Basic Statistical Values

Question

I wanted to make my own problem and solve it to learn some new stuff --

Let's say we have some type of standardized test, where a rough data curve depicts $P(x)=x^2e^{-\frac{x}{14}}$ which is the number of people who got that score.

Now, we have $\displaystyle \int_0^{100}P(x)\,dx = 5342.0$ approximately.

So, we scale our probabilities so we have $\displaystyle p(x)=\frac{1}{5342}x^2e^{-\frac{x}{14}}$.

Then, to get the mean, we evaluate $\displaystyle \int_0^{100}xp(x)\,dx$, right? I don't think we have to divide by $100$ because it's a weighted average and we have $p(x)$ being the weight of $x$.

This would give us $39.928$, which kind of makes sense from the graph!

Then, I think that variance would be $\displaystyle \int_0^{100}(x-39.928)^2p(x)\,dx=434.55$

Then, standard deviation would equal to $\sqrt{434.55}=\boxed{20.84}$ Are all my steps correct? What can I do with a standard deviation? Does the $68-95-99.7$ rule apply?

UPDATE

I now let $X$ range from $0$ to $+\infty$, and the results are really nice!

We have the density $p(x)=\displaystyle \frac{1}{5488}x^2e^{-\frac{x}{14}}$

We have mean $42$.

We have variance $588$.

We have standard deviation $\sqrt{588}$.

@spaceisdarkgreen Because what if I let it go all the way to infinity, then the mean.. OH !!!!! Omg ur right, I can go all the way to infinity!!! I was thinking originally that I would be dividing a finite integral over an infinite domain and then I forgot what I was doing omg thankssss — Saketh Malyala, Jun 17 '17 at 02:51
Oh, after I wrote that I actually figured it was cause it was scores on a test go from 0 to 100. But yes, the distribution go to infinity and still be well-defined. — spaceisdarkgreen, Jun 17 '17 at 02:54
See, I forgot, that we already redefined mean as integral of weight x value — Saketh Malyala, Jun 17 '17 at 03:07

spaceisdarkgreen · Accepted Answer · 2017-06-17T02:40:32.513

3

Yes, you have computed the quantities correctly. (But you made a minor typo in the denominator of $p(x)$).

As for the 68-95-99.7 rule, that applies to the Gaussian distribution and you can't always apply it to other distributions. You can compute the quantiles of this distribution, though. If you want the $95$-th percentile, just solve $$ \int_0^xp(t)dt = 0.95 $$ for $x.$ As for whether or not it will be approximately two standard deviations from the mean, that depends how much the distribution looks like a Gaussian. In this case it comes pretty close, but the relation has to break down for three standard deviations... that would be over 100.

edited Jun 17 '17 at 02:40

answered Jun 17 '17 at 02:26

spaceisdarkgreen

58,508

Ah I see. I will adjust my calculations to fix that error. Thanks!! – Saketh Malyala Jun 17 '17 at 02:49
It didn't go through to your calculations... I think it was just when you typed it up (the correct denominator is 5342) – spaceisdarkgreen Jun 17 '17 at 02:52

score 3 · Answer 2 · answered Jun 17 '17 at 02:41

3

Your formulas are correct and your results should also be correct should you make all your computations correct.

68-95-99.7 rule does not apply here because it is defined for the Gaussian distribution. Here you have a single tail..

answered Jun 17 '17 at 02:41

Seyhmus Güngören

7,848

Basic Statistical Values

2 Answers2