probability distribution function

[winter wing]

Here i need to integrate this function 0 to infinity 1/4xe^(-x/2) which is 1?

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[tall cape]

Yep, integration by parts will work to find it.

[winter wing]

So here I got the integration value is 1

So mean will be 1/2?

[tall cape]

You'll have to do another integral to find the mean: $\int_0^\infty xf(x)\mathrm{d}x = \int_0^\infty x\left(\frac{1}{4}xe^{-x/2}\right)\mathrm{d}x$

[deep spade]

MicMac

[winter wing]

Ohh i got it now

So here it is 4

[tall cape]

There you go!

[winter wing]

And variance is double?

[tall cape]

Variance is $E[x^2] - \mu^2= \int_0^\infty x^2f(x)\mathrm{d}x - \mu^2$, where $\mu$ is the mean you found, 4.

[deep spade]

MicMac

[winter wing]

Could you tell me what books you read for statistics and probability?