Average probability

I have a scenario where people work in a call centre doing market research surveys. Have used a Poisson distribution to work out P(X<=x) where x is the number of surveys they complete. The parameter for the distribution for shift 1 is s1. For different shifts, the individuals will work different hours and on different projects. The average will therefore change. Given that I can work out the probability for each shift for P(X<=x), I wanted to know how one can work out the average probability over these shifts.

I have been working out the geometric mean of the probabilities from different shifts. Wanted to know how you think this average probability should be worked out.

Thanks.

Ludwik said:

I have a scenario where people work in a call centre doing market research surveys. Have used a Poisson distribution to work out P(X<=x) where x is the number of surveys they complete. The parameter for the distribution for shift 1 is s1. For different shifts, the individuals will work different hours and on different projects. The average will therefore change. Given that I can work out the probability for each shift for P(X<=x), I wanted to know how one can work out the average probability over these shifts.

I have been working out the geometric mean of the probabilities from different shifts. Wanted to know how you think this average probability should be worked out.

Thanks.

Click to expand...

Just what is the question? Do you want a probability that you will receive at least an average of x calls per shift or possible the average expected number of calls per shift or ...

A simple case with the set of probabilities of receiving at least x calls {p_j: j = 1, 2, 3, 4, ...,n}: Suppose you take 1000 samples per set (for each j). Then you would expect 1000*p_j calls per set and you would expect the total number of calls to be
S = \(\displaystyle \Sigma\,\, 1000\,\, p_j \)
You would have a total of 1000*n calls so you would expect the average probability of receiving at least x number of calls for each set to be
A = \(\displaystyle \dfrac{\Sigma\,\, 1000 p_j}{1000 n} = \dfrac{\Sigma\,\, p_j}{n} \)

Note that if x changed for each set, say, for example, if the x were the average number of calls for each individual set, A would be something like the expected probability that each set would have at least the average number of calls per set.