Question

A small hair salon in Denver, Colorado, averages about 50 customers on weekdays with a standard deviation of 8. 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 $2 discount on 5 consecutive weekdays. She reports that her strategy has worked since the sample mean of customers during this 5 weekday period jumps to 57. [You may find it useful to reference the z table.]

a. What is the probability to get a sample average of 57 or more customers if the manager had not offered the discount? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

Answer 1

Solution:

Given:  A small hair salon in Denver, Colorado, averages about 50 customers on weekdays with a standard deviation of 8.

n= Sample size = 5

Sample mean =

We have to find: the probability to get a sample average of 57 or more customers if the manager had not offered the discount

That is:

Find z score:

Thus we get:

Look in z table for z = 1.9 and 0.06 and find corresponding area.

P( Z< 1.96 )= 0.9750

thus