Question

According to a study conducted in one​ city, 3838​% of adults in the city have credit card debts of more than​ $2000. A simple random sample of n equals 200n=200 adults is obtained from the city. Describe the sampling distribution of ModifyingAbove p with caretp​, the sample proportion of adults who have credit card debts of more than​ $2000. Round to three decimal places when necessary.

Answer 1

Answer:

The sampling distribution of the sample proportion of adults who have credit card debts of more than​ $2000 will be normally distributed with mean = 0.38 and standard deviation of 0.034.

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

In this problem, we have that:

$n = 200, p = 0.38$

So

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.38*0.62}{200}} = 0.034$

The sampling distribution of the sample proportion of adults who have credit card debts of more than​ $2000 will be normally distributed with mean = 0.38 and standard deviation of 0.034.