Hi I have some issue to understand the solution you proposed for the exercise in the topic name.
You use the exponential distribution to solve it with the following parameters x=1 mean=1/3 and cumulative = TRUE.
I though that this probability tell us that in the first minute we'll have x probability that an event occurs.
Apparently you use this value for the probability that 2 events occur within the first minute and I don't understand why.
I hope you can help me with that.
Thank you
Stefano
Hi Stefano,
Good question - this is a bit tricky & might take a minute to think about:
The probability of 2 arrivals within a minute basically means the probability of an arrival time of less than 1 minute (i.e. if a call arrives in less than a minute after this previous call, then the condition of 2 arrivals within a minute is satisfied).
The worksheet 'Neg. Exponential Distribution' shows the formula for calculating the cumulative probability of a service time up to a particular value.
The formula is =EXPONDIST(ServiceTime,1/MeanServiceTime,TRUE)
In this case, the first parameter is 1 (the particular value of ServiceTime being tested) and the second parameter is 1/MeanServiceTime, so with the MeanServiceTime of 3 minutes, that's why it is 1/3.
The formula EXPONDIST(1,1/3,TRUE) equates to 28.3%.
Cheers
Harold
Hi Harold,
First of all thanks for your reply.
I'm sorry but I don't really get it this part:
"The probability of 2 arrivals within a minute basically means the probability of an arrival time of less than 1 minute (i.e. if a call arrives in less than a minute after this previous call, then the condition of 2 arrivals within a minute is satisfied)."
Indeed if a call arrives after 30 seconds and the second one arrives after 40 seconds I would have both of the calls arriving in less than a minute but the condition that the two calls together should arrive in less than a minute shouldn't be satisfy isn't it?.
Furthermore I don't understand why a Lambda of 0.33333 per minute for you means that you'll have an occurrence in less than a minute.
Hope you can help me.
Thanks
Stefano
Hi Stefano,
Thanks for following up on this. I've prepared a visual simulation example that I hope will clarify and close out the subject . . .
It's in the attached workbook EC_Poisson Distribution_HG_CHECK.
Tab Inter-arrival CHECK contains a simulation of 1500 pairs of calls. [There is nothing magic about 1500, I just chose it as a reasonable sized sample.]
The arrival times are determined by the arrival formula at the left of the tab. The times of arrival are in cols L and M. Then we check (in col N) whether they arrive within a minute of each other. There is a "Y" in col N when they arrive within a minute of each other.
Then if you count the number of "Y"s out of 1500, you get the result. Every time you refresh the sample [press F9 to refresh], you will of course get a slightly different result based on the random numbers used to generate the arrival times.
Anyway, the results tend to cluster around 28.3%, so I'm confident that the analytical approach discussed previously is OK.
If you are interested, there are some worked examples [not mine] you can look at:
https://courses.lumenlearning.com/introstats1/chapter/the-exponential-distribution/
Cheers
Harold
