# 6.3: The Central Limit Theorem for Sample Proportions

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If $$X$$ is a binomial random variable, then

$X \sim B(n, p)\nonumber$

where $$n$$ is the number of trials and $$p$$ is the probability of a success.

To form a proportion, take $$X$$, the random variable for the number of successes and divide it by $$n$$, the number of trials (or the sample size). The random variable $$\hat{P}$$ (read "P hat") is that proportion,

$\hat{P} = \dfrac{X}{n}\nonumber$

When $$n$$ is large and $$p$$ is not close to zero or one, we can use the normal distribution to approximate the binomial.

$X \sim N(np, \sqrt{npq})\nonumber$

If we divide the random variable, the mean, and the standard deviation by $$n$$, we get a normal distribution of proportions with $$\hat{P}$$, called the estimated proportion, as the random variable. (Recall that a proportion as the number of successes divided by $$n$$.)

$\dfrac{X}{n} = \hat{P} \sim N\left(\dfrac{np}{n}, \dfrac{\sqrt{npq}}{n}\right)\nonumber$

Using algebra to simplify:

$\dfrac{\sqrt{npq}}{n} = \sqrt{\dfrac{pq}{n}}\nonumber$

$$\hat{P}$$ follows a normal distribution for proportions:

$\dfrac{X}{n} = \hat{P}\ \sim N\left(p, \sqrt{\dfrac{pq}{n}}\right)\nonumber$

## Review

The Central Limit Theorem for Proportions:

If

1. Sample is random with independent observations.

2. Sample is large:  The sample size, n, is considered large enough when the sample expects at least 10 successes (yes) and 10 failures (no); i.e.

$$n\cdot\ p$$≥10 and $$n\cdot(1-p)$$≥10

3. Big population:  If sampling is done without replacement, the population must be at least 10 times larger than the sample size.

then $$\hat{P}$$ follows a normal distribution with

1. The mean of sampling distribution of sample proportions $$\mu_{\hat{P}}$$ is the population proportion, p, $$\mu_{\hat{P}}=p$$
2. The standard deviation of the sampling distribution of sample proportions, $$\sigma_{\hat{P}}$$=$$\sqrt{\dfrac{pq}{n}}=\sqrt{\dfrac{p(1-p)}{n}}$$

## Formula Review

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