Help with finding median of continous probability density function?

[aahmedov]

f(x) = x/5 0 ≤ x ≤ 2

8/(5x^2) 2 ≤ x ≤ 8

Thanks

[Sammie Backman]

I'm not doing that option, but I'm bored and will therefore give it a try anyway.

Let X be the random variable whose probability density is given by f(x).

The probability that a≤X≤b is equal to the integral of f(x) from a to b.

The probability that X is between 0 and the median value is equal to 0.5.

The integral of f(x) is (x²/10) when 0≤x≤2 and -8/5x when 2≤x≤8.

The integral of f(x) from 0 to 2 is then equal to (2²/10)-(0/10) = 0.4

Let u be the median value. We now know that the integral of f(x) from 2 to u is equal to 0.1 (as 0.1+0.4 = 0.5).

Hence (-8/5u) - (-8/10) = 0.1

Solving this equation gives u = 80/35.

So the median is 80/35 which is 2.29 to three significant figures.

I hope I didn't mess up there

Edited July 20, 2012 by Sammie Backman

[aahmedov]

Thanks Sammie Backman, ur a genius. Not only did you get the correct answer for this question, you taught me how to solve these double continous density functions in general.

[The Economist]

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