variance

**skhan** · May 24th 2006, 08:53 PM

A student wants to calculate the variance of a set of 10 scores. Unfortunately, he doesn't have the raw scores but only has the deviation of each raw score from the mean. Worse yet, he only has 9 of these deviation scores, as listed below. Compute the variance for him.

-5, +11, -4, -2, +7, -8, -6, +1, -3

I have no idea how to do this problem

**CaptainBlack** · May 24th 2006, 09:48 PM

Originally Posted by skhan

A student wants to calculate the variance of a set of 10 scores. Unfortunately, he doesn't have the raw scores but only has the deviation of each raw score from the mean. Worse yet, he only has 9 of these deviation scores, as listed below. Compute the variance for him.

-5, +11, -4, -2, +7, -8, -6, +1, -3

I have no idea how to do this problem

The last deviation can be found as the mean of the deviations from the mean
is zeros. So let $x$ be the missing deviation, then:

$ 0=(-5+11-4-2+7-8-6+1-3+x)/10, $

so

$ x=9$ .

Now the variance:

$ v=\frac{1}{10}\sum_{i=1}^{10} (x_i)^2 $

where the $x_i$ s are the deviations from the mean, so:

$ v=40.6 $

RonL

**skhan** · May 25th 2006, 09:41 AM

thanks
the correct answer given to this assignment question was 45.11 though

**CaptainBlack** · May 25th 2006, 10:57 AM

Originally Posted by skhan

thanks
the correct answer given to this assignment question was 45.11 though

That is because this given answer is using the unbiased population variance
estimator which is:

$ v_1=\frac{1}{n-1}\sum_{i=1}^n (x_i-\bar x)^2 $ .

which in this case is $\sim 45.11$ .

(Now I would argue that this is the wrong answer to the problem as asked,
which was what is the variance of the given set of deviations, but what
are you gonna do?)

RonL

**ThePerfectHacker** · May 25th 2006, 03:58 PM

I have a simpler version.
"The varaince is the squared-mean minus mean-squared"
The 'squared-mean' is the mean of the sum of squares.
The 'mean-squared' is the mean itself squared.

**CaptainBlack** · May 25th 2006, 10:01 PM

Originally Posted by ThePerfectHacker

I have a simpler version.
"The varaince is the squared-mean minus mean-squared"
The 'squared-mean' is the mean of the sum of squares.
The 'mean-squared' is the mean itself squared.

Will give you the first of the forms of variance (with weighting 1/n, not the
second form with weighting 1/(n+1)).

RonL

**ThePerfectHacker** · May 26th 2006, 07:46 AM

Originally Posted by CaptainBlack

Will give you the first of the forms of variance (with weighting 1/n, not the
second form with weighting 1/(n+1)).

RonL

What are you talking about?

**CaptainBlack** · May 26th 2006, 08:00 AM

Originally Posted by ThePerfectHacker

What are you talking about?

There are two common definitions in use for variance:

$ v_0=\frac{1}{n}\sum_{i=1}^n (x_i-\bar x)^2 $

$ v_1=\frac{1}{n-1}\sum_{i=1}^n (x_i-\bar x)^2 $

The first of these simplifies to:

$ v_0=\left[\frac{1}{n}\sum_{i=1}^n x_i^2 \right]-\bar x^2 $
.. the mean square minus the square mean,

and the second does not

RonL

**ThePerfectHacker** · May 26th 2006, 08:07 AM

Originally Posted by CaptainBlack

.. the mean square minus the square mean,

I see what you are saying.

But it should be,
squared-mean minus mean-squared.
The easy way to remember is the the word "mean" means middle. Thus, it needs to be in the middle of the sentence.

I use this mnemonic because last year in Math B class the students very insane about the variance formula and the math teacher said people have to memorize it. This is what some of them learned.

This is my 12

th Post!!!

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