What is the variance of the data set? 63 54 62 59 52 (Points : 1) 11.6 14.5 18.8 23.5
@rishavraj
id subtract 52 from them all and then work it .. if i had to do it by hand
hmmmmm u know how to calculate the variance.... u need to find the average of the numbers first
i dont know how to calculate the variance
63 54 62 59 52 11 2 10 7 0 subtract another 6 5 -4 4 1 -6 this set is simpler to work by hand, and has the same variance
i guess 57 was the mean of the larger set, cause its 0 now
Why 6? I still am entirely confused
\[mean = \frac{ 63 + 54 + 62 + 59 + 52 }{ 5 }\]
all i did was shift the data set to smaller values, the varaince doesnt change since we still have the same internal measures
mean was 58 now, variance is determined by subtracting the mean from the data points anyways, to zero things out.
i just happened to get it zeroed out on a whim :) take the average of the squared results
5 -4 4 1 -6 square these out?
yep
we would have subtracted the mean of that smaller set from its values, but the mean just happened to be 0 ... soo, square, sum, divide by 5
2.23 -2 2 1 -2.44 ?
not sqrt, squares
25 16 16 1 36
-_- pardon my stupidity. So now I add em up and divide by 5 right?
variance is calcualted as the average of the squared differences from the mean. sum of [x-mean]^2 divided by the number of datapoints yes, add them up and divide by 5
18.8
yep
X = (Xi........ mean.........Xf) subtract the mean from the set of X X-mean = (Xi-mean........ 0.........Xf-mean) we have zeroed the set, square the differences from the mean and take their average
if we do this by hand, the sooner we can get the set near its zeroed mark, the simpler the calculations will be.
otherwise, stick them in the stats calculator and hit a button :)
as an aside, spose we got it down to this: 6 -3 5 2 -5 the mean is 5/5 = 1 we would subtract 1 from the set, and then square add average
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