# Thread: Standard Deviation & Mean Confusion

1. ## Standard Deviation & Mean Confusion

Hi,

Doing this question and working off an example in my book. The example is not explained properly, so I don't know why I am doing certain working out. Please help!

I have a random sample of 50 weights of boxes, with normally distributed mean of 750g and standard deviation of 20g.

I have worked out the mean of the sample...
I have worked out the standard deviation of the sample...

***In the book it then goes to squareroot by 50 and multiply by the standard deviation. What is the purpose of this and what does it do?***

Thanks

2. ## Re: Standard Deviation & Mean Confusion

Variance of the sample is usually denoted by $\displaystyle \sigma^2$, in your example $\displaystyle \sigma^2=50$, where $\displaystyle \sigma$ denotes the standard deviation of the sample: $\displaystyle \sigma^2= 50$, hence $\displaystyle \sigma=\sqrt{50}$.

So $\displaystyle \sqrt{50}\cdot \sigma = \sqrt{50}\cdot\sqrt{50}=50=\sigma^2$.

Hope it helps a bit.

3. ## Re: Standard Deviation & Mean Confusion

Many thanks MathoMan...cleared that up perfectly!

4. ## Re: Standard Deviation & Mean Confusion

If you have a sample of size n from a distribution with standard deviation s then the distribution of the means of samples has standard deviation s/root n

(note that for the population standard deviation I should really use the Greek letter sigma )