# Math Help - Exponential Distributions

1. ## Exponential Distributions

I am having a hard time distinguishing the difference between Poisson and Exponential distributions.

Here is the question I am on.

The time between calls to a plumbing supply business is exponentially distributed with a mean time between calls of 15 minutes

a.
What is the probability that there are no calls within a 30-minute interval?
b.
What is the probability that at least one call arrives within a 10-minute interval?
c.
What is the probability that the first call arrives within 5 and 10 minutes after opening?

I am confused about the part when the question asks, for example, "no calls within a 30 minute interval" or "at least one call in a 10 min interval" thanks for any help.

2. 1 The Poisson is discrete, in this case it's the number of calls per unit of time
2 The Exponential is continuous, it's the time between calls.

Let $X\sim Poisson(\lambda t)$

So $P(X=x)={e^{-\lambda t}(\lambda t)^x\over x!}$

Then $P(X=0)=e^{-\lambda t}$ means that we had zero arrivals by time t.

Let Y be the time of the first arrival (this is the exponential rv).

$P(Y>y)$ means the same, we have had no arrivals by time y.

So $P(Y>y)=e^{-\lambda y}$

But that is the distribution of an exponential, with parameter $\lambda$

So $F_Y(y)=P(Y\le y)=1-e^{-\lambda y}$

and $f_Y(y)= \lambda e^{-\lambda y}$

So in (a) you want no calls in any 30 minute period which is twice the mean

giving you $P(Y>2)=e^{-2}$

3. thanks for the clarification, what if I had the question.

The distance between major cracks in a highway follows an exponential distribution with a mean of 8 miles.

a) What is the probability that there are 2 major cracks in a 10-mile stretch of the highway?

what would I do about that "2 major cracks" part? Thanks.

could I just do (e^-1.25 * 1.25^2)/2! or would it be 1 - e^-1.25?

4. Originally Posted by Oblivionwarrior
thanks for the clarification, what if I had the question.

The distance between major cracks in a highway follows an exponential distribution with a mean of 8 miles.

X ~ Poisson $(\lambda = 10/8 = 1.25)$ where the mean is got by understanding the relationship between the Poisson distribution and the exponential distribution.