Asmall 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?

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 and standard deviation , the zscore of a measure X is given by

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

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:

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.

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step-by-step explanation: