SOLUTION: Given a binomial distribution with n = 21 and p = 0.77, would the normal distribution provide a reasonable approximation? Hi, I feel like I've missed a class. Everything I'm re

Question 632677: Given a binomial distribution with n = 21 and p = 0.77, would the normal distribution provide a reasonable approximation?
Hi, I feel like I've missed a class. Everything I'm reading says that to do this I need np and n(1-p) = nq and I have no idea what this means!! can someone help please? I don't want the answer I want to know how to get the answer. Thanks.
Found 2 solutions by jim_thompson5910, John10:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
np = 21*0.77 = 16.17

So np = 16.17

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n(1-p) = 21*(1-0.77) = 4.83

So n(1-p) = 4.83 or nq = 4.83 since q = (1-p)

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Since nq = 4.83 is NOT greater than 5, this means that the normal distribution would NOT provide a reasonable approximation

It would only provide a reasonable approximation if BOTH np AND nq are greater than 5.

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Answer by John10(297) (Show Source): You can put this solution on YOUR website!
Given a binomial distribution with n = 21 and p = 0.77, would the normal distribution provide a reasonable approximation?
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Hello,
In order using normal distribution to calculate binomial distribution:
p * n > 5
AND
(1 - p) * n > 5
If both are satisfied, you are GOOD TO GO:)
John10
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