Homework Help: 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.