Suppose a subdivision on the southwest side of Denver, Colorado, contains 1,500 houses. The subdivision was built in 1983. A sample of 120 houses is selected randomly and evaluated by an appraiser. If the mean appraised value of a house in this subdivision for all houses is $228,000, with a standard deviation of $8,500, what is the probability that the sample average is greater than $229,500?

0.0268 = 2.68% probability that the sample average is greater than $229,500

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use the z-score formula.

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

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 normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Single house:

The mean appraised value of a house in this subdivision for all houses is $228,000, with a standard deviation of $8,500, which means that

Sample:

Sample of 120, so, by the central limit theorem,

What is the probability that the sample average is greater than $229,500?

This is 1 subtracted by the pvalue of Z when X = 229.5. So

By the Central Limit Theorem

has a pvalue of 0.9732

1 - 0.9732 = 0.0268

0.0268 = 2.68% probability that the sample average is greater than $229,500

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step-by-step explanation:

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