Question

The owner of a pet store is trying to decide whether to discontinue selling specialty clothes for pets. She suspects that only 4% of the customers buy specialty clothes for their pets and thinks that she might be able to replace the clothes with more profitable items. Before making a final decision, she decides to keep track of the total number of customers for a day and whether they purchase specialty clothes. a. The owner had 275 customers that day. Assuming this was a typical day for her store, what would be the mean and standard deviation of the number of customers who buy specialty clothes for their pet each day? b. Surprised by the high number of customers who purchased specialty pet clothing that day, the owner decided that her 4% estimate must have been too low. How many clothing sales would it have taken to convince you? Justify your answer

Answer 1

Solution

Proportion of customer buy specialty clothes for their pets, p = 0.04

(a) n = 275, it follows binomial distribution with mean = np and variance = np(1-p)
Mean = np = 275*0.04 = 11
Standard deviation = (np(1-p))^0.5 =( 275*0.04*0.96)^0.5 = 10.56 = 3.25

(b) np = 11 , nq = 264
over 2 standard deviation = 11+ 3.25*2 = 17.5 = 18 approx
It would be unusual to see the number of customers who purchased specialty pet clothing more than 2 or 3 standard deviations above the mean
I would conclude that her 4% estimate must have been too low if more than 18 customers purchased specialty clothing for their pet.