Question

The length of time a person takes to decide which shoes to purchase is normally distributed with a mean of 8.21 minutes and a standard deviation of 1.90. Find the probability that a randomly selected individual will take less than 5 minutes to select a shoe purchase. Is this outcome unusual?

Answer 1

4.55% probability that a randomly selected individual will take less than 5 minutes to select a shoe purchase.

Since Z > -2 and Z < 2, this outcome is not considered unusual.

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.

If
`Z \leq 2` or
`Z \geq 2`, the outcome X is considered to be unusual.

In this question:

`\mu = 8.21, \sigma = 1.9`

Find the probability that a randomly selected individual will take less than 5 minutes to select a shoe purchase.

This is the pvalue of Z when X = 5. So

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

`Z = (5 - 8.21)/(1.9)`

`Z = -1.69`

`Z = -1.69` has a pvalue of 0.0455.

4.55% probability that a randomly selected individual will take less than 5 minutes to select a shoe purchase.

Since Z > -2 and Z < 2, this outcome is not considered unusual.