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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.

enter image description here

Given D = 0.07

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1  
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

1 Answer 1

You need to interpolate between those values and extrapolate beyond them.

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 $1732.5-39^2=211.5.$

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Thanks alot Henry. I have posted the graph, please take a look :) –  ben Oct 21 '12 at 0:40
    
@ben: My first calculation (a uniform distribution) would have a straight diagonal line, while my second would look like your graph, but with additional vertical steps at 60 and 65 as part of the extrapolation, both otherwise is the same as your graph. –  Henry Oct 21 '12 at 7:39

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