Answer:
[tex]Mean = \$15[/tex]
[tex]Standard\ Deviation = \$53.85[/tex]
Step-by-step explanation:
Given
See attachment for probability distribution
Required
Determine the mean and the standard deviation
The mean of the distribution is:
[tex]Mean = E(x)[/tex]
and
[tex]E(x) = \sum x * P(x)[/tex]
So, we have:
[tex]Mean = \$500 * 0.01 + \$100 * 0.05 + \$25 * 0.20 + \$0 * 0.74[/tex]
[tex]Mean = \$5 + \$5 + \$5 + \$0[/tex]
[tex]Mean = \$15[/tex]
So:
[tex]E(x) = 15.00[/tex]
The standard deviation of the distribution is:
[tex]S = \sqrt{Var(x)}[/tex]
Where:
[tex]Var(x) = E(x^2) - (E(x))^2[/tex]
[tex]E(x^2) = \sum x^2 * P(x)[/tex]
So, we have:
[tex]E(x^2) = 500^2 * 0.01 + 100^2 * 0.05 + 25^2 * 0.20 +0^2 * 0.74[/tex]
[tex]E(x^2) = 2500 + 500 + 125 +0[/tex]
[tex]E(x^2) = 3125[/tex]
So:
[tex]Var(x) = E(x^2) - (E(x))^2[/tex]
Where
[tex]E(x^2) = 3125[/tex]
[tex]E(x) = 15.00[/tex]
[tex]Var(x) = 3125 - 15^2[/tex]
[tex]Var(x) = 3125 - 225[/tex]
[tex]Var(x) = 2900[/tex]
Recall that:
[tex]S =\sqrt{Var(x)[/tex]
[tex]S =\sqrt{2900[/tex]
[tex]S =53.85[/tex]