# sigma/summation problem

• Feb 14th 2011, 08:24 PM
math321
simple problem
$\sum xy -\frac { (\sum x) (\sum y)}{n}$
x= 1, 3, 7, 10
y= 2, 0, 4, 8
• Feb 14th 2011, 08:32 PM
pickslides
These are bivariate observations, then

$\displaystyle \sum xy -\frac { (\sum x) (\sum y)}{n} =1\times 2 + 3\times 0 +\dots +10\times 8 - \frac{(1+3+\dots +10)(2+0+\dots +8)}{4}= \dots$
• Feb 14th 2011, 08:38 PM
math321
why did u put n = 4
• Feb 14th 2011, 08:53 PM
pickslides
Well n=4 if the sample is counted as one per (x,y)
• Feb 15th 2011, 06:08 PM
topsquark

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

-Dan
• Feb 15th 2011, 06:12 PM
math321
n represents the number of points of data

so what i suppose to do
• Feb 15th 2011, 06:17 PM
topsquark
Quote:

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
• Feb 15th 2011, 06:20 PM
math321
so therefore n=4
• Feb 15th 2011, 06:21 PM
math321
got 36.5
• Feb 16th 2011, 04:35 AM
HallsofIvy
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.
$\sum_{n=1}^4 x_ny_n- \frac{\left(\sum_{n=1}^4 x_n\right)\left(\sum_{n=1}^4 y_n\right)}{4}$
$= (1)(2)+ (3)(0)+ (7)(4)+ (10)(8)- \frac{(1+ 3+ 7+ 10)(2+ 0+ 4+ 8)}{4}$