Question

A study found that the mean amount of time cars spent in drive-thrus at Burger King's fast food restaurant was 132.8 seconds. Assuming drive-thru times are normally distributed with a standard deviation of 25 seconds.

A.) What is the probability that a randomly selected car will get through Burger King's drive-thru in less than 102 seconds?

B.) What is the probability that a randomly selected car will spend more than 175 seconds in the Burger King drive-thru?

C.) Would it be unusual for a car to spend more than 3 minites in the Burger King drive-thru? Why?

Answer 1

Given = 132.8, 25

To find the probability, we need to find the z scores.

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(a) For P( X < 102)

Z = (102 – 132.8)/25 = -1.23

The required probability from the normal distribution tables is = 0.1093

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(b) For P (X > 175) = 1 - P (X < 175), as the normal tables give us the left tailed probability only.

For P( X < 175)

Z = (175 – 132.8)/25 = 1.69

The probability for P(X < 175) from the normal distribution tables is = 0.9545

Therefore the required probability = 1 – 0.9545 = 0.0455

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(c) P(More than 3 minutes) = P(X > 180)

For P (X > 180) = 1 - P (X < 180), as the normal tables give us the left tailed probability only.

For P( X < 180)

Z = (180 – 132.8)/25 = 1.89

The probability for P(X < 180) from the normal distribution tables is = 0.9706

Therefore the required probability = 1 – 0.9706 = 0.0294

Yes, the probability that car spends more than 3 minutes is unusual as the probability is 0.0294 (2.94%) which is less than 0.05 (5%).

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