The probability of n telephone calls received at an exchange in a time frame (t_i, t_f ) is given by the Poisson point distribution:

P(n) = [Lambda(t_f-t_i)]^n/n! e^􀀀(t_f - t_i)

where Lambda is a constant called the rate parameter.

I want to:

  1. Find the expected number of calls in 6 hours (take the unit of time as hours) for a rate parameter of 5.
  2. What is the smallest rate parameter such that the probability of 8 or fewer calls in 2 hours is no more than 0.9?

For part 1 I tried: Solve[p /. {tf -> 6, ti -> 0, \[Lambda] -> 5}, n] and for part 2 I tried Solve[p /. {n -> 8, tf -> 2, ti -> 0} == 0.9, \[Lambda]] but I can't seem to figure it out. Some help would be much appreciated.

  • $\begingroup$ Have you seen PoissonDistribution[]? $\endgroup$ – J. M.'s technical difficulties Apr 15 '16 at 6:00
  • $\begingroup$ No I've never heard of it. How would I use it in this case? $\endgroup$ – Ccyan Apr 15 '16 at 6:03
  • 2
    $\begingroup$ Ah, that's where reading about it in the docs would come in... $\endgroup$ – J. M.'s technical difficulties Apr 15 '16 at 6:04
  • $\begingroup$ @Ccyan as J.M. has expressed look at documentation of PoissonDistribution ,esp applications. Also look at PoissonProcess. Also your P(n) appears incorrect (exponential time) and the round brackets are incorrect syntax for Mathematica. $\endgroup$ – ubpdqn Apr 15 '16 at 6:20
  • $\begingroup$ So I think I would have to use PDF with PoissonDistribution to get the probability? But what would be the mean in my case? $\endgroup$ – Ccyan Apr 15 '16 at 6:20

This question was largely dealt with in the comments. I post this for illustrative purposes (but mainly fun):

The following shows use of PoissonProcess and its use with RandomFunction, Expectation, Probability. The last line is probability at t=6 that there had been only 1 customer for rate 5/hour:

pp = PoissonProcess[5];
rf = RandomFunction[pp, {0, 6}, 1000];
Show[DistributionChart[{, , , , , , rf["LastValues"]}], 
 ListPlot[rf["Paths"], Joined -> True], GridLines -> {None, {30}}]
Expectation[v[6], v \[Distributed] pp]
Probability[v[6] == 1, v \[Distributed] pp]

enter image description here

Visually confirmation of Solve:

sol = {r, 8, CDF[PoissonDistribution[2 r], 8]} /. 
   First[Solve[CDF[PoissonDistribution[2 r], 8] == 0.9, r]];
 Plot[CDF[PoissonDistribution[2 r], x], {x, 0, 10}, 
  GridLines -> {{8}, {0.9}}, Exclusions -> None, 
  Epilog -> {Red, Point[sol[[{2, 3}]]], 
    Text[sol[[1]], {8, 0.8}]}], {r, 1, 5, Appearance -> "Labeled"}]

enter image description here

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