A certain vending machine offers 20 -ounce bottles of soda for \(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 Yequal the total revenue from this machine on a randomly selected 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.localid="1654251787401" μY=\)75,σY=\(22.50

e. μY=\)75,σY=$33.75

Short Answer

Expert verified

(d) The mean and standard deviation of YisμY=$75,σY=$22.50

Step by step solution

01

Given Information

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=15is the standard deviation

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

The following concept was used:

E(aX+b)=aE(X)+bV(aX+b)=a2V(X)

02

Explanation for correct option

Consider that

Y=1.50X

Mean of Y

localid="1654252088340" E(Y)=E(1.50X)E(Y)=$1.50E(X)E(Y)=$1.50×50=$75

Standard deviation of $Y$

Sd(Y)=Sd(1.50X)

Sd(Y)=$1.50Sd(X)

localid="1654252035154" Sd(Y)=$1.50×15=$22.50

So,

μy=75andσy=22.50

Therefore option (d) is the correct option.

03

Explanation for incorrect option

a. The mean and standard deviation ofYwill not be μY=$1.50,σY=$22.50

b. The mean and standard deviation of Y will not be μY=$1.50,σY=$33.75

c. The mean and standard deviation of Y will not be μY=$75,σY=$18.37

e. The mean and standard deviation of Y will not be μY=$75,σY=$33.75

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