Thread: Hey guys just looking for help with this question dealing with population variance.

1. Hey guys just looking for help with this question dealing with population variance.

I don't quite know how to write this problem down here, so I'll just do it in paint and post up a link.

imgur: the simple image sharer

Thanks, it would be helpful if you could explain it to me through paint also, as I don't really understand math displayed on a computer. I just don't know how to proceed on the question, and i don't even know what exactly it is asking me to do.

2. Re: Hey guys just looking for help with this question dealing with population varianc

Originally Posted by Joyeux
I don't quite know how to write this problem down here, so I'll just do it in paint and post up a link.

imgur: the simple image sharer

Thanks, it would be helpful if you could explain it to me through paint also, as I don't really understand math displayed on a computer. I just don't know how to proceed on the question, and i don't even know what exactly it is asking me to do.
Your equality is not true. The mean $\mu$ is missing an exponent of 2. See below (by the way, this is "math displayed on a computer", and I'm not quite sure how this is worse than "numbers by paint"):

$\sigma^2=\sum_{i=1}^n \frac{(x_i-\mu)^2}{n} = \sum_{i=1}^n \frac{x_i^2}{n} - 2\sum_{i=1}^n \frac{x_i \mu}{n} + \sum_{i=1}^n \frac{\mu^2}{n}$
$= \sum_{i=1}^n \frac{x_i^2}{n} - 2\mu\left(\sum_{i=1}^n \frac{x_i}{n}\right) + n\frac{\mu^2}{n}$
$= \sum_{i=1}^n \frac{x_i^2}{n} - 2\mu^2 + \mu^2$
$= \sum_{i=1}^n \frac{x_i^2}{n} - \mu^2$

3. Re: Hey guys just looking for help with this question dealing with population varianc

Oops sorry I edited it, I still don't know how to solve it though. I don't think it's asking to change from one form to another. Your solution didn't have the substituted version of mu. Still really confused.

4. Re: Hey guys just looking for help with this question dealing with population varianc

I think this is the question it's asking, imgur: the simple image sharer. Still not 100% sure though.