Probability Poisson Distribution 90 and 110 calls in a 100 second period

Question

How can i get the probability using poisson distribution of this question:

Telephone calls arrive at an exchange at an average rate of one every second. Find the probabilities of the following events, explaining briefly your assumptions.

a) No calls arriving in a given five-second period.

b) Between four and six calls arriving in the five-second period.

c) Between $90$ and $110$ calls arriving in a 100-second period. (Give answer as a decimal.)

I got a) and b) but i dont know the way to do c).

I know I can use poisson distribution on $$P(X=90) + P(X=91) + \cdots + P(X=110)$$ but I feel like there is an easier way to do this. Thanks

Answer 1

You can use normal approximation to the poisson distribution.

Let $X\sim \mathsf {pois}(100)$. Then the exact probability is

$$\mathsf P(90\leq X\leq110)=\sum_{x=90}^{110}\frac{e^{-100}100^x}{x!}\approx0.7065$$

In R statistical software

> sum(dpois(90:110,100))
[1] 0.7065165

The normal approximation would give that $X\sim \mathsf N(100,100)$ and so using continuity correction, we get

$$\mathsf P(89.5\lt X\lt 110.5)=\Phi\left(\frac{110.5-100}{10}\right)-\Phi\left(\frac{89.5-100}{10}\right)\approx 0.7063$$

In R statistical software

> pnorm((110.5-100)/10)-pnorm((89.5-100)/10)
[1] 0.7062819

so the normal approximation did very well.

Answer 2

The Poisson distribution has a nice property when it comes to time intervals. If the mean rate of occurrence is $\lambda$ per second, then given $t$ seconds, the rate of occurrence in a $t$-second period is $\lambda' = t$.

This means that your "new" distribution is $\sim \operatorname{POI}(\lambda = 100)$

So, using this distribution, you calculate:

$\operatorname{P}(90 \ge X \ge 110) = \operatorname{P}(X \ge 110) - \operatorname{P}(X \gt 90) \approx 0.85286 - 0.14635 = 0.70651$