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

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.

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

In this problem, we have that:

`n = 200, p = 0.38`

So

`s = \sqrt{(p(1-p))/(n)} = \sqrt{(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.