Question

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 $(\backslash (500,\backslash )100,\backslash (25)$ or the statement "This ticket is not a winner." Monetary prizes can be redeemed for groceries at the store. Here 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?

a. $\backslash )15.00,\backslash (2900.00$

b. $\backslash )15.00,\backslash (53.85$

c. $\backslash )15.00,\backslash (26.93$

d. $\backslash )156.25,\backslash (53.85$

e. $\backslash )156.25,\$26.93$

Short answer

(b) The mean and standard deviation, respectively, of the winnings are$\$15.00,\$53.85$.

Step-by-step solution

Given information

Given,

Amount won	$\$500$	$\$100$	$\$100$	$\$0$
Probability	$0.01$	$0.05$	$0.20$	$0.74$

Below is the formulae used to get the result,

$$\begin{gathered} \mu =\sum x\times p\left(x\right) \\ {\sigma }^2=\sum (x-\mu {)}^2\times p(x) \\ \sigma =\sqrt{{\sigma }^2} \end{gathered}$$

Explanation for correct option

Let's consider,

$\mu =500\times 0.01+100\times 0.05+25\times 0.20+0\times 0.74=15$

Variance

$\begin{array}{rcl}{\sigma }^2& =& (500-15{)}^2\times 0.01+(100-15{)}^2\times 0.05+(25-15{)}^2\times 0.20+(0-15{)}^2\times 0.74\\ & =& 2900\end{array}$

The standard deviation (SD)

$\sigma =\sqrt{2900}=53.85$

Therefore, the correct option is (b)

Explanation for incorrect option

(a) The mean and standard deviation, respectively, of the winnings will not be$\$15.00,\$2900.00$.

(c)The mean and standard deviation, respectively, of the winnings will not be$\$15.00,\$26.93$.

(d)The mean and standard deviation, respectively, of the winnings will not be$\$156.25,\$53.85$.

(e)The mean and standard deviation, respectively, of the winnings will not be$\$156.25,\$26.93$.