A grocery chain runs a prize game by giving each customer a ticket that may win a prize when the box is scratched off. Printed on the ticket is a dollar value ($500, $100, $25) or the statement "This ticket is not a winner." Monetary prizes can be redeemed for groceries at the store. Below is the probability distribution of the amount won on a randomly selected ticket. Which of the following are the mean and standard deviation, respectively, of the winnings? * 2 points Captionless Image $156.25 $53.85 $15.00, $2900.00 $15.00, $26.93 $15.00, $53.85 $156.25, $26.93

Respuesta :

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]

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