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.

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

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 [tex]\sqrt{\frac{p(1-p)}{n}}[/tex].

In this problem:

[tex]p = \frac{54}{100} = 0.54[/tex]

So

Mean 0.54

Standard deviation [tex]\sqrt{\frac{0.54*0.46}{100}} = 0.05[/tex].

The mean is 0.54 and the standard deviation is 0.05.