Question

A study of drive-thru wait times for fast-food restaurants found that the average time spent Wendy's drive-thru was 138.5 seconds. Assume that drive-thru wait times are normally distributed with a standard deviation of 29 seconds, A Wendy's owner wants to give a free Frosty to customers who have an unusually long wait in the drive-thru.
The length of time the owner should choose so that only 0.75% of customers get a free Frosty i.e. only 0.75% wait longer than ________ seconds) is _________ seconds (give your answer to 2 decimal places).

Answer 1

Answer:

207.72 seconds.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean waiting time of 138.5 seconds, standard deviation of 29 seconds.

This means that $\mu = 138.5, \sigma = 29$

The length of time the owner should choose so that only 0.75% of customers get a free Frosty i.e. only 0.75% wait longer than

The 100 - 0.75 = 99.15th percentile, which is X when Z has a pvalue of 0.9915, so X when Z = 2.387.

$Z = \frac{X - \mu}{\sigma}$

$2.387 = \frac{X - 138.5}{29}$

$X - 138.5 = 2.387*29$

$X = 207.72$

So the answer is 207.72 seconds.