# Thread: need to solve summation equation to solve sum(x2)

1. ## need to solve summation equation to solve sum(x2)

So I need to calculate an estimated variance, the info given to me is:
A sample of 140 bags of flour. The masses of x grams of the contents are summarized by $\displaystyle \sum (x - 500) = -266$ and $\displaystyle \sum (x-500)^2=1178$ 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 $\displaystyle \sum x^2$

Someone suggested :
$\displaystyle \sum_{i = 1}^{140}(x_i - 500)^2 = 1178$
$\displaystyle \Rightarrow \sum_{i = 1}^{140}x_i^2 -2\sum_{i = 1}^{140} 500*x_i + \sum_{i = 1}^{140}500^2 = 1178$

But unfortunately I dont even know how to solve the above equations, I did google and read this : A-level Mathematics/FP1/Summation of Series - Wikibooks, collection of open-content textbooks

But It doesnt apply here since we're dealing with a random variable xi? Right? How do I simplify the second and third summations on the left

I dont need the answer, I just a hint or maybe how to solve summations like this? Could someone point me in the right direction?
(Im studying for my A levels on my own =S)

2. Originally Posted by giddy
So I need to calculate an estimated variance, the info given to me is:
A sample of 140 bags of flour. The masses of x grams of the contents are summarized by $\displaystyle \sum (x - 500) = -266$ and $\displaystyle \sum (x-500)^2=1178$ 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 $\displaystyle \sum x^2$

Someone suggested :
$\displaystyle \sum_{i = 1}^{140}(x_i - 500)^2 = 1178$
$\displaystyle \Rightarrow \sum_{i = 1}^{140}x_i^2 -2\sum_{i = 1}^{140} 500*x_i + \sum_{i = 1}^{140}500^2 = 1178$

But unfortunately I dont even know how to solve the above equations, I did google and read this : A-level Mathematics/FP1/Summation of Series - Wikibooks, collection of open-content textbooks

But It doesnt apply here since we're dealing with a random variable xi? Right? How do I simplify the second and third summations on the left

I dont need the answer, I just a hint or maybe how to solve summations like this? Could someone point me in the right direction?
(Im studying for my A levels on my own =S)
You know that:

$\displaystyle \displaystyle {-266 = \sum_{i = 1}^{140}(x_i - 500) = \sum_{i = 1}^{140} x_i - (500)(140) = \sum_{i = 1}^{140} x_i -70000 \Rightarrow \sum_{i = 1}^{140} x_i = .....}$

You also know that $\displaystyle \displaystyle {1178 = \sum_{i = 1}^{140}(x_i - 500)^2 = \sum_{i = 1}^{140} x_i^2 - 1000 \sum_{i = 1}^{140} x_i + 500^2 (140) }$.

Substitute the value of $\displaystyle \displaystyle {\sum_{i = 1}^{140} x_i}$ and make $\displaystyle \displaystyle {\sum_{i = 1}^{140} x_i^2}$ the subject.

Note: $\displaystyle \displaystyle {\sum_{i = 1}^{140} 1 = 140 (1) = 140 ....}$

3. Darn! That was simple I just didn't see it! =S

Thanks so much! On my way now to hypothesis testing of discrete variables!!