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Posts: 232
2 weeks ago
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?
Textbook 

Stats: Modeling the World


Edition: 4th
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Answer verified by a subject expert
wrote...
Posts: 255
2 weeks ago
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Expected value of Geometric model: μ = = = 25
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wrote...
2 weeks ago
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.

What is the probability that she does not sell a garment until the 7th customer? Show work.
wrote...
2 weeks ago
P(6 no sales, sale on 7th) = (0.96)6(0.04) = 0.0313
wrote...
2 weeks ago
Helps a lot... Now I'm ready for my quiz
wrote...
2 weeks ago
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.

What is the probability that exactly 3 of the first 10 customers buy specialty clothes for their pet? Show work.
wrote...
2 weeks ago
P(exactly 3 out of 10) = P(X = 3) = (0.04)3(0.96)7 = 0.00577
wrote...
2 weeks ago
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.

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?
wrote...
2 weeks ago
Using the Binomial model, mean: μ = np = (275)(0.04) = 11
Standard deviation: σ= = = 3.25
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2 weeks ago
TY
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2 weeks ago
Welcome Slight Smile
wrote...
2 weeks ago
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.

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.
wrote...
2 weeks ago
Since np = 11 and nq = 264, we expect at least 10 successes and at least 10 failures. The sample size is large enough to apply a Normal model.  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.  Since the standard deviation = 3.25, it would be unusual to see more than 18 (11 + 2(3.25) = 17.5) customers who purchased specialty clothing.  So, I would conclude that her 4% estimate must have been too low if more than 18 customers purchased specialty clothing for their pet.
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