# How to calculate the mean and variance of a cumulative probability graph?

I'm given a cumulative probability graph, however i can't post any images. Thus, the X and Y coordinates of the graph are:

X:

2.5 7.5 12.5 17.5 22.5 27.5 32.5 37.5 42.5 47.5 52.5 57.5

Y:

0 0 0 0.07 0.17 0.27 0.37 0.47 0.57 0.67 0.77 0.87

How do i calculate the mean and variance of this cumulative probability graph?

Thanks alot!!!!

EDIT:

Now that i have enough rep points, i can post the picture of the graph.

Given D = 0.07

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Hint: For your problem, where $X$ takes on only positive values, $E[X]$ equals the area between the curve $F(x)$ and the line $y = 1$ for nonnegative $x$ from $0$ to $\infty$. Thus, area of rectangular region with opposing corners at $(0,0)$ and $(2.5,1)$ plus area of rectangular region with opposing corners $(2.5,0)$ and $(7.5, 1)$ plus... Note that what the area is between $x=7.5$ and $x=12.5$ depends on the graph of $F(x)$ which may increase linearly or as some smooth rise from $(7.5,0)$ to $(12.5,0.07)$, or in a sudden jump at $x=12.5$, etc. –  Dilip Sarwate Oct 19 '12 at 20:49
As well as Dilip's points, you need to know about the right tail of the distribution (or be able to estimate it), i.e. when the cumulative probability reaches $1$ –  Henry Oct 19 '12 at 22:05

One possibility is that what you actually have is a uniform distribution on the interval $[14,64]$ in which case the mean would be $\dfrac{14+64}{2}=39$ and variance $\dfrac{(64-14)^2}{12}\approx 208.3333$.
In most cases life will not be as simple as this, and you will need numerical methods. You have incomplete data, so suppose your $X_i$ finish with 62.5 67.5 and your $Y_i$ finish with 0.97 1. Then you can estimate, by putting the probability at the centre of each interval, the first moment with $$\sum_i (Y_i-Y_{i-1})\left(\frac{X_{i-1}+X_i}{2}\right)$$ which in this case would be $39$, and estimate the second moment with $$\sum_i (Y_i-Y_{i-1})\left(\frac{X_{i-1}+X_i}{2}\right)^2$$ which in this case would be $1732.5$, giving an estimate of the variance of $1732.5-39^2=211.5.$