Question

A large auto dealership keeps track of sales made during each hour of the day. Let $X$= the number of cars sold during the first hour of business on a randomly selected Friday. Based on previous records, the probability distribution of $X$is as follows:

The random variable $X$has mean $\mu X=1.1$and standard deviation $\sigma X=0.943$.

Suppose the dealership’s manager receives a $500$bonus from the company for each car sold. Let $Y$ = the bonus received from car sales during the first hour on a randomly selected Friday. Find the mean and standard deviation of $Y$.

Short answer

From the given information, the required mean and standard deviation are$550$and$471.50$respectively.

Step-by-step solution

Given Information

It is given in the question that, Bonus on each day is $500$

mean is $1.1$

Standard deviation is$0.943$

Step 2: Explanation

The mean and standard deviation of $Y$can be calculated as :

localid="1649912763196" $$\begin{gathered} \mu Y=500\mu X \\ =500\left(1.1\right) \\ =550 \\ \sigma Y=500\sigma X \\ =500\left(0.943\right) \\ =471.50 \end{gathered}$$