Question

Suppose that the average weekly earnings for employees in general automotive repair shops is $450, and that the standard deviation for the weekly earnings for such employees is $50. A sample of 100 such employees is selected at random. Find the standard deviation of the sampling distribution of the means of average weekly earnings for samples of size 100

Answer 1

Answer: 5

Explanation:

We know that the standard deviation of the sampling distribution of the means is given by :_

`SE=\sigma_x=(\sigma)/(√(n))`

Given : The average weekly earnings for employees in general automotive repair shops is
`\mu=\ $450`.

The standard deviation for the weekly earnings for such employees =
`\sigma= \$50.`

Now, the standard deviation of the sampling distribution of the means of average weekly earnings for samples of size
`n=100`:-

`SE=\sigma_x=(50)/(√(100))\\\\\Rightarrow\ SE=(50)/(10)\\\\\Rightarrow\ SE=5`

Hence, the standard deviation of the sampling distribution of the means of average weekly earnings for samples of size 100 = 5