Solution :
n 0.005 N
200 0.05 * 25000
200 1250 .........(1)
np(1-p) = 200 * 0.2 * 0.8 = 32 10 ..............(2)
From 1 and 2
The mean of the sampling distribution of proportion is ,
= p = 0.8
The standard deviation of the proportion is ,
= [p( 1 - p ) / n = [(0.8 * 0.2) / 200 ] = 0.028
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