Question

According to the latest financial reports from a sporting goods store, the mean sales per customer was $75 with a population standard deviation of $6. The store manager believes 39 randomly selected customers spent more per transaction. What is the probability that the sample mean of sales per customer is between $76 and $77 dollars? You may use a calculator or the portion of the z -table given below. Round your answer to two decimal places if necessary.

Answer 1

13.04% probability that the sample mean of sales per customer is between $76 and $77 dollars.

Explanation:

To solve this problem, we have to understand the Normal Probability Distribution and the Central Limit Theorem.

Normal probability distribution:

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, a large sample size, of at least 30, can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 75, \sigma = 6, n = 39, s = (6)/(√(39)) = 0.96`

What is the probability that the sample mean of sales per customer is between $76 and $77 dollars?

This is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 76. So

X = 77

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

By the Central Limit Theorem

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

`Z = (77 - 75)/(0.96)`

`Z = 2.08`

`Z = 2.08` has a pvalue of 0.9812

X = 76

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

`Z = (76 - 75)/(0.96)`

`Z = 1.04`

`Z = 1.04` has a pvalue of 0.8508

0.9812 - 0.8508 = 0.1304

13.04% probability that the sample mean of sales per customer is between $76 and $77 dollars.