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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 82, \sigma = 16[/tex]
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:
[tex]s = \frac{\sigma}{\sqrt(7)} = \frac{16}{\sqrt(7)} = 6.05[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{93-82}{6.05}[/tex]
[tex]Z = 1.82[/tex]
[tex]Z = 1.82[/tex] 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.
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.
