Question

According to loyalty card data, the amount of money spent per shopper at the grocery store is normally distributed with a mean of $75 per trip and a standard deviation of $25 per trip. How much money does a shopper spend that is at the 95th percentile of this distribution

Answer 1

A shopper that is at the 95th percentile of this distribution spends $116.125.

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 of $75 per trip and a standard deviation of $25 per trip.

This means that
`\mu = 75, \sigma = 25`

How much money does a shopper spend that is at the 95th percentile of this distribution?

This is X when Z has a p-value of 0.95, so X when Z = 1.645.

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

`1.645 = (X - 75)/(25)`

`X - 75 = 1.645*25`

`X = 116.125`

A shopper that is at the 95th percentile of this distribution spends $116.125.