sigma/summation problem

**math321** · February 14th 2011, 07:24 PM

$\sum xy -\frac { (\sum x) (\sum y)}{n}$
x= 1, 3, 7, 10
y= 2, 0, 4, 8

**pickslides** · February 14th 2011, 07:32 PM

These are bivariate observations, then

$\displaystyle \sum xy -\frac { (\sum x) (\sum y)}{n} =1\times 2 + 3\times 0 +\dots +10\times 8 - \frac{(1+3+\dots +10)(2+0+\dots +8)}{4}= \dots$

**math321** · February 14th 2011, 07:38 PM

why did u put n = 4

**pickslides** · February 14th 2011, 07:53 PM

Well n=4 if the sample is counted as one per (x,y)

**topsquark** · February 15th 2011, 05:08 PM

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

**math321** · February 15th 2011, 05:12 PM

n represents the number of points of data

so what i suppose to do

**topsquark** · February 15th 2011, 05:17 PM

Originally Posted by math321

n represents the number of points of data

so what i suppose to do

Your data set is {(x, y)} = (1, 2), (3, 0), (7, 4), (10, 8).

-Dan

**math321** · February 15th 2011, 05:20 PM

so therefore n=4

**math321** · February 15th 2011, 05:21 PM

got 36.5

**HallsofIvy** · February 16th 2011, 03:35 AM

What exactly is your question? Do you not understand what the $\sum$ means? It just means "add them up". I assume that n is 4 here.
$\sum_{n=1}^4 x_ny_n- \frac{\left(\sum_{n=1}^4 x_n\right)\left(\sum_{n=1}^4 y_n\right)}{4}$
$= (1)(2)+ (3)(0)+ (7)(4)+ (10)(8)- \frac{(1+ 3+ 7+ 10)(2+ 0+ 4+ 8)}{4}$

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