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

Short Answer

Expert verified

The mean and standard deviation is μY=$75,σY=$22.50. So, the option(d) is correct.

Step by step solution

01

Given information

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

02

Explanation

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

μaX+b=aμX+bσaX+b=|a|σX
The mean and standard deviation of the total revenue is then:μY=μ1.5X=1.5μX=1.5(50)=$75

σY=σ1.5X=1.5σX=1.5(15)=$22.50

Hence, option (d) μY=$75,σY=$22.50is correct.

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