SOLUTION: I need HELP: Given a binomial distribution with n=20 and p=0.76, would the normal distribution provide a reasonable approximation? Why or Why Not?
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Question 659927: I need HELP: Given a binomial distribution with n=20 and p=0.76, would the normal distribution provide a reasonable approximation? Why or Why Not?
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Now is this the type of problem where the n represents the number of tries?
like this
n=20 tries
p=0.76 success rate
or something like that? Please help me....... Answer by Edwin McCravy(20055) (Show Source):
Some books say:
Calculate np and n(1 - p) and if both are
greater than 5, the approximation is good.
np > 5 and n(1 - p) > 5
np = 20(0.76) = 15.2
n(1 =- p) = 20(1 - 0.76) = 4.8
The second one is not > 5 so it would not be a good approximation.
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Other books say:
Calculate np(1 - p) and if this is >= 10 then the approximation is good.
np(1 - p) = (20)(0.76)(1 - 0.76) = 3.648
That is not greater than or equal to 10
So the normal approximation would not be good in either case.
Edwin
