Chapter 7: Q. 2.20 (page 491)

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. 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. $$ 15.00, $2900.00$
b. $$ 15.00, $53.85$
c. $$ 15.00, $26.93$
d. $$ 156.25, $53.85$
e. $$ 156.25, $ 26.93$

Expert verified

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

Step by step solution

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,

μ = \sum x \times p (x) σ^{2} = \sum (x - μ)^{2} \times p (x) σ = \sqrt{σ^{2}}

Let's consider,

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

Variance

\begin{array}{rcl} σ^{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)

σ = \sqrt{2900} = 53.85

Therefore, the correct option is (b)

(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$ .

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