Question

According to a study conducted in one city, 39% of adults in the city have credit card debts of more than $2000. A simple random sample of 100 adults is obtained from the city. Describe the sampling distribution of the sample proportion of adults who have credit card debts of more than $2000.

Answer 1

Answer:

The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean $\mu = 0.39$ and standard deviation $s = 0.0488$

Step-by-step explanation:

Central Limit Theorem

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 in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

$p = 0.39, n = 100$

Then

$s = \sqrt{\frac{0.39*0.61}{100}} = 0.0488$

By the Central Limit Theorem:

The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean $\mu = 0.39$ and standard deviation $s = 0.0488$