Thread: 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

Variance of the sample is usually denoted by , in your example , where denotes the standard deviation of the sample: , hence .

So .

Hope it helps a bit.

Many thanks MathoMan...cleared that up perfectly!

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 )