Question

Given n=1010 and p= 0.84, determine whether this is approxamlty a normal distrubution. if it is find the standard deviation of the sampling distribution of proportion.

Answer 1

Solution

Given that,

p = 0.84

1 - p =1-0.84=0.16

n = 1010

standard deviation =$\sigma$$\hat p$ =  $\sqrt$[p ( 1 - p ) / n] =  $\sqrt$ [(0.84*0.16) / 1010 ] = 0.0115

(*)when n = 10

than

standard deviation =$\sigma$$\hat p$ =  $\sqrt$[p ( 1 - p ) / n] =  $\sqrt$ [(0.84*0.16) / 10 ] = 0.1159