Question

A credit card company decides to study the frequency with which its cardholders charge for items from a certain change of retail stores. The data values collected in the study appear to be normally distributed with a mean of 25 charged purchases and a standard distribution of 2 charged purchases. Out of the total number of cardholders about hw many would you expect are charging 27 or more in the study

Answer 1

15.87% of the total number of cardholder would be expected to be charging 27 or more in the study.

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 25 charged purchases and a standard distribution of 2

This means that
`\mu = 25, \sigma = 2`

Proportion above 27

1 subtracted by the pvalue of Z when X = 27. So

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

`Z = (27 - 25)/(2)`

`Z = 1`

`Z = 1` has a pvalue of 0.8413

1 - 0.8413 = 0.1587

Out of the total number of cardholders about how many would you expect are charging 27 or more in the study?

0.1587*100% = 15.87%

15.87% of the total number of cardholder would be expected to be charging 27 or more in the study.