SOLUTION: Assuming you have a binomial distribution with n=1000 and p=0.91, what is the probability of randomly getting a score less than or equal to 925?

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Question 1142506: Assuming you have a binomial distribution with n=1000 and p=0.91, what is the probability of randomly getting a score less than or equal to 925?
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
The mean is np=910
the variance is 910*0.09 (np(1-p))=81.9
sd is sqrt (V)=9.05
z would be (x-mean/sd) and use 925.5 as top score (continuity correction factor)
z < (925.5-910)/9.05
z< 1.71
probability is 0.9564