Chapter 6, Problem T6.5SPT

To determine

To find: Mean and standard deviation of Y.

Answer to Problem T6.5SPT

Option d is the correct option.

Explanation of Solution

Given information:

X is the number of bottles bought from the machine

Y is the total revenue earned from this machine on a day.

Mean of X $=\text{ 5}0$

Standard deviation of X $=\text{ 15}$

Price of 20 ounces bottles of soda $=\$\text{1}.\text{5}0.$

Concept used:

  $\begin{array}{l}\text{E }\left(\text{aX }+\text{ b}\right)=\text{ aE}\left(\text{X}\right)+\text{ b}\\ \text{V }\left(\text{aX }+\text{ b}\right)={\text{ a}}^{\text{2}}\text{V }\left(\text{X}\right)\end{array}$

Calculation:

  $\text{Y }=\text{ 1}.\text{5}0\text{X}$

Mean of Y

  $\begin{array}{l}\text{E }\left(\text{Y}\right)=\text{ E}\left(\text{1}.\text{5}0\text{X}\right)\\ \text{E }\left(\text{Y}\right)=\$\text{1}.\text{5}0\text{E }\left(\text{X}\right)\\ \text{E }\left(\text{Y}\right)=\$\text{1}.\text{5}0\times \text{50=\$75}\end{array}$

Standard deviation of Y

  $\begin{array}{l}\text{Sd }\left(\text{Y}\right)=\text{ Sd}\left(\text{1}.\text{5}0\text{X}\right)\\ \text{Sd }\left(\text{Y}\right)=\$\text{1}.\text{5}0\text{Sd }\left(\text{X}\right)\\ \text{Sd }\left(\text{Y}\right)=\$\text{1}.\text{5}0\times 15=\$22.50\end{array}$

So,

  ${{\displaystyle \mu }}_y=75\text{ and }{{\displaystyle \sigma }}_y=22.50$