Question

A small hair salon in Denver, Colorado, averages about 30 customers on weekdays with a standard deviation of 6. 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 35. Use Table 1.

a.	What is the probability to get a sample average of 35 or more customers if the manager had not offered the discount? (Round your intermediate calculations to 4 decimal places, "z" value to 2 decimal places, and final answer to 4 decimal places.)

Probability

b.	Do you feel confident that the manager’s discount strategy has worked?
	Yes, there is good chance (more than 5%) of getting 35 or more customers without the discount. Yes, there is only a small chance (less than 5%) of getting 35 or more customers without the discount. No, there is only a small chance (less than 5%) of getting 35 or more customers without the discount. No, there is good chance (more than 5%) of getting 35 or more customers without the discount.

Answer 1

Solution

given : M = 30, 0 = 6, n = 5

1 - formula : z=

a) PI > 35) =?

P - 30 시

— P(2 > 1.86)

1-P(Z < 1.86)

Refer Z-table to find the probability or use excel formula "=NORM.S.DIST(1.86, TRUE)" to find the probability.

+ 1 -0.9686

合0.0314

:. Probability = 0.0314

----------------------------------------------------------------------------------------------------------------

_{${\color{Blue} \mathbf{b)}}\;$} No, there is only a small chance (less than 5%) of getting 35 or more customers without the discount.