Question

The duration of shoppers' time in Browse Wrld's new retail outlets is normally distributed with a mean of 27.8 minutes and a standard deviation of 11.4 minutes. How long must a visit be to put a shopper in the longest 10 percent

Answer 1

Answer:

A visit must be of at least 42.39 minutes to put a shopper in the longest 10 percent.

Step-by-step 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:

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

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 of 27.8 minutes and a standard deviation of 11.4 minutes.

This means that $\mu = 27.8, \sigma = 11.4$

How long must a visit be to put a shopper in the longest 10 percent?

The 100 - 10 = 90th percentile, which is X when Z has a p-value of 0.9, so X when Z = 1.28.

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

$1.28 = \frac{X - 27.8}{11.4}$

$X - 27.8 = 1.28*11.4$

$X = 42.39$

A visit must be of at least 42.39 minutes to put a shopper in the longest 10 percent.