Alternative formula for variance problem

[Biosyn]

Homework Statement

Which of the following formulas represents the variance of the data set {x1,x2,x3,...,x10}?
(μ denotes the mean of the data set)

Here is a photo that I took of the problem for better understanding.

http://i.imgur.com/PqEKajx.jpg?1?7734

I understand why the answer I chose is wrong.
What I don't understand is how e) is the answer. I did the calculations by hand with that formula and it is correct.

Would someone please show me how that formula is derived from the variance formula we usually see:

Homework Equations

Formula for variance

The Attempt at a Solution

 

[SammyS]

Homework Statement

Which of the following formulas represents the variance of the data set {x1,x2,x3,...,x10}?
(μ denotes the mean of the data set)

Here is a photo that I took of the problem for better understanding.

http://i.imgur.com/PqEKajx.jpg?1?7734

I understand why the answer I chose is wrong.
What I don't understand is how e) is the answer. I did the calculations by hand with that formula and it is correct.

Would someone please show me how that formula is derived from the variance formula we usually see:

Homework Equations

Formula for variance

The Attempt at a Solution

In your case this becomes

\displaystyle \sigma^2=\frac{\sum_{i=1}^{10}(x_i-\mu)^2}{10}​

Expand the square.

\displaystyle (x_i-\mu)^2=x_i^2-2x_i\mu+\mu^2

Now, take the sum of each term. Then recall how you calculate μ .

 

[Biosyn]

In your case this becomes

\displaystyle \sigma^2=\frac{\sum_{i=1}^{10}(x_i-\mu)^2}{10}​

Expand the square.

\displaystyle (x_i-\mu)^2=x_i^2-2x_i\mu+\mu^2

Now, take the sum of each term. Then recall how you calculate μ .

I get \frac{{x_1 + x_2 + x_3...+x_10} + { 2x_1μ + 2x_2μ...+2x_10μ} + 10μ^2}{10}

I can see where \displaystyle \sigma^2=\frac{\sum_{i=1}^{10}(x_i)^2}{10} comes from and that 10μ^2 cancels out.

What about the {2x_1\mu + 2x_2μ...+2x__10\mu} ?p.s. Sorry, having a hard time with Latex. I hope you understand what I mean! :O

 

[SammyS]

I get \frac{{x_1 + x_2 + x_3...+x_10} + { 2x_1μ + 2x_2μ...+2x_10μ} + 10μ^2}{10}

I can see where \displaystyle \sigma^2=\frac{\sum_{i=1}^{10}(x_i)^2}{10} comes from and that 10μ^2 cancels out.

What about the {2x_1\mu + 2x_2\mu...+2x_10\mu} ?

Divide that by N, i.e. 10 .

That's \displaystyle \ \ 2\mu\frac{\sum_{i=1}^{10}x_i}{10}

 

[Biosyn]

Divide that by N, i.e. 10 .

That's \displaystyle \ \ 2\mu\frac{\sum_{i=1}^{10}x_i}{10}

Oh, stupid me. >.>
Thanks for your help!

I understand now:
http://i.imgur.com/NpdcN86.jpg

 

[Ray Vickson]

I get \frac{{x_1 + x_2 + x_3...+x_10} + { 2x_1μ + 2x_2μ...+2x_10μ} + 10μ^2}{10}

I can see where \displaystyle \sigma^2=\frac{\sum_{i=1}^{10}(x_i)^2}{10} comes from and that 10μ^2 cancels out.

What about the {2x_1\mu + 2x_2μ...+2x__10\mu} ?


p.s. Sorry, having a hard time with Latex. I hope you understand what I mean! :O

You should not get \frac{{x_1 + x_2 + x_3...+x_10} + { 2x_1μ + 2x_2μ...+2x_10μ} + 10μ^2}{10}, which is wrong. You should not get \displaystyle \sigma^2=\frac{\sum_{i=1}^{10}(x_i)^2}{10} because that is also wrong unless μ = 0. Start over, and proceed carefully!

 

[Biosyn]

You should not get \frac{{x_1 + x_2 + x_3...+x_10} + { 2x_1μ + 2x_2μ...+2x_10μ} + 10μ^2}{10}, which is wrong. You should not get \displaystyle \sigma^2=\frac{\sum_{i=1}^{10}(x_i)^2}{10} because that is also wrong unless μ = 0. Start over, and proceed carefully!

Sorry, I was a bit lazy typing out my work. But in the post before this I solved it!
http://i.imgur.com/NpdcN86.jpg
Thanks anyways. :P