Question

according to a recent publication, the mean price of new mobile homes is $64,500. assume a standard deviation of $6600. let \\( \bar { x } \\) denote the mean price of a sample of new mobile homes.
a. for samples of size 25, find the mean and standard deviation of \\( \bar { x } \\). interpret your results in words.
b. repeat part (a) with \\( n = 50 \\).
a. for samples of 25 mobile homes, the mean and standard deviation of all possible sample mean prices are \\( \\$ 64500 \\) and \\( \\$ 1320 \\), 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 \\( \\$ \\) and \\( \\$ \\), respectively.
(round to the nearest cent as needed.)

Explanation:

Step1: Recall the formula for the mean of the sample mean

The mean of the sample mean ($\mu_{\bar{x}}$) is equal to the population mean ($\mu$). Given that the population mean price of new mobile homes ($\mu$) is $\$64500$. So, for any sample size $n$, $\mu_{\bar{x}}=\mu$.
For part (a) with $n = 25$:
$\mu_{\bar{x}}=64500$

Step2: Recall the formula for the standard deviation of the sample mean

The formula for the standard deviation of the sample mean (also known as the standard error) is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size.
Given that $\sigma = 6600$.
For $n = 25$:
$\sigma_{\bar{x}}=\frac{6600}{\sqrt{25}}=\frac{6600}{5}=1320$

Step3: Repeat for $n = 50$

For the mean of the sample mean (using the same property as in Step 1)
$\mu_{\bar{x}}=64500$
For the standard deviation of the sample mean with $n = 50$:
$\sigma_{\bar{x}}=\frac{6600}{\sqrt{50}}\approx\frac{6600}{7.071}\approx933.4$

Answer:

a. The mean of all possible sample mean prices is $\$64500$ and the standard deviation is $\$1320$.
b. The mean of all possible sample mean prices is $\$64500$ and the standard deviation is $\$933.4$