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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.

Sagot :

MrRoyalidn

Answer:

[tex]E(x) = 177.130[/tex]

[tex]Var(x) = 484.551[/tex]

[tex]\sigma = 22.013[/tex]

Step-by-step explanation:

Given

The attached table

Solving (a): The mean

This is calculated as:

[tex]E(x) = \sum x * p(x)[/tex]

So, we have:

[tex]E(x) = 182.11 * 13\% + 163.88 * 19\% + 180.30 * 33\% + 216.08 * 17\% + 144.92 * 18\%[/tex]

Using a calculator, we have:

[tex]E(x) = 177.1297[/tex]

[tex]E(x) = 177.130[/tex] --- approximated

The average opening price is $177.130

Solving (b): The Variance

This is calculated as:

[tex]Var(x) = E(x^2) - (E(x))^2[/tex]

Where:

[tex]E(x^2) = \sum x^2 * p(x)[/tex]

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

[tex]E(x^2) = 31859.482249[/tex]

So:

[tex]Var(x) = E(x^2) - (E(x))^2[/tex]

[tex]Var(x) = 31859.482249 - 177.1297^2[/tex]

[tex]Var(x) = 31859.482249 - 31374.9306221[/tex]

[tex]Var(x) = 484.5516269[/tex]

[tex]Var(x) = 484.551[/tex] --- approximated

Solving (c): standard deviation

The standard deviation is:

[tex]\sigma = \sqrt{Var(x)}[/tex]

[tex]\sigma = \sqrt{484.5516269}[/tex]

[tex]\sigma = 22.0125418796[/tex]

Approximate

[tex]\sigma = 22.013[/tex]

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