Question

According to data from the 2010 United States Census, 64.2% of Americans in Susanville, CA were male. Suppose Maria, a researcher, takes a random sample of 100 Americans in Susanville and finds that 54 are male. Let ^ p represent the sample proportion of Americans in Susanville that were male. What are the mean and standard deviation of the sampling distribution of ^ p ? Provide your answer rounded to three decimal places.

Answer 1

Answer:

The mean is 0.54 and the standard deviation is 0.05.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a sample proportion p in a sample of size n, the mean of the sampling distribution is p and the standard deviation is $\sqrt{\frac{p(1-p)}{n}}$.

In this problem:

$p = \frac{54}{100} = 0.54$

So

Mean 0.54

Standard deviation $\sqrt{\frac{0.54*0.46}{100}} = 0.05$.

The mean is 0.54 and the standard deviation is 0.05.