1. Exponential Distribution
‘waiting time density’
the time until the next event for a Poisson distribution.
The mean number of events per unit time is represented by λ.
X ~ Exp(λ)

2. In Poisson – the variable is the number of events in an interval (discrete)
In the Exp dist – the variable is the waiting time until the next event (time is continuous).
The pdf:
x ≥ 0 as time cannot be negative.
Exp dist is consider ‘memoryless’ – the mean waiting time can start at any moment. If you have waited 30 mins without the next event occuring, the mean waiting time is still 10 mins.

3. The exp dist is always a decreasing function - The mode of the exp dist is always 0 - λ is a parameter affecting the decay rate.

4. The mean =
If the mean number of events per hour is 5 then λ = 5 and the mean waiting time will be 1/5 so 12 minutes.

5. Variance =
The std dev = σ =

6. The probability that the waiting time is a minutes or less when X ~ Exp(λ) is
P(X ≤ a) =
- Thus the probability of waiting at least a minutes is
P(X ≥ a) = 1 – P(X ≤ a) = 1 – ( ) =

7. Median waiting time So

8. An online statistics forum gets 3 postings per randomly distributed per hour.
A) If a posting was just made, find the mean waiting time to the next posting.
Soln: the mean psoting time is 1/λ which is 1/3. this has to be converted to minutes. It is 1/3rd of an hour which is 20 mins.
B) If a posting was made 10 minutes ago, find the mean waiting time to the next posting.
Soln: ‘memoryless’, so 20 mins.

9. C) Find the standrad deviation of the waiting time to the next posting.
Soln: μ = σ so also 20 mins.
D) Find the probability that the waiting time will be 30 minutes or less.
Soln: P(X≤ 30) =
One can also use P(X≤a) = 1 –
which is = 0.777

10. E) Find the median waiting time to the next posting
Soln: this can be solved using:

11. We could also use the formula
Here