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The Question is here

A life insurance salesman sells on the average 3 life insurance policies per week. Use Poisson's law Pr ( = ) =

−33 / ! to calculate the probability that in a given week he will sell 2 or more policies but less than 5 policies

I answered this question by substituting the r=2,r=3,r=4 as 0.6160 and when I used the Table of Poisson Probabilities also I got the same answer.

I have an issue using the Cumulative Poisson Probability Table, because the answer is given as

 P(x>=2) - P(x>=5)
 0.80085 - 0.18774
 0.61611

But in the question it is given that the sell 2 or more policies but less than 5 policies. Then why should we use P(x>=5). When I tried the question first I used P(x>=4) and I got a wrong answer.

 P(x>=2) - P(x>=4)
 0.80085 - 0.35277
 0.44808

Can anybody please explain what I've got wrong! Thankyou in advance!!!

laish138
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1 Answers1

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In your answer you have subtracted $\Pr (X = 4)$. Your answer is therefore too small. You could use $\Pr (X \ge 2) - \Pr (X > 4)$. Alternatively, you want the probability given by $$\Pr (2 \le X < 5) = \Pr (X < 5) - \Pr (X < 2)$$

But $\Pr (X < 5) = 1 - \Pr (X \ge 5)$ and $Pr (X < 2) = 1 - \Pr (X \ge 2)$. Substituting into the RHS in the equation above yields $$\Pr (2 \le X < 5) = 1-\Pr (X \ge 5) - (1 - \Pr (X \ge 2))$$ which is $$\Pr (X \ge 2) - \Pr (X \ge 5).$$ The first probability in this expression is the CDF from 2, 3, ...

while the second is the CDF from 5, 6, ...

which is a more intuitive way of thinking about it.

In other words you are left with the probability of X=2, 3, 4.