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

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`

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 = (\sigma)/(√(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{(p(1-p))/(n)}`

In this question:

`p = 0.39, n = 100`

Then

`s = \sqrt{(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`