Question

Fast-food restaurants spend quite a bit of time studying the amount of time cars spend in their drive-throughs. Certainly, the faster the cars get through the drive-through, the more the opportunity for making money. QSR Magazine studied drive-through times for fast-food restaurants and found Wendy’s had the best time, with a mean time spent in the drive-through of 138.5 seconds. Assuming drive-through times are normally distributed with a standard deviation of 29 seconds, answer the following.
(a) Management wants to set a wait time over which a customer will not have to pay for their food. Of course, management wants to set that wait time to the 99th percentile. Find the X for the 99th percentile. (Hint: 99% = 0.9900). Round to the nearest whole second. Sullivan, Michael III. Fundamentals of Statistics: #40 (Ch 7.2). Pearson Education. Kindle Edition.

Answer 1

Answer:

For the 99th percentile, we have X = 206 seconds.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 138.5, \sigma = 29$

99th percentile:

Value of X when Z has a pvalue of 0.99. So we use $Z = 2.325$

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

$2.325 = \frac{X - 138.5}{29}$

$X - 138.5 = 29*2.325$

$X = 205.92 = 206$

For the 99th percentile, we have X = 206 seconds.