A phone company is looking to build a tower near a rural town to provide service to the community. The local government will approve the project only if the tower can handle at least 90% of the calls placed at noon. Each antenna placed on the tower can handle 3 simultaneous calls.

(a) The expected number of calls simultaneously placed at noon is 2. How many antennae should the cell phone company place on the tower to ensure that all calls placed at noon can be handled with at least 90% probability? (Hint: assume that each call placed in the rural community is independent of any other call placed. Also, assume the size of the rural community is between 20 and 50 inclusive, and the probability that any individual makes a call is less than 0.05. Notice that we do not know the exact number of people living in the community, nor do we know the exact probability of any one of them placing a call at noon.)

(b) A discovery of mineral deposits in the area has caused the population of the town to double. The probability that each resident makes a call at noon remains the same. What is the probability that a tower with only one antenna can handle all calls placed at noon?

Not sure where to start with this one, but I think it has to do with the Poisson distribution? I have done part b) using the Poisson distribution pmf using E(x)=4 and solved for P(X<=3). For part a), I used the given E(X)=2 and used it in the Poisson distribution pmf and solved for P(X>=3m) where m is the number of antennae. I'm sure I'm doing part a) wrong as the calculations seem a bit tedious.


2 Answers 2


It indeed looks as if the question is about approximating Binomial with Poisson distribution. The wiki on this topic mentions that the approximation is good if $n\geq20$ and $p\leq0.05$, which is the numbers you have.

You have $n\in[20,50]$ and $p\leq\frac{1}{20}$. So that the number of calls placed on noon is a binomial distribution with $n$ and $p$, which has mean $np=2$. Call this random variable $X$. Approximating binomial with Poisson, $X$ converges to $Y$ that is Poisson with $\lambda=np=2$.

At this point I am confused by your question since it asks for the project to handle at least $90\%$ of the calls. But if there are $n$ people there is always a strictly positive probability of all of them making calls and hence there is need for making sure $0.9n$ of calls can be handled.

If the question asked for 'being able to handle all calls with probability $90\%$', then the question would be about $x$ such that $\mathbb{P}[X\leq x]=0.9$. Using the approximation, the answer then follows since cdf of Poisson with $\lambda=2$ equals $0.857123$ at $x=3$ and $0.947347$ at $x=4$.

  • $\begingroup$ You seem to be using $n = 40$ people to get your 'mean'. In that case capacity to handle 4 calls would suffice. $\endgroup$
    – BruceET
    Commented Feb 6, 2017 at 19:44
  • $\begingroup$ @BruceET I was using 'The expected number of calls simultaneously placed at noon is 2.' to get that. $\endgroup$
    – Jan
    Commented Feb 6, 2017 at 20:50
  • $\begingroup$ OK. Good point (+1). So they are really assuming a population of size 40. Either way, it seems they need 2 antennas. $\endgroup$
    – BruceET
    Commented Feb 7, 2017 at 1:25

You have $X \sim \mathsf{Pois}(\lambda = 2.5)$ and you seek the smallest $k$ such that $P(X \le k) \ge .9.\,$ I am using $\lambda = 50(.05) = 2.5$ to cover the most demanding case with 50 residents. In R statistical software:

qpois(.9, 2.5)  # 'qpois' is quantile fcn of Pois (i.e, inverse CDF)
## 5

So the company needs to provide capacity for 5 simultaneous calls. For verification, here is a partial CDF table of $\mathsf{Pois}(2).$ Notice that 5 is indeed the smallest number that gives CDF greater than .9.

x = 0:6;  cdf.P = ppois(x, 2.5)
cbind(x, cdf.P)
## x     cdf.P
## 0 0.0820850
## 1 0.2872975
## 2 0.5438131
## 3 0.7575761
## 4 0.8911780
## 5 0.9579790
## 6 0.9858127

To explore what would happen if we used binomial distributions with $p = .05$ and $n = 20$ and $50$ we make CDF tables for both binomial distributions, to compare with the Poisson results.

y = 0:6;  p = .05;  cdf.20 = pbinom(y, 20, p);  cdf.50 = pbinom(y, 50, p)
cbind(x, cdf.20, cdf.50, cdf.P)
## x    cdf.20     cdf.50     cdf.P
## 0 0.3584859 0.07694498 0.0820850
## 1 0.7358395 0.27943175 0.2872975
## 2 0.9245163 0.54053312 0.5438131
## 3 0.9840985 0.76040796 0.7575761
## 4 0.9974261 0.89638319 0.8911780
## 5 0.9996707 0.96222383 0.9579790
## 6 0.9999661 0.98821355 0.9858127

Notice that the probabilities for $n = 50$ are not far from the the Poisson probabilities, so the Poisson approximation is working fairly well. (If we could be sure there are only 20 people in the area, then capacity for only two simultaneous calls, provided by one antenna, would be enough.)

In part (b), if the population doubles, then $\lambda$ doubles. You want $P(X \le 3).$

I don't know how you are expected to do the computations. Some probability books have CDF tables for a few Poisson distributions in an appendix. There are many kinds of software that will do the computations. In this particular case, using a simple calculator would not be difficult, for example:

$$P(X \le 3; \lambda=2.5) = e^{-2.5}(1 + 2.5 + 2.5^2/2! + 2.5^3/3!) = 0.5438.$$

So you could quickly see that $k = 3$ is not enough, and start trying larger values with more terms in the series.

Below is a plot comparing the PDF's of $\mathsf{Pois}(2.5)$ and $\mathsf{Binom}(50,.05).$ Binomial probabilities are at the centers of the small red circles.

enter image description here


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