Probability distribution of a moving particle

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.

• Note that the probability density you're given is the probability density of a variable $v/v_{avg}$, not just $v$. So you need to calculate $P(0<v/v_{avg}<2) = \int_0^2 e^{-x} dx$. – Adam Latosiński Apr 18 at 11:41
• Please notice that your distribution P() is not a probability distribution function (PDF) of $v$ as it is not normalized. The PDF is rather $f(v)=\frac{1}{v_{avg}} Exp\left(- \frac{v}{v_{avg}}\right)$. The average of $v$ is then $<v> = \int_0^\infty v f(v) \,dv = v_{avg}$, the variance is $<(v-v_{avg})^2>= v_{avg}^2$. – Dr. Wolfgang Hintze Apr 18 at 12:25

The result of your calculation is correct, but the result itself requires the normalization factor $$a=v_{avg}$$, since $$\int_0^{\infty}e^{-\frac{t}{a}}\;dt = a$$
You have in general $$\frac {1}{a}\int_0^{2a}e^{-\frac{t}{a}}\;dt= \int_0^{2}e^{-x}\;dx = 1-\frac{1}{e^2}$$
• In general $e^{-t/a}$ isn't a probability density, because it's not normalized. And if you do normalize it, the dependence of the final result on $a$ vanishes. – Adam Latosiński Apr 21 at 21:31
• @james2018 : I had to adjust the answer as the distribution in you question seems to miss a normalization factor depending on how one interprets it. If the distribution refers to $v$, then the distribution needs normalization. If it refers to $v/v_{avg}$, then all is good but integration limits need to be from 0 to 2. – trancelocation Apr 22 at 7:15