Question

A line for tickets to a Broadway show had a mean waiting time of 20 minutes with a standard deviation of 5 minutes.

What percentage of the people in line waited for more than 28 minutes?

Answer 1

5.48% of the people in line waited for more than 28 minutes

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 = (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 20 minutes with a standard deviation of 5 minutes.

This means that
`\mu = 20, \sigma = 5`

What percentage of the people in line waited for more than 28 minutes?

The proportion is 1 subtracted by the p-value of Z when X = 28. So

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

`Z = (28 - 20)/(5)`

`Z = 1.6`

`Z = 1.6` has a p-value of 0.9452.

1 - 0.9452 = 0.0548.

As a percentage:

0.0548*100% = 5.48%

5.48% of the people in line waited for more than 28 minutes