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wrote...
Posts: 230
2 weeks ago
The owner of a small convenience store is trying to decide whether to discontinue selling magazines. He suspects that only 5% of the customers buy a magazine and thinks that he might be able to use the display space to sell something more profitable. Before making a final decision he decides that for one day he'll keep track of the number of customers and whether or not they buy a magazine.

Assuming the owner is correct in thinking that 5% of the customers purchase magazines, how many customers should he expect before someone buys a magazine?
Textbook 

Stats: Modeling the World


Edition: 4th
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wrote...
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2 weeks ago
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Expected value of Geometric model: μ = = = 20
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wrote...
2 weeks ago
The owner of a small convenience store is trying to decide whether to discontinue selling magazines. He suspects that only 5% of the customers buy a magazine and thinks that he might be able to use the display space to sell something more profitable. Before making a final decision he decides that for one day he'll keep track of the number of customers and whether or not they buy a magazine.

What is the probability that he does not sell a magazine until the 8th customer? Show work.
wrote...
2 weeks ago
Using the Geometric model, Geom(0.05):
P(number of customers = 8) = (0.95)7(0.05) = 0.035
wrote...
2 weeks ago
The owner of a small convenience store is trying to decide whether to discontinue selling magazines. He suspects that only 5% of the customers buy a magazine and thinks that he might be able to use the display space to sell something more profitable. Before making a final decision he decides that for one day he'll keep track of the number of customers and whether or not they buy a magazine.

What is the probability that exactly 2 of the first 10 customers buy magazines? Show work.
wrote...
2 weeks ago
Using the Binomial model, Binom(10, 0.05)
P(2 of 10 customers buy magazines) = (0.05)2(0.95)8 = 0.075
wrote...
2 weeks ago
Thank you, thank you, thank you!
wrote...
2 weeks ago
You're welcome
wrote...
2 weeks ago
The owner of a small convenience store is trying to decide whether to discontinue selling magazines. He suspects that only 5% of the customers buy a magazine and thinks that he might be able to use the display space to sell something more profitable. Before making a final decision he decides that for one day he'll keep track of the number of customers and whether or not they buy a magazine.

He had 280 customers that day. Assuming this day was typical for his store, what would be the mean and standard deviation of the number of customers who buy magazines each day?
wrote...
2 weeks ago
μ = (280)(0.05) = 14
 σ= = 3.65
wrote...
2 weeks ago
The owner of a small convenience store is trying to decide whether to discontinue selling magazines. He suspects that only 5% of the customers buy a magazine and thinks that he might be able to use the display space to sell something more profitable. Before making a final decision he decides that for one day he'll keep track of the number of customers and whether or not they buy a magazine.

Surprised by the high number of customers who purchased magazines that day, the owner decided that his 5% estimate must have been too low. How many magazine sales would it have taken to convince you? Justify your answer.
wrote...
2 weeks ago
Since np = 14 and nq = 266, 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 sales more than 2 (or 3) standard deviations above the anticipated mean.  Since 14 + 2(3.65) = 21.3 (or 14 + 3(3.65) = 24.95), I would conclude the 5% estimate was probably too low if 22 (25) customers or more bought magazines.
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