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

Answer:

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

Step-by-step 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

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

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:

$s = \frac{\sigma}{\sqrt(7)} = \frac{16}{\sqrt(7)} = 6.05$

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

$Z = \frac{93-82}{6.05}$

$Z = 1.82$

$Z = 1.82$ 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.