I am having a issue with the wording of this question.
Find the probability of the following.
The velocity $v$ of a randomly selected particle, whose distribution obeys the probability density function
$$P(\frac{v}{v_{avg}})=exp(-\frac{v}{v_{avg}})$$ will lie between 0 and $2_{v_{avg}}$ where $v_{avg}$ is the average velocity.
I have been given the density function so I just integrated between the limits to find the probability between the given points.
which came to be $P(0\leq v \leq 2v_{avg})=0.86 v_{avg}$
However I am not happy with this as to me the probability can change given a different average velocity which dose not make sense.
I have looked at various probability distribution information in a text book am I using by John E.Frenuds and I cant see to find anything that jumps out then only thing that did is the use of 'probability density' where they define a given function say
$$f(x)=\left\{\begin{matrix} kx^2 >0 \\0 \: everywhere \: else \end{matrix}\right.$$
So then it get normalized and integrated between given limits, which make me think the above equation given is maybe a 'probability density function' as been used in the book.
Could someone maybe explain if I have use the correct method and if not, expand where I have gone wrong.
The result of your calculation is correct, but the result itself requires the normalization factor $a=v_{avg}$, since $$\int_0^{\infty}e^{-\frac{t}{a}}\;dt = a$$
The probability is then independent of the average velocity because of the specific properties of the exponential distribution:
You have in general $$\frac {1}{a}\int_0^{2a}e^{-\frac{t}{a}}\;dt= \int_0^{2}e^{-x}\;dx = 1-\frac{1}{e^2}$$
With other probability distributions you may not see such a result in general.