Question

T6.4. A certain vending machine offers $20$-ounce bottles of soda for $\backslash (1.50.$The number of bottles$X$bought from the machine on any day is a random variable with mean $50$and standard deviation $15.$Let the random variable $Y$equal the total revenue from this machine on a given day. Assume that the machine works properly and that no sodas are stolen from the machine. What are the mean and standard deviation of $Y$?
(a) $\mu Y=\backslash )1.50,\sigma Y=\backslash (22.50$
(b) $\mu Y=\backslash )1.50,\sigma Y=\backslash (33.75$
(c)$\mu Y=\backslash )75,\sigma Y=\backslash (18.37$
(d)$\mu Y=\backslash )75,\sigma Y=\backslash (22.50$
(e) $\mu Y=\backslash )75,\sigma Y=\$33.75$

Short answer

The mean and standard deviation is $\mu Y=\$75,\sigma Y=\$22.50$. So, the option(d) is correct.

Step-by-step solution

Given information

The number of bottles $X$ is a random variable with mean $50$and standard deviation $15$. Let the random variable $Y$equal the total revenue from the machine.

Explanation

Given: ${\mu }_X=50$, and ${\mu }_X=50$
The total revenue is the income of $\$1.50$per bottle multiplied by the number of bottles $X:\$Y=1.50X$
Properties mean and standard deviation:

$$\begin{gathered} {\mu }_{aX+b}=a{\mu }_X+b{\sigma }_{aX+b} \\ =\left|a\right|{\sigma }_X \end{gathered}$$
The mean and standard deviation of the total revenue is then:$$\begin{gathered} {\mu }_Y={\mu }_{1.5X} \\ =1.5{\mu }_X \\ =1.5\left(50\right) \\ =\mathrm{\$}75 \end{gathered}$$

$$\begin{gathered} {\sigma }_Y={\sigma }_{1.5X} \\ =1.5{\sigma }_X \\ =1.5\left(15\right) \\ =\mathrm{\$}22.50 \end{gathered}$$

Hence, option (d) $\mu Y=\$75,\sigma Y=\$22.50$is correct.