Question

according to a recent publication, the mean price of new mobile homes is $65,300. assume a standard deviation of $8000. let x denote the mean price of a sample of new mobile homes.
a. for samples of size 25, find the mean and standard deviation of x. interpret your results in words.
b. repeat part (a) with n = 50.
a. for samples of 25 mobile homes, the mean and standard deviation of all possible sample mean prices are $65300 and $1600, respectively.
(round to the nearest cent as needed.)
b. for samples of 50 mobile homes, the mean and standard deviation of all possible sample mean prices are $65300 and $\square, respectively.
(round to the nearest cent as needed.)

Explanation:

Step1: Recall the formula for the mean and standard deviation of the sample mean

The mean of the sample mean \(\mu_{\bar{x}}\) is equal to the population mean \(\mu\). The standard deviation of the sample mean (also known as the standard error) is given by \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size.

Given \(\mu = 65300\) and \(\sigma=8000\)

Part a: \(n = 25\)

• Mean of the sample mean:

\(\mu_{\bar{x}}=\mu\)
Since \(\mu = 65300\), \(\mu_{\bar{x}}=65300\)

• Standard deviation of the sample mean:

\(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\)
Substitute \(\sigma = 8000\) and \(n = 25\)
\(\sigma_{\bar{x}}=\frac{8000}{\sqrt{25}}=\frac{8000}{5}=1600\)

Part b: \(n = 50\)

• Mean of the sample mean:

\(\mu_{\bar{x}}=\mu\)
Since \(\mu=65300\), \(\mu_{\bar{x}} = 65300\)

• Standard deviation of the sample mean:

\(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\)
Substitute \(\sigma = 8000\) and \(n = 50\)
\(\sigma_{\bar{x}}=\frac{8000}{\sqrt{50}}\approx\frac{8000}{7.071}\approx1131\)

Answer:

a. The mean of all possible sample mean prices is \(\$65300\) and the standard deviation is \(\$1600\)
b. The mean of all possible sample mean prices is \(\$65300\) and the standard deviation is \(\$1131\)