1.045

(c) 2.000 – 1.045 = 0.955

Long-term income is 95.5cents per ticket

19.5 (a) P(N) = (0.5)N N = 1,2,3,...

N 1 2 3 4 5 etc.

P(N) 0.5 0.25 0.125 0.0625 0.03125

F(N) 0.5 0.75 0.875 0.9375 0.96875

Plot P(N) and F(N); N is discrete.

P(L) is triangular like the distribution in Figure 19-5 with the mode at 5.

f(mode) = f(M) = 2 = 2

5-2 3

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

5-2

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

19.6 First cost, P

PP = first cost to purchase

PL = first cost to lease

Use the uniform distribution relations in Equation [19.3] and plot.

f(PP) = 1/(25,000–20,000) = 0.0002

f(PL) = 1/(2000–1800) = 0.005

Salvage value, S

SP is triangular with mode at $2500.

The f(SP) is symmetric around $2500.

f(M) = f(2500) = 2/(1000) = 0.002 is the probability at $2500.

There is no SL distribution

AOC

f(AOCP) = 1/(9000–5000) = 0.00025

f(AOCL) is triangular with:

f(7000) = 2/(9000–5000) = 0.0005

Life, L

f(LP) is triangular with mode at 6.

f(6) = 2/(8-4) = 0.5

The value LL is certain at 2 years.

19.7 (a) Determine several values of DM and DY and plot.

DM or DY f(DM) f(DY)

0.0 3.00 0.0

0.2 1.92 0.4

0.4 1.08 0.8

0.6 0.48 1.2

0.8 0.12 1.6

1.0 0.00 2.0

f(DM) is a decreasing power curve and f(DY) is linear.

(b) Probability is larger that M (mature) companies have a lower debt percentage and that Y (young) companies have a higher debt percentage.

19.8 (a) Xi 1 2 3 6 9 10

F(Xi) 0.2 0.4 0.6 0.7 0.9 1.0

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

or

P(X = 6, 9 or 10) = 0.1 + 0.2 + 0.1 = 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

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

19.9 Plot the F(Xi) from Problem 19.8 (a), assign the RN values, use Table 19.2 to obtain 25 sample X values; calculate the sample P(Xi) values and compare them to the stated probabilities in 19.8.

(Instructor note: Point out to students that it is not correct to develop the sample F(Xi) from another sample where some discrete variable values are omitted).

19.10 (a) X 0 .2 .4 .6 .8 1.0

F(X) 0 .04 .16 .36 .64 1.00

Take X and p values from the graph. Some samples are:

RN X p

18 .42 7.10%

59 .76 8.80

31 .57 7.85

29 .52 7.60

(b) Use the sample mean for the average p value. Our sample of 30 had p = 6.3375%; yours will vary depending on the RNs from Table 19.2.

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

0.30 G=A

0.40 G=B

P(G = g) = 0.20 G=C

0.10 G=D

0.00 G=F

0.00 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

...