Distribution Function Probability

[boneill3]

Hi Guy's
I was wondering if anyone can check my work to see if I'm on the right track.

I was given the question

The random variable X has density f(x)= 1/2 e^(-x/2) for x>0


Find a number x0 such that P(X > x0) = 1/2


What I did was find the integral of f(x)= 1/2 e^(-x/2)
which equals

-e^(-x/2)

and solved for x

-e^(-x/2) = 1/2

x = 1.386

so I integrate

1/2 e^(-x/2) between [1.386, infinity] which gives 1/2

Is this right?

regards
Brendan

[jemidiah]

Yeah, that works, though you either omitted a step or got lucky when solving your integral for x. Notice that your integral is negative everywhere--plugging ~1.386 in you get -1/2, not 1/2.

We know P(X > x0) = Integral of f from x0 to infinity, basically by definition of the probability density. This integral is -e^(-x/2) *evaluated between x0 and infinity*. Then P(X > x0) = [-e^(-inf/2) + e^(-x0/2)] = e^(-x0/2)=1/2. You had -e^(-x0/2)=1/2. The exact solution happens to be 2ln2.