Question

Suppose that in one region of the country the mean amount of credit card debt perhousehold in households having credit card debt is $15,250 with standard deviation $7,125.

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly______.

Answer 1

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly 0.907 = 90.7%.

Explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, a large sample size can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 15250, \sigma = 7125, n = 1600, s = (7125)/(√(1600)) = 178.125`

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly

This probability is the pvalue of Z when X = 1600 + 300 = 1900 subtracted by the pvalue of Z when X = 1600 - 300 = 1300. So

X = 1900

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (1900 - 1600)/(178.125)`

`Z = 1.68`

`Z = 1.68` has a pvalue of 0.9535.

X = 1300

`Z = (X - \mu)/(s)`

`Z = (1300 - 1600)/(178.125)`

`Z = -1.68`

`Z = -1.68` has a pvalue of 0.0465.

0.9535 - 0.0465 = 0.907.

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly 0.907 = 90.7%.