The owner of a pet store is trying to decide whether to discontinue selling spec

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 interesting and profitable items on the shelves. 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 for their pet. Assuming the pet store owner is correct in thinking that only 4% of her customers purchase specialty clothes for their pets, how many customers should she expect before someone buys a garment for their pet? What is the probability that she does not sell a garment until the 7^th customer? Show work. The owner had 275 customers that day. Assuming this was a typical day for his store, what would be the mean and standard deviation of the number of customers who buy specialty clothes for their pet each day? Surprised by a 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.

Explanation / Answer

1) 1/4% = 25

2) X - number of customer which purchase the garment first

X-Geometric disribution with parameter p = (0.04)

P(X> 7) = P(X =8) +P(X=9) + ..

= q^7*p + q^8*p+ ...

= q^7*p/(1 - q) = q^7 = (1-0.04)^7 = 0.75144747

c) n = 275 ,p = 0.04

X - number of customer who buy cloth

X -Binomial (n,p)

E(X ) = np = 275*0.04 = 11

sd(X) =sqrt( npq) = sqrt(275*0.04*0.96) = 3.2496153

d) z = (X -11)/3.2496153

z-critical for 5 % level is 1.645

hence

X* = 11 + 1.645*10.56 = 16.345617

hence n = 17

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