Question

The selling price for houses in a neighborhood in West Richland, WA have an approximately normal distribution with a mean selling price of $275,000 and a standard deviation of $10,500. If 35 homes sell in this neighborhood, what is the probability that the average selling price of the 35 homes will be within $2000 of the true mean (that is, what is the probability the sample mean will be between $273,000 and $277,000)

Answer 1

The probability that the average selling price of the 35 homes will be within $2000 of the true mean is 0.5753

First, we need to find the z-score and this is expressed according to the formula;

$z=\dfrac{x-\mu}{\sigma}$

x is the sample space

$\mu$ is the mean value

$\sigma$ is the standard deviation

Given the following parameters

Mean = $275,000

Standard deviation = $10,500

If the true mean is $273,000, then;

$z=\dfrac{273,000-275000}{10,500} \\z = \frac{-2000}{10500}\\z= -0.1905$

If the true mean is $277,000

$z=\dfrac{277,000-275000}{10,500} \\z = \frac{2000}{10500}\\z= 0.1905$

This means that the z-score is between -0.1095 and 0.1095

Pr( -0.1095 ≤ z ≤ 0.1095)

According to the z table, the probability that the average selling price of the 35 homes will be within $2000 of the true mean is 0.5753

Learn more here: brainly.com/question/17436641