# 4.4 Poisson distribution  (Page 3/18)

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On May 13, 2013, starting at 4:30 PM, the probability of low seismic activity for the next 48 hours in Alaska was reported as about 1.02%. Use this information for the next 200 days to find the probability that there will be low seismic activity in ten of the next 200 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close?

Let X = the number of days with low seismic activity.

Using the binomial distribution:

• $P\left(x=10\right)=\frac{200!}{10!\left(200-10\right)!}×{.0102}^{10}=0.000039$

Using the Poisson distribution:

• Calculate μ = np = 200(0.0102) ≈ 2.04
• $P\left(x=10\right)=\frac{{\mu }^{x}{e}^{\mathrm{-\mu }}}{\mathrm{x!}}=\frac{{2.04}^{10}{e}^{-2.04}}{10!}=0.000045$

We expect the approximation to be good because n is large (greater than 20) and p is small (less than 0.05). The results are close—both probabilities reported are almost 0.

## Estimating the binomial distribution with the poisson distribution

We found before that the binomial distribution provided an approximation for the hypergeometric distribution. Now we find that the Poisson distribution can provide an approximation for the binomial. We say that the binomial distribution approaches the Poisson. The binomial distribution approaches the Poisson distribution is as n gets larger and p is small such that np becomes a constant value. There are several rules of thumb for when one can say they will use a Poisson to estimate a binomial. One suggests that np , the mean of the binomial, should be less than 25. Another author suggests that it should be less than 7. And another, noting that the mean and variance of the Poisson are both the same, suggests that np and npq , the mean and variance of the binomial, should be greater than 5. There is no one broadly accepted rule of thumb for when one can use the Poisson to estimate the binomial.

As we move through these probability distributions we are getting to more sophisticated distributions that, in a sense, contain the less sophisticated distributions within them. This proposition has been proven by mathematicians. This gets us to the highest level of sophistication in the next probability distribution which can be used as an approximation to all of those that we have discussed so far. This is the normal distribution.

A survey of 500 seniors in the Price Business School yields the following information. 75% go straight to work after graduation. 15% go on to work on their MBA. 9% stay to get a minor in another program. 1% go on to get a Master's in Finance.

What is the probability that more than 2 seniors go to graduate school for their Master's in finance?

This is clearly a binomial probability distribution problem. The choices are binary when we define the results as "Graduate School in Finance" versus "all other options." The random variable is discrete, and the events are, we could assume, independent. Solving as a binomial problem, we have:

Binomial Solution

$n*p=500*0.01=5=µ$
$P\left(0\right)=\frac{500!}{0!\left(500-0\right)!}{0.01}^{0}\left(1-0.01{\right)}^{{500}^{{-}^{0}}}=0.00657$
$P\left(1\right)=\frac{500!}{1!\left(500-1\right)!}{0.01}^{1}\left(1-0.01{\right)}^{{500}^{{-}^{1}}}=0.03318$
$P\left(2\right)=\frac{500!}{2!\left(500-2\right)!}{0.01}^{2}\left(1-0.01{\right)}^{{500}^{{-}^{2}}}=0.08363$

Adding all 3 together = 0.12339

$1-0.12339=0.87661$

Poisson approximation

$n*p=500*0.01=5=\mu$
$n*p*\left(1-p\right)=500*0.01*\left(0.99\right)\approx 5={\sigma }^{2}=\mu$
$P\left(X\right)=\frac{{e}^{{\mathrm{-np}}^{}}\left(np{\right)}^{x}}{x!}=\left\{P\left(0\right)=\frac{{e}^{-5}*{5}^{0}}{0!}\right\}+\left\{P\left(1\right)=\frac{{e}^{-5}*{5}^{1}}{1!}\right\}+\left\{P\left(2\right)=\frac{{e}^{-5}*{5}^{2}}{2!}\right\}$
$0.0067+0.0337+0.0842=0.1247$
$1-0.1247=0.8753$

An approximation that is off by 1 one thousandth is certainly an acceptable approximation.

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