Question

Fast-food restaurants spend much time studying the amount of time cars spend in their drive-thrus. Certainly, the faster the cars get service, the more opportunity for making money. According to a recent study by QSR magazine, Wendy’s has the best time, with a mean time spent in the drive thru of 138.5 seconds. Assuming drive-thru time is normally distributed with a standard deviation of 29 seconds, what proportion of cars spends between 120 and 180 seconds in Wendy's drive-thru? Write answer as decimal rounded to the thousandth.

Answer 1

0.663

Explanation:

Problems of normally distributed samples are 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`

What proportion of cars spends between 120 and 180 seconds in Wendy's drive-thru?

This is the pvalue of Z when X = 180 subtracted by the pvalue of Z when X = 120. So

X = 180

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

`Z = (180 - 138.5)/(29)`

`Z = 1.43`

`Z = 1.43` has a pvalue of 0.924

X = 120

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

`Z = (120 - 138.5)/(29)`

`Z = -0.64`

`Z = -0.64` has a pvalue of 0.261

0.924 - 0.261 = 0.663