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

Answer:

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

Step-by-step 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 $s = \sqrt{\frac{p(1-p)}{n}}$

So

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