Question

1. (4 points) from the set of numbers {4,6,8}, a random sample of size 2 is selected with replacement. calculate the mean e(\bar{x}) and the standard deviation sd(\bar{x}) of the sampling distribution of the sample mean \bar{x}.
2. (4 points) consider a population with the mean \mu = 51,000 and the standard deviation \sigma = 5,000. for a random sample of size n = 100, find the probability that the sample mean \bar{x} will be between 50,000 and 51,500.

Explanation:

Step1: Calculate the population mean and standard deviation

The population mean \(\mu=\frac{4 + 6+8}{3}=6\)
The population variance \(\sigma^{2}=\frac{(4 - 6)^{2}+(6 - 6)^{2}+(8 - 6)^{2}}{3}=\frac{4+0 + 4}{3}=\frac{8}{3}\)
The population standard deviation \(\sigma=\sqrt{\frac{8}{3}}\)

Step2: Use the properties of the sampling distribution of the sample mean

For a sample of size \(n = 2\) selected with replacement:
The mean of the sampling distribution of the sample mean \(E(\bar{X})=\mu = 6\)
The standard deviation of the sampling distribution of the sample mean \(sd(\bar{X})=\frac{\sigma}{\sqrt{n}}\)
Since \(\sigma=\sqrt{\frac{8}{3}}\) and \(n = 2\), then \(sd(\bar{X})=\sqrt{\frac{\frac{8}{3}}{2}}=\sqrt{\frac{4}{3}}=\frac{2}{\sqrt{3}}\approx1.15\)

Answer:

The mean \(E(\bar{X}) = 6\) and the standard deviation \(sd(\bar{X})\approx1.15\)