﻿ Econ 900 Solutions: Homework # 2, Dr. Usip, Economics

## Econ. 6900: Solutions, Homework # 2       Dr. Usip

Part I: Probability Problems
1. A survey of subscribers to Forbes showed that 72% have investment in money market funds (MMFs) and 36.4% have investments in certificates of deposit (CDs) ( Forbes 1993 Subscriber Study).  If 20% have investments in both MMFs and CDs

Let MMFs denote the event that a randomly selected subscriber has investment in money market funds - with    P(MMFs) = .72.
Let CDs denote the event that a randomly selected subscriber has investment in certificates of deposits - - with P(CDs) = .364
Let MMF and CDs denote the event that a randomly selected subscriber has investment in both the money market funds and the CDs - with P(MMFs and CDs) = .20
a) what is the probability that a subscriber has investments in either MMFs or CDs?
Ans
: P(She/he holds either MMFs or CDs or both) = P(MMFs) + P(CDs) - P(MMFs and CDs) = .884

b) what is the probability that a subscriber does not have investments in either MMFs or CDs?
Ans: P(She/ he has no investment in either MMFs or CDs) = 1- P(She/he does have ...) = 1 - .884 = .116

2.
Text # 39, p. 148

Ans:
a) A and B are not mutually exclusive events because there is not enough information to determine P(A and B).
b) B and C are mutually exclusive events assuming because both could not be determined as the primary cause of death since the occurrence of death by cancer precludes the occurrence of death by heart disease and vice versa. Thus P(B or C or Both) = P(B) + P(C) - P(B and C) = .25 + .20 - 0 = .45
c) Let Cc denote the event that a person dies from causes other than cancer. Thus, P(Cc) = 1- P(C) = 1- .20 = .80

Part I1:  Expected Value problem
The corporate planning group of AB&C is considering two investment alternatives for expanding its telephone service. Investment A limits its geographic expansion to three big Midwest cities, while investment B covers five major cities on the East Coast. The net profits for identical periods and probabilities of success for investments A and B are given in the following table:

 PROBABILITIES OF RETURN NET PROFITS (in \$000) Investment A Investment B 8,000 0.0 0.1 9,000 0.3 0.2 10,000 0.4 0.4 11,000 0.3 0.2 12,000 0.0 0.1

a.) Which investment option yields a higher expected return?
Ans: Let X denote the returns from investment option A
Let Y denote the returns from investment option B
E(X) = \$10,000 and E(Y) = \$10,000   using the expected value formula. On the basis of the expected value, it is not clear which option should be selected. The variance must thus be used to assess the risk associated with each option in terms of the variability about the respective means of the two investment options.

b.) Compute the variance of each investment alternative. Can you make a decision on which
investment alternative is better, given this additional information?
Ans: Var(x) = 600,000 and Var(Y) = 1200,000. Thus, A is a better option to select since the variability about the expected return of \$10,000 is smaller.

Part III: Binomial Distribution problem
Text # 33, p. 196

Ans: Let X denote the number of wells that yield oil, and 'n' denote the number of wells drilled (Bernoulli trails) such that n = 12 wells, and P = .15 is the probability of finding oil each time a well is drilled. Thus, X is a Binomial RV that takes on the following values: 0, 1, 2, ..., 12.
a) P(X =12) = f(12) = .0000 from table 5, p. A-18.  This means that it is impossible to drill 12 wells and find oil in all of them.

b) P(X = 0) = f(0) = .1422 this suggests not finding oil is highly unlikely if 12 well are drilled.

c) P(X=1) = f(1) = .3012 suggests a relatively likely occurrence of finding oil in one of the 12 wells.

d) P(A venture will be profitable) implies P(X takes on a value at the least 3) =  1 - ( X takes on a value at the most 2) = 1 - [f(0) + f(1) + f(2)] = 1- [.1422+.3012+.2924] =  1- .7358 = .2642 which is not very encouraging.

Part IV: Uniform Probability Distribution Problem
The decision Dilemma states that the average U. S. household spends \$2100 a year on all types of insurance. Suppose the figures are uniformly distributed between the values \$400 and \$3800.
a) What is the random variable of interest X?
Ans: X denotes the household spending on all types of insurance.
b) What is the probability distribution of X denoted as f(X)?
Ans: f(X) = 1/(3800-400) = 1/3400 = .000294
c) What is the standard deviation of X?
Ans: Ó = (3800-400)/{square root of 12} = \$981.524
d) What proportion of households spend
i) more than \$3000 a year on insurance.
Ans: P(X > 3000) = (3800-3000)xf(X) = 800/3400 = .23529 = .2353
ii) between \$700 and \$1500 a year on insurance?
Ans: P(700 < Xi < 1500) =  (1500 - 700)/3400 = .23529 = .2353

Part V: Normal Probability Distribution Problem
Text # 19, p. 232
Ans: Let X denote annual housing provisions of college presidents such that it has a normal probability distribution with the mean µ = \$26,234 and the standard deviation ó = \$5000.
a) P(X > 35,000) = P(Zi > ZLL) where ZLL= (35,000-26,234)/5000 = 1.75. Thus P(Zi > 1.75) = .5000 - Area from 0 to 1.75 = .5000-.4599 = .0401. Thus, there is a 4% chance that a college president will receive over \$35,000 in annual housing provisions.

b) P(X < 20.000) = P(Zi < ZUL) where ZUL= (20,000 - 26234)/5000 = -1.25. Thus, P(Zi < -1.25) =  .5000 - Area from -1.25 to 0 = .5000 - .3944 = .1056. Thus, there is a 10.56% chance that a college president will receive less than \$20,000 in annual housing provisions.

c) Note that we are asked to find the value of X corresponding to the top 10% college presidents. Thus, given P(X > XLL) = .10 our task is to find the value XLL from the relation ZLL = (XLL- µ)/ó =  +1.28 = (XLL- 26,234)/5000. Solving for the unknown yields the result XLL= 26,234 + 1.28x5000 = \$32,634.  Thus, the top 10% of college presidents receive \$32,634 or more in annual housing provisions.

Part VI: Group Projects from the Manual: Use the Computer
A: Probability Problems
Exercise # 1.2, p. 44
Exercise # 1.4, p. 44
Exercise # 1.6, p. 44
B: Binomial Distribution Problems
Exercise # 1, p. 48.
Ans:

Exercise # 2, p. 48.
Exercise # 4, p. 48
Exercise # 6, p. 48
C: Normal Distribution Problems
Exercise #2, p. 49
Exercise #4, p. 49