A small gym in Biloxi, Mississippi, averages about 50 customers on weekdays with a standard deviation of 10. 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 $10 membership discount on 5 consecutive weekdays. She reports that her strategy has worked since the sample mean of customers during this weekday period jumps to 75. a. How unusual would it be to get a sample average of 75 or more customers if the manager had not offered the discount? b. Do you feel confident that the manager's discount strategy has worked? Explain.

Answer:It is very unusual as the probability to get a sample average of 75 or more customers if the manager had not offered the discount is 0.006     Step-by-step explanation:We are given the following information in the question: Mean, μ = 50Standard Deviation, σ = 10We are given that the distribution of number of customers is a bell shaped distribution that is a normal distribution. In an attempt to increase the number of weekday customers, the manager offers a $10 membership discount on 5 consecutive weekdays and  the sample mean of customers during this weekday period jumps to 75. Formula: [tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex] a) P(sample average of 75 or more customers if the manager had not offered the discount) P(x > 75) [tex]P( x > 75) = P( z > \displaystyle\frac{75 - 50}{10}) = P(z > 2.5)[/tex] [tex]= 1 - P(z \leq 2.5)[/tex] Calculation the value from standard normal z table, we have,  [tex]P(x > 610) = 1 - 0.994 = 0.006= 0.6\%[/tex]It is very unusual as the probability to get a sample average of 75 or more customers if the manager had not offered the discount is 0.006b) Yes, the managers discount strategy worked as it increased the average number of customers from 50 customers to 75 customers.