Question

The duration of shoppers' time in BrowseWorld's new retail outlets is normally distributed with a mean of 41.8 minutes and a standard deviation of 17.3 minutes. How long must a visit be to put a shopper in the longest 20 percent

Answer 1

A visit must be of at least 56.33 minutes to put a shopper in the longest 20 percent.

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:

$Z=\frac{X-\mu}{\sigma}$

How long must a visit be to put a shopper in the longest 20 percent?

Here $\mu=41.8 , \sigma=17.3$

The 100-20=80th percentile, which is X when Z has p-value of 0.8, so Z=0.84

$Z=\frac{X-\mu}{\sigma}$

$0.84=\frac{X-41.8}{17.3}$

14.532=X-41.8

X=14.532+41.8

X=56.33

A visit must be of at least 56.33 minutes to put a shopper in the longest 20 percent.

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