- #1
kuahji
- 394
- 2
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
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?
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?