1. ## exponential

The number of minutes between trucks arriving at a checking station are shown to follow an exponential distribution, with an average of fifty minutes between trucks arriving. The administrators of the station are concerned about congestion. It requests information on the probability that at least twenty minutes will occur between truck arrivals. You calculate this to be:

and i keep trying to get the answer but i can't

here's my working:

$\displaystyle \lambda = 0.4, x = 20$
$\displaystyle (.4) e^{-.4 \times 20} = 0.000134185$

and that's wrong, can someone please show me how to do this?

2. Originally Posted by gconfused

The number of minutes between trucks arriving at a checking station are shown to follow an exponential distribution, with an average of fifty minutes between trucks arriving. The administrators of the station are concerned about congestion. It requests information on the probability that at least twenty minutes will occur between truck arrivals. You calculate this to be:

and i keep trying to get the answer but i can't

here's my working:

$\displaystyle \lambda = 0.4, x = 20$
$\displaystyle (.4) e^{-.4 \times 20} = 0.000134185$

and that's wrong, can someone please show me how to do this?
Read Exponential distribution - Wikipedia, the free encyclopedia: $\displaystyle f(x) = \frac{1}{50} e^{-x/50}$.

$\displaystyle \Pr(X \geq 20) = \int_{20}^{+ \infty} f(x) \, dx$.

3. hmmm i dont think that we're meant to integrate coz i havent seen an integral sign at all in this subject.. but thanks though

4. Originally Posted by gconfused
hmmm i dont think that we're meant to integrate coz i havent seen an integral sign at all in this subject.. but thanks though
Then how are you expected to calculate the probability?

5. well i was thinking

$\displaystyle e^{-.4}$

that works but then i dont get why the $\displaystyle \lambda = 1$ and $\displaystyle x = .4$.

but yeahhh we havent done integration at all in this subject.. this question is a multiple choice question and they told us that the answer is .67

coz like for the average is 1/50 and we want 20 minutes right, so i divided the 1 by 2.5 (50/20 = 2.5) so i thought $\displaystyle \lambda = .4$ and x = 20

6. Originally Posted by gconfused
well i was thinking

$\displaystyle e^{-.4}$

that works but then i dont get why the $\displaystyle \lambda = 1$ and $\displaystyle x = .4$.

but yeahhh we havent done integration at all in this subject.. this question is a multiple choice question and they told us that the answer is .67

coz like for the average is 1/50 and we want 20 minutes right, so i divided the 1 by 2.5 (50/20 = 2.5) so i thought $\displaystyle \lambda = .4$ and x = 20
Sorry but your logic is incomprehensible to me. I suggest you ask your teacher how s/he expects you to do it at this current stage of your learning. What I posted is how it has to be done.

7. Here's a formula you were probably given maybe:

$\displaystyle P(x>=x_0) = e^{-\lambda x_0}$

This always yields the right tail tail of the distribution - which is what you want since youre calculating x>20..

So, the value of $\displaystyle \lambda$ is 1 per 50 minutes so 1/50 = 0.02 per minute. The question is to determine the probability of there being at least 20 minutes between truck arrivals, so:

$\displaystyle P(x >= 20) = e^{-(0.02)(20)} = 0.67$