Question

A small hair salon in Denver, Colorado, averages about 82 customers on weekdays with a standard deviation of 16. It is safe to assume that the underlying distribution is normal. In an attempt to increase the number of weekday customers, the manager offers a $4 discount on 7 consecutive weekdays. She reports that her strategy has worked since the sample mean of customers during this 7 weekday period jumps to 93. What is the probability to get a sample average of 93 or more customers if the manager had not offered the discount?

Answer 1

There is a 3.44% probability to get a sample average of 93 or more customers if the manager had not offered the discount.

Answer 2

There is a 3.44% probability to get a sample average of 93 or more customers if the manager had not offered the discount.

Explanation:

Problems of normally distributed samples can be 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)`

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

A small hair salon in Denver, Colorado, averages about 82 customers on weekdays with a standard deviation of 16. This means that
`\mu = 82, \sigma = 16`

She reports that her strategy has worked since the sample mean of customers during this 7 weekday period jumps to 93. What is the probability to get a sample average of 93 or more customers if the manager had not offered the discount?

This is 1 subtracted by the pvalue of Z when X = 93.

We also have to find a standard deviation of the sample(that is, the 7 days), so:

`s = (\sigma)/(\sqrt(7)) = (16)/(\sqrt(7)) = 6.05`

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

`Z = (93-82)/(6.05)`

`Z = 1.82`

`Z = 1.82` has a pvalue of 0.9656.

This means that there is a 1-0.9656 = 0.0344 = 3.44% probability to get a sample average of 93 or more customers if the manager had not offered the discount.