# Thread: Stats Question

1. ## Stats Question

A simple random sample of 200 employees out of the 1000 employees of the corporation XY is conducted and each person in the sample is asked how many hours a week he spends at the gym.

Given that population standard deviation is 2 hours, what is the standard deviation of sampling distribution of the means?

2. Originally Posted by tikimasala
A simple random sample of 200 employees out of the 1000 employees of the corporation XY is conducted and each person in the sample is asked how many hours a week he spends at the gym.

Given that population standard deviation is 2 hours, what is the standard deviation of sampling distribution of the means?
What formula have you been given for the standard deviation of sampling distribution of the means? Where are you stuck in applying it? (Note: n = 200, $\sigma = 2$).

3. this is the formula ive got so far...

σx = σ * sqrt( 1/n - 1/N )

so if n= 200
N= 1,000
σ= 2

=2 * sqrt (1/200 - 1/1,000)

this is where im stuck so far

4. Originally Posted by tikimasala
this is the formula ive got so far...

σx = σ * sqrt( 1/n - 1/N )

so if n= 200
N= 1,000
σ= 2

=2 * sqrt (1/200 - 1/1,000)

this is where im stuck so far
Read this: Sampling Distribution

Substitute the appropriate values.

Use a calculator to do the arithmetic.