Question

Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5. (a) what are the mean and standard deviation of the x sampling distribution?

Answer 1

The mean of the sampling distribution will be 40, which is the same as the population mean, and the standard deviation (standard error) will be 0.625, which is found by dividing the population standard deviation by the square root of the sample size.

Step-by-step explanation:

When a random sample of size 64 is selected from a population with a mean of 40 and a standard deviation of 5, the sampling distribution of the sample mean will have its own mean and standard deviation. According to the Central Limit Theorem, the mean of the sampling distribution of the sample mean will be the same as the population mean. Therefore, the mean of the sampling distribution is also 40.

The standard deviation of the sampling distribution of the sample mean, often referred to as the standard error (SE), is calculated by dividing the population standard deviation by the square root of the sample size (n). Therefore, the standard error for this scenario is 5 / √64, which simplifies to 5 / 8 or 0.625.

Answer 2

The sample mean is equal to the population mean.

`\bar{X}=\mu=40`
The standard deviation of X-bar (the standard error) is:

`S.E.=(\sigma)/( √(n) )=(5)/(√(64))=0.625`