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Chapter 6: Q. 4 (page 409)

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

Short Answer

Expert verified

The mean and standard deviation is 渭Y=$75,蟽Y=$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: 渭X=50, and 渭X=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:

渭aX+b=a渭X+b蟽aX+b=|a|蟽X
The mean and standard deviation of the total revenue is then:渭Y=渭1.5X=1.5渭X=1.5(50)=$75

蟽Y=蟽1.5X=1.5蟽X=1.5(15)=$22.50

Hence, option (d) 渭Y=$75,蟽Y=$22.50is correct.

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