Question

The quality control manager of at Long John Silver's restaurant wants to analyze the length of time that a car spends at the drive-thru window waiting for an order. It is determined that the mean time spent at the window is 59.3 seconds with a standard deviation of 13.1 seconds. The distribution of time at the window is skewed right. The quality control manager wishes to use a new delivery system designed to get cars through the drive-thru system faster. A random sample of 40 cars results in a sample mean time spent at the window of 56.8 seconds. There is a 15% chance that the mean time spent at the window of a random sample of 50 cars will exceed what value?

Answer 1

Since the population standard deviation is known and the sample size is large (n ≥ 30), we can use the z-distribution to find the probability of the sample mean exceeding a certain value.

First, we need to find the standard error of the sample mean:

SE = σ/√n = 13.1/√40 ≈ 2.07

Then, we can find the z-score corresponding to a 15% probability in the right tail of the distribution:

z = invNorm(0.85) ≈ 1.036

Finally, we can use the z-score formula to find the sample mean that corresponds to this z-score:

z = (x - μ) / SE

x = z * SE + μ

x = 1.036 * 13.1/√50 + 59.3 ≈ 61.8

Therefore, there is a 15% chance that the mean time spent at the window of a random sample of 50 cars will exceed 61.8 seconds.