Question

If "np is greater than or equal to 15" and "n(1-p) is greater than or equal to 15", what is the approximate shape of the sampling distribution of the sample proportion?

Answer 1

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Answer 2

The sampling distribution of the sample proportion will be approximately normally distributed with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`

Explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

If np >= 15 and n(1-p) >= 15

Can be approximated to the normal distribution, with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`

So

The sampling distribution of the sample proportion will be approximately normally distributed with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`