Question

Among all monthly bills from a certain credit card company, the mean amount billed was $465 and the standard deviation was $300. In addition, for 15% of the bills, the amount billed was greater than $1000. A sample of 900 bills is drawn. What is the probability that the average amount billed on the sample bills is greater than $500? (Round the final answer to four decimal places.)

Answer 1

Answer:

0.02% probability that the average amount billed on the sample bills is greater than $500.

Step-by-step explanation:

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 $\frac{\sigma}{\sqrt{n}}$.

Normal probability distribution

Problems of normally distributed samples can be 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:

$Z = \frac{X - \mu}{\sigma}$

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.

In this problem, we have that:

$\mu = 465, \sigma = 300, n = 900, s = \frac{300}{\sqrt{900}} = 10$.

What is the probability that the average amount billed on the sample bills is greater than $500?

This probability is 1 subtracted by the pvalue of Z when $X = 500$. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{500 - 465}{10}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998.

So there is a 1-0.9998 = 0.0002 = 0.02% probability that the average amount billed on the sample bills is greater than $500.