Question

A small hair salon in Denver, Colorado, averages about 62 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 $3 discount on 9 consecutive weekdays. She reports that her strategy has worked since the sample mean of customers during this 9 weekday period jumps to 74. Use Table 1. a. What is the probability to get a sample average of 74 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.) Probability b. Do you feel confident that the manager’s discount strategy has worked? No, there is good chance (more than 5%) of getting 74 or more customers without the discount. No, there is only a small chance (less than 5%) of getting 74 or more customers without the discount. Yes, there is good chance (more than 5%) of getting 74 or more customers without the discount. Yes, there is only a small chance (less than 5%) of getting 74 or more customers without the discount.

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	62.000
std deviation   =σ=	16.0000
sample size       =n=	9
std error=σx̅=σ/√n=	5.3333

  probability to get a sample average of 74 or more customers if the manager had not offered the discount:

probability =

P(X>74)

=

P(Z>2.25)=

1-P(Z<2.25)=

1-0.9878=

0.0122

b)

Yes, there is only a small chance (less than 5%) of getting 74 or more customers without the discount.