Question

A small hair salon in Denver, Colorado, averages about 85 customers on weekdays with a standard deviation of 17. 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 $5 discount on 6 consecutive weekdays. She reports that her strategy has worked since the sample mean of customers during this 6 weekday period jumps to 99.

a. What is the probability to get a sample average of 99 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.)

b. Do you feel confident that the manager’s discount strategy has worked?

• No, there is good chance (more than 5%) of getting 99 or more customers without the discount.

• No, there is only a small chance (less than 5%) of getting 99 or more customers without the discount.

• Yes, there is good chance (more than 5%) of getting 99 or more customers without the discount.

• Yes, there is only a small chance (less than 5%) of getting 99 or more customers without the discount.

Answer 1

(A) it is given that mean = 85, standard deviation = 17

we have to find the probability of 99 or more for the sample average

sample size is n = 6

sample standard deviation = sd/sqrt{n}

= 17/sqrt{6}

= 6.94

using normalcdf(lower bound, upper bound, mean, standard deviation)

setting lower bound = 99, upper bound = E99, mean = 85 and standard deviation = 6.94

= normalcdf(99,E99, 85, 6.94)

= 0.0218

(B) yes, manager's strategy has worked good because the probability is 0.0218, which is less than 0.05. Therefore, result is significant and the strategy has worked well

option D is correct