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svetoff [14.1K]
3 years ago
9

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.
Mathematics
1 answer:
Butoxors [25]3 years ago
5 0

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

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