Consider a large population with p = .63. Assuming n/N ≤ .05, find the mean and standard deviation of the sample proportion for a sample size of a. 100 b. 900

SOLUTION
As given in the question n/N is <0.05 we have to use binomial distribution to normal approximation.
Now we have to find mean and standard deviation which are equal to np and sqrt np respectively.
Therefore
(a) n = 100
E(x) = np and p is 0.63 as given in the question...