Question

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?

Answer 1

Answer:

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 $\mu$ 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

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 $\mu = 228, \sigma = 8.5$

Sample:

Sample of 120, so, by the central limit theorem, $n = 120, s = \frac{8.5}{\sqrt{120}} = 0.7759$

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

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

By the Central Limit Theorem

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

$Z = \frac{229.5 - 228}{0.7759}$

$Z = 1.93$

$Z = 1.93$ 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