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Vladimir [108]
3 years ago
11

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.)
Mathematics
1 answer:
Fynjy0 [20]3 years ago
8 0

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.

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