Q. 20 - The Practice of Statistics for AP

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 $(\backslash (500, \backslash )100, \backslash (10)$ or the statement, “This ticket is not a winner.” Monetary prizes can be redeemed for groceries at the store. Here are the distribution of the prize values and the associated probabilities for each prize:

Which of the following are the mean and standard deviation, respectively, of the winnings?
$\backslash )15.00, \backslash (2900.00$
(b) $\backslash )15.00, \backslash (53.85$
(c) $\backslash )15.00, \backslash (26.9$

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

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

Step-by-Step Solution

Answer

The mean and standard deviation are $\$15$ and $\$53.83$. So, the correct answer is option(b).

Step 1: Given information

A grocery chain runs a prize game by giving tickets. Printed on the ticket is a dollar value $(\$500, \$100, \$10)$. To determine the mean and standard deviation of the winnings.

Step 2: Explanation

The mean is determined by:

 $$\begin{gathered} Mean=\sum x\times P\left(x\right) \\ =500(0.01)+100(0.05)+25(0.20)+0(0.74) \\ =15 \end{gathered}$$
The standard deviation is calculated by:
$$\begin{gathered} \sigma =\sqrt{\sum x^2\times P\left(x\right)-{\left(\sum x\times P\left(x\right)\right)}^2} \\ =\sqrt{\left({500}^2\times 0.01+{100}^2\times 0.05+\dots .+0^2\times 0.74\right)-(15{)}^2} \\ =\sqrt{2900} \\ =53.83 \end{gathered}$$
Hence, the mean and standard deviation are $\$15$ and $\$53.83$.

So, option (b) is correct answer.