$\displaystyle \sum xy -\frac { (\sum x) (\sum y)}{n}$

x= 1, 3, 7, 10

y= 2, 0, 4, 8

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- Feb 14th 2011, 07:24 PMmath321simple problem
$\displaystyle \sum xy -\frac { (\sum x) (\sum y)}{n}$

x= 1, 3, 7, 10

y= 2, 0, 4, 8 - Feb 14th 2011, 07:32 PMpickslides
These are bivariate observations, then

$\displaystyle \displaystyle \sum xy -\frac { (\sum x) (\sum y)}{n} =1\times 2 + 3\times 0 +\dots +10\times 8 - \frac{(1+3+\dots +10)(2+0+\dots +8)}{4}= \dots$ - Feb 14th 2011, 07:38 PMmath321
why did u put n = 4

- Feb 14th 2011, 07:53 PMpickslides
Well n=4 if the sample is counted as one per (x,y)

- Feb 15th 2011, 05:08 PMtopsquark
Please see rule #2 here. Please do not do this again.

Both sums are right. What's n? (That's the question you were last on in the other thread, too.)

-Dan - Feb 15th 2011, 05:12 PMmath321
n represents the number of points of data

so what i suppose to do - Feb 15th 2011, 05:17 PMtopsquark
- Feb 15th 2011, 05:20 PMmath321
so therefore n=4

- Feb 15th 2011, 05:21 PMmath321
got 36.5

- Feb 16th 2011, 03:35 AMHallsofIvy
What exactly is your question? Do you not understand what the $\displaystyle \sum$ means? It just means "add them up". I assume that n is 4 here.

$\displaystyle \sum_{n=1}^4 x_ny_n- \frac{\left(\sum_{n=1}^4 x_n\right)\left(\sum_{n=1}^4 y_n\right)}{4}$

$\displaystyle = (1)(2)+ (3)(0)+ (7)(4)+ (10)(8)- \frac{(1+ 3+ 7+ 10)(2+ 0+ 4+ 8)}{4}$