Question

In a recent national survey, the mean price for a 2000 sq ft home in Florida is $240,000 with a standard deviation of $16,000. The mean price for the same sized home in Ohio is $170,000 with a standard deviation of $12,000. In which city would a home priced at $200,000 be closer to the mean price, compared to the distribution of prices in the city? Find the z-score corresponding to each city.

Answer 1

Answer:

In both Florida and Ohio the house is 2.5 standard deviations from the mean.

Step-by-step explanation:

Z-score

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:

The 200,000 house would be closer to the mean in the state in which the absolute value of the z-score is smaller.

Florida:

$\mu = 240000, \sigma = 16000$

So

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

$Z = \frac{200000 - 240000}{16000}$

$Z = -2.5$

In Florida, a home priced at $200,000 is 2.5 standard deviations from the mean.

Ohio:

$\mu = 170000, \sigma = 12000$

So

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

$Z = \frac{200000 - 170000}{12000}$

$Z = 2.5$

In Ohio, it is also 2.5 standard deviations from the mean.

In both Florida and Ohio the house is 2.5 standard deviations from the mean.