"According to a survey, only 20% of customers who visited the web site of a major retail store made a purchase. Random samples of size 100 are selected.
a. What is the standard deviation of all the sample proportions of customers who will make a purchase after visiting the web site?
b. 80% of the samples will have more than what percentage of customers who will make a purchase after visiting the web site?"
For part a I did the following
sqrt((.20)*(1-.20)/100) = .04 (I don't know if this is right)
For part b I am lost for how to approach the problem.
Any help is greatly appreciated, I changed the numbers from the ones on my work so I am looking more for how to solve these than the answers themselves, thanks!
a. What is the standard deviation of all the sample proportions of customers who will make a purchase after visiting the web site?
b. 80% of the samples will have more than what percentage of customers who will make a purchase after visiting the web site?"
For part a I did the following
sqrt((.20)*(1-.20)/100) = .04 (I don't know if this is right)
For part b I am lost for how to approach the problem.
Any help is greatly appreciated, I changed the numbers from the ones on my work so I am looking more for how to solve these than the answers themselves, thanks!
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a. ok !
b. z-score for top 80% = - 0.8416, % of customers = 20 - 0.04*0.8416 = 19.966%
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b. z-score for top 80% = - 0.8416, % of customers = 20 - 0.04*0.8416 = 19.966%
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