If all possible samples of the same (large) size are selected from a population, what percentage of all sample proportions will be within 2.1 standard deviations of the population proportion? Round your answer to two decimal places.
Solution:
We know that the sampling distribution of the sample proportions follow approximately normal distribution.
Here, we have to find P(-2.1<Z<2.1)
P(-2.1<Z<2.1) = P(Z<2.1) - P(Z<-2.1)
P(-2.1<Z<2.1) = 0.982136 - 0.017864
(by using z-table)
P(-2.1<Z<2.1) = 0.964272
Required percentage = 96.43%
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