Historically it is known that the mean amount of time a car

Historically, it is known that the mean amount of time a car spends at the drive through window of Taco Queen to wait for an order is approximately normal with mean 59.3 seconds and standard deviation 13.1 seconds. The quality control manager of Taco Queen wants to use a new delivery system designed to get cars through the drive through system faster. He takes a random sample of 40 cars and records their time spending at the drive through window.

Is the sampling distribution normal? Why? What is the mean and what is the standard deviation of this sampling distribution?

What is the probability that a randomly selected car spends at least 55 seconds at the drive through window for its order?

What is the probability that a sample mean amount of time cars spend at the drive through window is more than 56.5 seconds?

What is the probability that a sample mean amount of time cars spend at the drive through window is between 57 seconds and 62 seconds?

Find the 20^th percentile of the sampling distribution of the sample mean.

Would it be unusual for a sample mean to be less than 54 seconds? Explain.

Would it be unusual for an individual car to spend less than 54 seconds at the drive through window? Explain.

Solution

Is the sampling distribution normal? Why? What is the mean and what is the standard deviation of this sampling distribution?

YES, BECAUSE YOU HAVE THE MEAN AND STANDARD DEVIATION YOU CAN SUPPOSE THAT IS NORMAL

MEAN = 59.3

STANDARD DEVIATION = 13.1 / srqt (40) = 2.07

What is the probability that a randomly selected car spends at least 55 seconds at the drive through window for its order?

P( z > (55 -59.3 ) / 2.07 ) = -2.08

P( z > -2.08) = 0.9812

What is the probability that a sample mean amount of time cars spend at the drive through window is more than 56.5 seconds?

P( z > (56.5 -59.3 ) / 2.07 ) = -1.35

P( z > -1.35) = 0.9115

What is the probability that a sample mean amount of time cars spend at the drive through window is between 57 seconds and 62 seconds?

P( [57-59.3] / 2.07 < z < [62-59.3]/ 2.07 )

P( -1.11 < z < 1.30 ) =0.7697

Find the 20^th percentile of the sampling distribution of the sample mean.

0.20 -- Z = -0.84

-0.84 = x -59.3 / 2.07

x = 57.5612

for the other literals

I can gladly help you but you should post it in a new question