A random variable X is gamma distributed with a=3, b=2 (alpha/beta). Determine the probability that 1[tex]\leq[/tex]X[tex]\leq[/tex]2(adsbygoogle = window.adsbygoogle || []).push({});

The first thing I did was plug a & b into the gamma distribution formula 1/(b^a [tex]\Gamma[/tex]a * x^(a-1) * e^(-x/b)

Which I ended up with

1/16(x^2 * e^(-x/2)

Which I then though I'd take the integral from 1 to 2

[tex]\int[/tex] 1/16(x^2 * e^(-x/2) dx = 1/16(-2x^2 * e^-.5 - 8x * e^-.5 - 16e^-.5) & then I evaluated that from 1 to 2. Which gave .06...

The answer in the book gives (b-a)^4/80

So where am I going wrong & am I just approaching this problem in totally the wrong way?

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# Homework Help: Gamma Distribution (Statistics)

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