Answer:
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%.
Step-by-step 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
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation ![s = \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In this problem, we have that:
![\mu = 15250, \sigma = 7125, n = 1600, s = \frac{7125}{\sqrt{1600}} = 178.125](https://tex.z-dn.net/?f=%5Cmu%20%3D%2015250%2C%20%5Csigma%20%3D%207125%2C%20n%20%3D%201600%2C%20s%20%3D%20%5Cfrac%7B7125%7D%7B%5Csqrt%7B1600%7D%7D%20%3D%20178.125)
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 = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{1900 - 1600}{178.125}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1900%20-%201600%7D%7B178.125%7D)
![Z = 1.68](https://tex.z-dn.net/?f=Z%20%3D%201.68)
has a pvalue of 0.9535.
X = 1300
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{1300 - 1600}{178.125}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1300%20-%201600%7D%7B178.125%7D)
![Z = -1.68](https://tex.z-dn.net/?f=Z%20%3D%20-1.68)
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%.