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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.)

1 Answer

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Answer:

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

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
(\sigma)/(√(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 = (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 = (300)/(√(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 = (X - \mu)/(s)


Z = (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.

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