The average full­sized front­loading Energy Star clothes washer uses 15 gallons

Question

The average full­sized front­loading Energy Star clothes washer uses 15 gallons of water per load (greenbuildingadvisor.com, December 4, 2015). Assume that the popula­ tion standard deviation for the number of gallons of water used to wash a load by these machines is 3.0. a. What is the probability that a random sample of 90 loads washed in a full­sized front­ loading Energy Star clothes washer will provide a sample mean gallons of water used that is within one­half gallon of the population mean of 15 gallons? b. What is the probability that a random sample of 50 loads washed in a full­sized front­ loading Energy Star clothes washer will provide a sample mean gallons of water used that is at least three­quarters of a gallon greater than the population mean of 15 gallons? C.What is the probability that a random sample of 75 loads washed in a full­sized front­ loading Energy Star clothes washer will provide a sample mean gallons of water used that is no more than one­quarter of a gallon less than the population mean of 15 gallons?

Explanation / Answer

mean = 15
std. dev. = 3

SE = sigma/sqrt(n)
Central limit theorem, z = (xbar - mu)/SE

a)
within one­half gallon of the population mean of 15 gallons" means probability that mean is between 14.5 and 15.5

P(14.5 < X 15.5)
= P((14.5 - 15)/(3/sqrt(90)) < z < (15.5 - 15)/(3/sqrt(90)))
= P(-1.5811 < z < 1.5811)
= P(z < 1.5811) - P(z < -1.5811)
= 0.9431- 0.0569 (using standard normal z table)
= 0.8862

b)
"at least three­quarters of a gallon greater than the population mean of 15 gallons" means 15+(3/4) = 15.75

P(X > 15.75)
= P(z > (15.75 - 15)/(3/sqrt(50)))
= P(z > 1.7678)
= 0.03855 (using standard normal z table)

c)
"no more than one­quarter of a gallon less than the population mean of 15 gallons" means 15 - 1/4 = 14.75

P(X < 14.75)
= P(z < (14.75 - 15)/(3/sqrt(75)))
= P(z < -0.7217)
= 0.2352 (using standard normal z table)

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