91Ó°ÊÓ

The number of bottles purchased from the machine is $X$.

The total revenue earned by this machine in a day is $Y$.

The mean of $X=50$

$X=15$is the standard deviation

The cost of a 20-ounce bottle of soda is $1.50.

The following concept was used:

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

Consider that

$Y=1.50X$

Mean of Y

localid="1654252088340" $$\begin{gathered} \mathrm{E}\left(\mathrm{Y}\right)=\mathrm{E}(1.50\mathrm{X}) \\ \mathrm{E}\left(\mathrm{Y}\right)=\$1.50\mathrm{E}\left(\mathrm{X}\right) \\ E\left(Y\right)=\$1.50×50=\$75 \end{gathered}$$

Standard deviation of $Y$

$\mathrm{Sd}\left(\mathrm{Y}\right)=\mathrm{Sd}(1.50\mathrm{X})$

$Sd\left(Y\right)=\$1.50Sd\left(X\right)$

localid="1654252035154" $\mathrm{Sd}\left(\mathrm{Y}\right)=\$1.50×15=\$22.50$

So,

${\mathrm{μ}}_y=75\text{and}{\mathrm{σ}}_y=22.50$

Therefore option (d) is the correct option.

a. The mean and standard deviation of$Y$will not be ${\mathrm{μ}}_Y=\$1.50,{\mathrm{σ}}_Y=\$22.50$

b. The mean and standard deviation of Y will not be ${\mathrm{μ}}_Y=\$1.50,{\mathrm{σ}}_Y=\$33.75$

c. The mean and standard deviation of Y will not be ${\mathrm{μ}}_Y=\$75,{\mathrm{σ}}_Y=\$18.37$

e. The mean and standard deviation of Y will not be ${\mathrm{μ}}_Y=\$75,{\mathrm{σ}}_Y=\$33.75$