# Exponential distribution, two exercises

1. Jul 14, 2013

### skrat

1. The problem statement, all variables and given/known data
Waiting time in a restaurant is exponentially distributed variable, with average of 4 minutes. What is the probability, that a student will in at least 4 out of 6 days get his meal in less than 3 minutes?

2. Relevant equations

3. The attempt at a solution
If I understand correctly I could say that $f(t)=e^{-\lambda t}$, meaning $P(t<3)=\int_{0}^{3}\omega (t)dt=\int_{0}^{3}(\frac{f(t)}{dt})dt=\int_{0}^{3}\lambda e^{-\lambda t}dt$

Ok? This is now probability that a student will get his meal in less than 3 minutes, so the probability that it would take longer is $1-P(t<3)$ and the final result should be something like $\sum_{i=4}^{6}\begin{pmatrix} 6\\ i \end{pmatrix}p^{i}(1-p)^{6-i}$

Does this sound right? How do I determine $\lambda$???

1. The problem statement, all variables and given/known data
Elapsed time until a device suddenly stops working is exponentially distributed with median 4h. Calculate the probability, that the device will work at least for 5 hours!

2. Relevant equations

3. The attempt at a solution
Very different as before, but still have no idea how to get $\lambda$???

2. Jul 14, 2013

### Ray Vickson

Your notation is unfortunate; the usual expression for the density of the exponential is
$$f(t) = \left\{ \begin{array}{l} \lambda e^{-\lambda t},\; t \geq 0\\ f(t) = 0, \;t < 0 \end{array} \right.$$ The (cumulative) distribution function is
$$F(t) = \int_0^t f(x) \, dx = 1 - e^{-\lambda t},$$
while the complementatry cumulative is $P\{ T > t \} = 1 - F(t) = e^{-\lambda t}.$

You get the mean by using standard formulas found in any textbook or on-line. Alternatively, you can perform the integration $E T = \int_0^{\infty} t f(t) \, dt$ and see what you get.

BTW: the easiest way to get a binomial coefficient in TeX is to use the command {n \choose m}, which gives ${n \choose m}.$

Last edited: Jul 14, 2013
3. Jul 14, 2013

### skrat

Hmmm... Just cheking:$F(t)=\int_{0}^{t }f(t)dt=\int_{0}^{t }e^{-\lambda t}dt=-\frac{1}{\lambda }\int_{0}^{u_1 }e^{u}du=\frac{1}{\lambda }(1-e^{-\lambda t})$
What does $\frac{1}{\lambda }$ tell me, if anything, because I don't see it in your calculation?

AHA! So $E(t)=\int_{0}^{\infty }tf(t)dt=\int_{0}^{\infty }te^{-\lambda t}dt=\frac{\lambda t+1}{\lambda ^{2}}=4$ Which gives me a quadratic equation for $\lambda$ where the right soultion is with + so $\lambda =\frac{1+\sqrt{5}}{8}$

Thanks for helping and this hint!

4. Jul 14, 2013

### Ray Vickson

The density should read as $f(t) = \lambda e^{-\lambda t}$; I have edited the post to correct this. So, subsequent expressions are OK: $F(t) = 1 - e^{-\lambda t}$, etc. Think of it this way: if t has dimensions of time, λ must have dimensions 1/time = rate per unit time; that makes the product λt dimensionless, which it must be if we are going to exponentiate it. Similarly, probabilities are dimensionless numbers, so F(t) cannot have λ or 1/λ outside the exponential.

Your computation of ET is incorrect; getting λ is much, much easier than what you wrote. Look in a book, or look on line!

5. Jul 14, 2013

### skrat

Ammm, this is written in my notebook, just as it is written on wikipedia: http://en.wikipedia.org/wiki/Expected_value#Univariate_continuous_random_variable

So expected value $E(t)=\int_{-\infty }^{\infty }tf(t)dt$ but $t$ can only be positive and $f(t)=\lambda e^{-\lambda t}$.

I am a bit confused at this moment? How is this incorrect?

6. Jul 14, 2013

### Ray Vickson

This part is correct, but you are not finished. You need to do the integral (or consult a book or a web page, as I have said now for the third---and last---time). In a previous post you gave some weird expression for the mean that was incorrect (for one thing, it involved 't, but that has already been integrated out). You had a quadratic equation to solve, which is totally false.

7. Jul 14, 2013

### skrat

According to wolfram alpha and my own calculations with the help of a book the integral is http://www.wolframalpha.com/input/?i=axe^(-ax)+integrate

$E(t)=\int_{0}^{\infty }t\lambda e^{-\lambda t}dt=-\frac{1}{\lambda }(e^{-\infty}(\lambda t+1))+\frac{1}{\lambda }(e^{0}(\lambda t+1))=\frac{1}{\lambda }(\lambda t+1)=4$
now $4\lambda=\lambda t+1$

What is wrong here?

8. Jul 14, 2013

### Ray Vickson

There should be no 't' in the final answer. As I said before, 't' has been integrated out---it is gone.