Chapter 5, Problem 59SE

The Barron’s Big Money Poll asked 131 investment managers across the United States about their short-term investment outlook (Barron’s, October 28, 2002). Their responses showed 4% were very bullish, 39% were bullish, 29% were neutral, 21% were bearish, and 7% were very bearish. Let x be the random variable reflecting the level of optimism about the market. Set x = 5 for very bullish down through x = 1 for very bearish.

1. a. Develop a probability distribution for the level of optimism of investment managers.
2. b. Compute the expected value for the level of optimism.
3. c. Compute the variance and standard deviation for the level of optimism.
4. d. Comment on what your results imply about the level of optimism and its variability.

To determine

Construct a probability distribution for the level of optimism of investment managers.

Answer to Problem 59SE

The probability distribution for the level of optimism of investment managersis given by,

x	$f\left(x\right)$
1	0.07
2	0.21
3	0.29
4	0.39
5	0.04

Explanation of Solution

Calculation:

The given information is that the responses of the invest managers showed 4% were very bullish, 39% were bullish, 29% were neutral, 21% were bearish and 7% were very bearish. The random variable x represents the level of optimism about the market. The random variable x takes the value 5 for very bullish and takes the value 1 for very bearish.

The corresponding probabilities are obtained by converting the percentages in to probabilities. That is, by dividing each value with 100.

The probability distribution for the random variable x can be obtained as follows:

x	f	$\frac{f}{N}$	$f\left(x\right)$
1	7	$\frac{7}{100}$	0.07
2	21	$\frac{21}{100}$	0.21
3	29	$\frac{29}{100}$	0.29
4	39	$\frac{39}{100}$	0.39
5	4	$\frac{4}{100}$	0.04
Total	100		1

To determine

Find the expected value for the level of optimism.

Answer to Problem 59SE

The expected value for thelevel of optimism is 3.12.

Explanation of Solution

Calculation:

The formula for the expected value of a discrete random variable is,

$\begin{array}{c}E\left(x\right)=\mu \\ ={\displaystyle \sum x\cdot f\left(x\right)}\end{array}$

The expected value for the random variable x is obtained using the following table:

x	f(x)	$x\cdot f\left(x\right)$
1	0.07	0.07
2	0.21	0.42
3	0.29	0.87
4	0.39	1.56
5	0.04	0.20
Total	1	3.12

Thus, the expected value for thelevel of optimism is 3.12.

To determine

Find the variance and standard deviationfor thelevel of optimism.

Answer to Problem 59SE

The variance for thelevel of optimismis 1.03.

The standard deviation for thelevel of optimismis 1.01.

Explanation of Solution

Calculation:

The formula for the variance of the discrete random variable is,

${\sigma }^2={\displaystyle \sum \left[{\left(x-\mu \right)}^2.f\left(x\right)\right]}$

The variance of the random variable x is obtained using the following table:

x	f(x)	$\left(x-\mu \right)$	${\left(x-\mu \right)}^2$	${\left(x-\mu \right)}^2.f\left(x\right)$
1	0.07	–2.12	4.49	0.31
2	0.21	–1.12	1.25	0.26
3	0.29	–0.12	0.01	0.00
4	0.39	0.88	0.77	0.30
5	0.04	1.88	3.53	0.14
Total	1	–0.60	10.07	1.03

Therefore,

${\sigma }^2=1.03$

Thus, the variance for thelevel of optimism is 1.03.

The formula for the standard deviation of the discrete random variable is,

$\sigma =\sqrt{{\displaystyle \sum \left[{\left(x-\mu \right)}^2.f\left(x\right)\right]}}$

Thus, the standard deviation is,

$\begin{array}{c}\sigma =\sqrt{1.03}\\ =1.01\end{array}$

Hence, the standard deviation for thelevel of optimism is 1.01.

To determine

Explain what the result implies about the level of optimism and its variability.

Explanation of Solution

The expected value for the level of optimism of the investment managers is 3.12 and it represents that the investment managers are somewhat optimistic. The variability for the level of optimism of the investment managers is 1.03.