Standard Deviation of a sample of a population's means

[2^Oscar]

Hey guys,

Just a question which has been puzzling me for some time.

I am told that means of samples of a population can be normally distributed with mean of \mu and with standard deviation of \sigma/\sqrt{}n

Can someone please explain to me how the standard deviation is derived or is it ismply as a result of experimentation?

Thanks,
Oscar

 

[ssd]

It is derived. The given result follows if the population is normal.

 

[statdad]

The mean is

<br /> \overline X = \frac{\sum X}{n}<br />

If all of the X variables are independent, and identically distributed, then

<br /> Var[X] = Var\left(\frac{\sum X}n\right) = \frac 1 {n^2} \sum Var(X) = \frac{n\sigma^2}{n^2} = \frac{\sigma} n<br />

so the standard deviation is as you state.

If the X values are themselves normally distributed, then the mean is as well (linear combinations of normally distributed normal random variables are normal)

If the X values are not normally distributed, the distribution of \overline X is approximately normal if certain conditions are satisfied.

 

[stewartcs]

<br /> Var[X] = Var\left(\frac{\sum X}n\right) = \frac 1 {n^2} \sum Var(X) = \frac{n\sigma^2}{n^2} = \frac{\sigma} n<br />

You forgot the root in the denominator.

CS

 

[statdad]

You forgot the root in the denominator.

CS

No, actually I forgot the square in the numerator, since I was writing out the variance.
Still a stupid mistake on my part.

The variance is

<br /> \frac{\sigma^2} n<br />

so the standard deviation is

<br /> \frac{\sigma}{\sqrt n}<br />

 

[stewartcs]

No, actually I forgot the square in the numerator, since I was writing out the variance.
Still a stupid mistake on my part.

The variance is

<br /> \frac{\sigma^2} n<br />

so the standard deviation is

<br /> \frac{\sigma}{\sqrt n}<br />

Sorry...it appeared as if you were answering the OP's question about the standard deviation directly which is why I assumed you had finished the derivation all the way out to the standard deviation and not just the variance.

CS

 

[statdad]

stewartcs; there is no need for you to apologize and I certainly did not mean to imply (in my earlier post) that there was. I am sorry if it seemed that way.

 

[2^Oscar]

Hey guys,

Thanks very much for showing me the derivation - it has helped make things a lot clearer :)

Thanks again,

Oscar