Probability Density Function

**Collegeboy110** · March 29th 2009, 12:49 PM

My college proffesor says its fine to use the calculator to find the answer so for a and b I did.. I got 20 and 13.862. However I want to learn to do these problems by hand.. for c and d Im stuck.. Thank you for your help.

Suppose that p(x) = 0.05e^(-0.05x), where x is greater than 0, is a probability density function for the number of people in line at a movie theater.
a. Find the mean number of people in line and interpret.
b. Find the median number of people in line and interpret.
c. Find the probability that there are between 10 and 20 people in line
d. Find the probaiblity that there are more than 30 people in line.

**matheagle** · March 30th 2009, 10:52 PM

This is an exponential density ....

${1\over \theta} e^{-x/\theta}$ , for x>0.
The mean is $\theta$ , which 20 for your problem.

you can compute that via parts...

$E(X)=\int_0^{\infty}{x\over \theta} e^{-x/\theta}dx$ .

For (b), you want the value that puts .5 probability to the right and left, let's call that m for median...

$.5=\int_0^b{1\over \theta} e^{-x/\theta}dx=\int_b^{\infty}{1\over \theta} e^{-x/\theta}dx=e^{-b/\theta}$

I do have a problem with (c) and (d). The number of people doing anything is a discrete rv and this is a continuous distibution. This does cause a problem with the wording. Between 10 and 20 isn't clear. I don't know if 10 and 20 are included. If so you just need to calculate...

$\int_{10}^{20}{1\over \theta}e^{-x/\theta}dx$ and $\int_{30}^{\infty}{1\over \theta}e^{-x/\theta}dx$

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