Hi

If out of 50,000 calls, 48,000 did not last longer than 20 minutes. How do we determine the mean holding time h?

Thanks in advanced

 

Get 50,000 friends with stopwatches and telephones.

If there were 50,000 operators, there would be no holding time. Is this ALL info avail on the problem?

 

The full unedited question is:

A company made a survey and found that out of 50,000 calls it made 48,000 did not last longer than 20 minutes. Determine the mean holding time h.

 

Get 50,000 friends with stopwatches and telephones.

If there were 50,000 operators, there would be no holding time. Is this ALL info avail on the problem?

retched

See my previous response for the answer to your question.

 

Average call duration is <20 min, but call rate is not listed, so I haven't an idea how to solve this. Im sure Erlang would have an idea. Hopefully someone else will have a formula to solve for h with so few stats.

I would think it would be 50000/48000 * 20 But I could be wrong. h=20.833

 

Average call duration is <20 min, but call rate is not listed, so I haven't an idea how to solve this. Im sure Erlang would have an idea. Hopefully someone else will have a formula to solve for h with so few stats.

I would think it would be 50000/48000 * 20 But I could be wrong. h=20.833

I disagree. Average call duration cannot be determined, since the 2000 calls that were >20 minutes could have averaged more than 8 hours each.

On the other hand, one can cay the median call time is 20 minutes or less.

John

BTW, When we studied our call center, duration of holding was not normally distributed. It seemed to be at least bi-modal. A lot of people gave up quickly (e.g., <2 minutes), but most of those who stayed seemed to be there for the duration.

 

I disagree. Average call duration cannot be determined, since the 2000 calls that were >20 minutes could have averaged more than 8 hours each.

On the other hand, one can cay the median call time is 20 minutes or less.

John

BTW, When we studied our call center, duration of holding was not normally distributed. It seemed to be at least bi-modal. A lot of people gave up quickly (e.g., <2 minutes), but most of those who stayed seemed to be there for the duration.

You think poisson probability distribution plays any role in this question?

 

jpanhalt, I agree. I was reaching. In most of the equations I have run into, mean hold time is already defined, or atleast in a period of time where it can be determined.

darkman12, Probability likely does play a role in this due to the simple fact that people are random. Most call centers will go by averages+ in an attempt to mitigate loss. If a call center averages 1000 calls an hour, they will want to be sure they are capable of 1100+

It has been a while since I have been involved in the call center environment, and I was not involved in design, but developing the hardware and software to handle routing. We would do estimates of data rates to be sure the hardware that served the prerecorded messages and menu systems could handle the projected load plus 50%. That was considered conservative. From what I was told, call centers wanted capacity of double the average calls. This would attempt to account for system crashes, personnel problems, etc.

 

Although average (aka mean) is frequently used for non-normal distributions, as it is a very robust statistical parameter, there are instances in which it can give a misleading view of what actually is happening, such as when there is a distinct bi-modal or non-continuous distribution.

If you can define the distribution, then the problem may become considerably simpler. But then, that is also a different question.

John