# Thread: Probability Distribution Function

1. ## Probability Distribution Function

Hi Every one. Can any one help me to start solving this problem?i am stocked in the beginning.

The range of a ballistic projectile is given by

r=(v^2)sin2x/g

where v is the initial velocity, x is the launch angle (relative to the horizontal) and g is the
gravitational acceleration. The launch angle cannot be controlled precisely so is treated as a random variable x with f x(x). It should be clear that the maximum range occurs for x=0.25pi
.

#Determine an expression for the PDF of the range (now a random variable,r~f (r))
assuming that x can never exceed the interval 0<x<0.5pi(0 and 0.5pi included)

thanks

2. ## Re: Probability Distribution Function

Do you mean now is a random variable with bounded support , and e.g. is a continuous random variable with a certain pdf, which you would like to know the pdf of ?

3. ## Re: Probability Distribution Function

actually i just found out i need to know the pdf os R which is equal to pdf of x in some point. i hope i did not make any one confuse
thanks for your concern

4. ## Re: Probability Distribution Function

I think you were saying that you figured it out? I don't know... I'm just confused...

no i did not

6. ## Re: Probability Distribution Function

Assume you want to obtain the pdf of in terms of the pdf of .

Then just use the Jacobian transform will do.

7. ## Re: Probability Distribution Function

so do you think if i start like: P(R<a)
it gives me the probability of R in point (a) which is equal to P(x). then derivative of that to get pdf
am i in right track?

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