Question

A retail store did research and determined that, on average, each customer spends 40 minutes in their store per visit, with a standard deviation of 2 minutes. What percentage of customers spend more than 46 minutes?

Answer 1

0.13% of customers spend more than 46 minutes

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 = 40, \sigma = 2`

What percentage of customers spend more than 46 minutes?

This is 1 subtracted by the pvalue of Z when X = 46. So

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

`Z = (46 - 40)/(2)`

`Z = 3`

`Z = 3` has a pvalue of 0.9987

1 - 0.9987 = 0.0013

0.13% of customers spend more than 46 minutes