151k views
3 votes
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.

1 Answer

4 votes

Answer:

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

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 = (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 = (X - \mu)/(\sigma)


2.325 = (X - 138.5)/(29)


X - 138.5 = 29*2.325


X = 205.92 = 206

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

User Snejame
by
7.4k points