Asked by Kim
The ratio of brown M&M’s in a bag is stated by the candy company to be
56%. If a random sample of 500 bags is selected, determine the probability that
the sample proportion of random bags of M&M’s selected has a percentage of
brown between 51% and 59%
p=.56
standard deviation=sqrt(.56(1-.56)/500)= .0222
56%. If a random sample of 500 bags is selected, determine the probability that
the sample proportion of random bags of M&M’s selected has a percentage of
brown between 51% and 59%
p=.56
standard deviation=sqrt(.56(1-.56)/500)= .0222
Answers
Answered by
Reiny
Since you have the mean and sd
plug in your range into
http://davidmlane.com/hyperstat/z_table.html
(one of the best sites for replacing all those nasty tables in the back of your text)
plug in your range into
http://davidmlane.com/hyperstat/z_table.html
(one of the best sites for replacing all those nasty tables in the back of your text)
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