Question

One invests 100 shares of IBM stocks today. He expects that there could be five possible opening prices with the respective probabilities at 9:30 a.m. in NYSE the next day. The following table lists these possible opening prices and their respective probabilities:
Outcome 1 Outcome 2 Outcome 3 Outcome 4 Outcome 5
Possible Opening
Price of IBM, Xi $182.11 $163.88 $180.30 $216.08 $144.92
Probability, pi 13% 19% 33% 17% 18%
Let X represent the five random opening prices of IBM the next day, calculate the mean, variance, and the standard deviation of X. Make your comments on the results you obtain.

Answer 1

Answer:

$E(x) = 177.130$

$Var(x) = 484.551$

$\sigma = 22.013$

Step-by-step explanation:

Given

The attached table

Solving (a): The mean

This is calculated as:

$E(x) = \sum x * p(x)$

So, we have:

$E(x) = 182.11 * 13\% + 163.88 * 19\% + 180.30 * 33\% + 216.08 * 17\% + 144.92 * 18\%$

Using a calculator, we have:

$E(x) = 177.1297$

$E(x) = 177.130$ --- approximated

The average opening price is $177.130

Solving (b): The Variance

This is calculated as:

$Var(x) = E(x^2) - (E(x))^2$

Where:

$E(x^2) = \sum x^2 * p(x)$

$E(x^2) = 182.11^2 * 13\% + 163.88^2 * 19\% + 180.30^2 * 33\% + 216.08^2 * 17\% + 144.92^2 * 18\%$

$E(x^2) = 31859.482249$

So:

$Var(x) = E(x^2) - (E(x))^2$

$Var(x) = 31859.482249 - 177.1297^2$

$Var(x) = 31859.482249 - 31374.9306221$

$Var(x) = 484.5516269$

$Var(x) = 484.551$ --- approximated

Solving (c): standard deviation

The standard deviation is:

$\sigma = \sqrt{Var(x)}$

$\sigma = \sqrt{484.5516269}$

$\sigma = 22.0125418796$

Approximate

$\sigma = 22.013$