Figure out summation(x^2) in summation equation[Simple]

[giddy]

Hi,
So this is just part of my problem but its got me stumped for days and I can't ignore it since its popping up too often in my problems.

Homework Statement

For A sample of 140 bags of flour. The masses of x grams of the contents are summarized by \sum (x - 500) = -266 and \sum (x-500)^2=1178 I need to find the mean and estimated variance. The mean is simple 140(x - 500) = -266; mean = 498.3 But how the heck do I figure out \sum x^2 with the above info? I need only \sum x^2

The Attempt at a Solution

Mostly I just doodled pages trying to get this one! =S I tried 140(x - 500)^2 = 1178 And solve it, comes out as x = -1.780 or - 998.22. Which isn't correct. I need \sum x^2 basically in the formula for estimated variance s^2 = \frac{1}{n-1}(\sum x^2 - \frac{(\sum x)^2}{n})
I tried reworking from the answer(variance=4.839) so sum of x2 should be 34773692.21 but I don't know how to get to this answer?

 

[diggy]

You have a sum(x_squared) in your second equation if you expand it. Just like you have a sum of x in your first.

 

[giddy]

Sorry I am not sure what you mean =S

If I do expand (x - 500)^2 it'll be x^2 - 2(500)(x) + 500^2 Right? So where would I get sum of x? How would I expand sum(x^2)

 

[diggy]

How would I expand sum(x^2)

You don't have to expand it, just solve for it.

 

[Mark44]

You have
\sum_{i = 1}^{140}(x_i - 500)^2 = 1178
You can expand the sum on the left, and solve for \sum x^2.

\sum_{i = 1}^{140}(x_i - 500)^2 = 1178
\Rightarrow \sum_{i = 1}^{140}x_i^2 -2\sum_{i = 1}^{140} 500*x_i + \sum_{i = 1}^{140}500^2 = 1178
The second and third summations on the left can be simplified and substituted for.

 

[giddy]

aha.. ok so i didn't even know how to really solve summation equations, but I looked it up.

So Sum(x^2) = 34735178! And its correct... =)