# Math Help - Another qn!

1. ## Another qn!

Jack drives down to town every morning. On 16 randomly chosen trips, he recorded his travel time xi (i is a subscript!) in minutes.

He finds that $\displaystyle\sum_{i=1}^{16} xi = 519$ and $\displaystyle\sum_{i=1}^{16} (xi)^2 = 16983$

Calculate the unbiased estimates of the variance of the driving times to town.
I could calculate the unbiased estimate of the mean, but I dont get how to get the variance from $\displaystyle\sum_{i=1}^{16} (xi)^2 = 16983$. What can I get from the sum of (xi)^2? I know I can easily get the mean from the sum of xi, but (xi)^2? Isn't the variance the sum of ((xi-mean)^2) / n so it has nothing to do with the sum of (xi)^2?

Thank you so much in advance, really appreciate this!

2. ## Re: Another qn!

Hi Tutu,

I think you know how to answer your own question, so I'm going to do my best to give you a hint without telling you what to do. You said,

Isn't the variance the sum of ((xi-mean)^2) / n so it has nothing to do with the sum of (xi)^2?
But $\frac{1}{n}\sum(x_{i}-\mu)^{2}$ does have something to do with $\sum x_{i}^{2}.$ Stare at $(x_{i}-\mu)^{2}$ for a minute and I think you'll see what needs to happen.

Let me know if anything is unclear. Good luck!

3. ## Re: Another qn!

Hi I stared. I can infer that I am to subtract the mean from every xi value. And I know the mean already. But how to, if they already give me the sum of (xi)^2? I am unsure about how to insert the equation..Is it (xi-Mean)^2 = 16983?

4. ## Re: Another qn!

I looked at a few videos that manipulated the variance formula. Decided to use (sum of (xi)^2)/N - mean^2 as the variance. However, I got 11.7, which is still not the answer..

5. ## Re: Another qn!

Apparently this formula will get me the answer, 9.84.
(N * sum of all x² - (sum of all x's)²) / (N * (N-1))

Can someone tell me how this formula is the variance? How is it derived algebraically from the variance? Please help me, really.