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# Poisson approximation to binomial distribution examples

## Poisson approximation to binomial distribution examples

Let $X$ be a binomial random variable with number of trials $n$ and probability of success $p$.

The mean of $X$ is $\mu=E(X) = np$ and variance of $X$ is $\sigma^2=V(X)=np(1-p)$.

The general rule of thumb to use Poisson approximation to binomial distribution is that the sample size $n$ is sufficiently large and $p$ is sufficiently small such that $\lambda=np$ (finite).

Thus, for sufficiently large $n$ and small $p$, $X\sim P(\lambda)$.

The probability mass function of Poisson distribution with parameter $\lambda$ is

 \begin{align*} P(X=x)&= \begin{cases} \dfrac{e^{-\lambda}\lambda^x}{x!} , & x=0,1,2,\cdots; \lambda>0; \\ 0, & Otherwise. \end{cases} \end{align*}

In general, the Poisson approximation to binomial distribution works well if $n\geq 20$ and $p\leq 0.05$ or if $n\geq 100$ and $p\leq 0.10$.

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## Poisson approximation to binomial Example 1

Suppose 1% of all screw made by a machine are defective. We are interested in the probability that a batch of 225 screws has at most one defective screw. Compute

b. the Poisson approximation.

#### Solution

Let $X$ denote the number of defective screw produced by a machine. Let $p$ be the probability that a screw produced by a machine is defective.

Given that $n=225$ (large) and $p=0.01$ (small). $X\sim B(225, 0.01)$.

a. Using Binomial Distribution: The probability that a batch of 225 screws has at most 1 defective screw is

 \begin{aligned} P(X\leq 1) & =\sum_{x=0}^{1} P(X=x)\\ & =P(X=0) + P(X=1) \\ & = 0.1042+0.2368\\ &= 0.3411 \end{aligned}

b. Using Poisson Approximation: If $n$ is sufficiently large and $p$ is sufficiently large such that that $\lambda = n*p$ is finite, then we use Poisson approximation to binomial distribution.

Here $\lambda=n*p = 225*0.01= 2.25$ (finite). Thus $X\sim P(2.25)$ distribution.

The probability mass function of $X$ is

 \begin{aligned} P(X=x) &= \frac{e^{-2.25}2.25^x}{x!}; x=0,1,2,\cdots \end{aligned}

The probability that a batch of 225 screws has at most 1 defective screw is

 \begin{aligned} P(X\leq 1) &= P(X=0)+ P(X=1)\\ &= \frac{e^{-2.25}2.25^{0}}{0!}+\frac{e^{-2.25}2.25^{1}}{1!}\\ &= 0.1054+0.2371\\ &= 0.3425 \end{aligned}

## Poisson approximation to binomial Example 2

On the average, 1 in 800 computers crashes during a severe thunderstorm. A certain company had 4,000 working computers when the area was hit by a severe thunderstorm.

a. Compute the expected value and variance of the number of crashed computers.
b. Compute the probability that less than 10 computers crashed.
c. Compute the probability that exactly 10 computers crashed.

#### Solution

Let $X$ be the number of crashed computers out of $4000$. Let $p=1/800$ be the probability that a computer crashed during severe thunderstorm. Thus $X\sim B(4000, 1/800)$.

Here $n=4000$ (sufficiently large) and $p=1/800$ (sufficiently small) such that $\lambda =n*p =4000*1/800= 5$ is finite. Thus we use Poisson approximation to Binomial distribution.

That is $X\sim P(5)$ distribution.

The probability mass function of $X$ is

 \begin{aligned} P(X=x) &= \frac{e^{-5}5^x}{x!}; x=0,1,2,\cdots \end{aligned}

a. The expected value of the number of crashed computers

 \begin{aligned} E(X)&= n*p\\ &=4000* 1/800\\ &=5 \end{aligned}

The variance of the number of crashed computers

 \begin{aligned} V(X)&= n*p*(1-p)\\ &=4000* 1/800*(1-1/800)\\ &=4.99 \end{aligned}

b. The probability that less than 10 computers crashed is

 \begin{aligned} P(X < 10) &= P(X\leq 9)\\ &= 0.9682\\ & \quad \quad (\because \text{Using Poisson Table}) \end{aligned}

c. The probability that exactly 10 computers crashed is

 \begin{aligned} P(X= 10) &= P(X=10)\\ &= \frac{e^{-5}5^{10}}{10!}\\ &= 0.0181 \end{aligned}

## Poisson approximation to binomial Example 3

Suppose that the probability of suffering a side effect from a certain flu vaccine is 0.005. If 1000 persons are inoculated, use Poisson approximation to binomial to find the probability that

a. at least 2 people suffer,
b. at the most 3 people suffer,
c. exactly 3 people suffer.

#### Solution

Let $X$ be the number of persons suffering a side effect from a certain flu vaccine out of $1000$. Let $p=0.005$ be the probability that a person suffering a side effect from a certain flu vaccine. Thus $X\sim B(1000, 0.005)$.

Here $n=1000$ (sufficiently large) and $p=0.005$ (sufficiently small) such that $\lambda =n*p =1000*0.005= 5$ is finite. Thus we use Poisson approximation to Binomial distribution.

That is $X\sim P(5)$ distribution.

The probability mass function of $X$ is

 \begin{aligned} P(X=x) &= \frac{e^{-5}5^x}{x!}; x=0,1,2,\cdots \end{aligned}

a. The probability that at least 2 people suffer is

 \begin{aligned} P(X \geq 2) &=1- P(X < 2)\\ &= 1- \big[P(X=0)+P(X=1) \big]\\ &= 1-0.0404\\ & \quad \quad (\because \text{Using Poisson Table})\\ &= 0.9596 \end{aligned}

b. The probability that at the most 3 people suffer is

 \begin{aligned} P(X \leq 3) &= P(X=0)+P(X=1)+P(X=2)+P(X=3)\\ &= 0.1247\\ & \quad \quad (\because \text{Using Poisson Table}) \end{aligned}

c. The probability that exactly 3 people suffer is

 \begin{aligned} P(X= 3) &= P(X=3)\\ &= \frac{e^{-5}5^{3}}{3!}\\ &= 0.1404 \end{aligned}

## Poisson approximation to binomial Example 4

If know that 5% of the cell phone chargers are defective. find the probability that 3 of 100 cell phone chargers are defective using

a) formula for binomial distribution
b) Poisson approximation to binomial distribution

#### Solution

Let $X$ denote the number of defective cell phone chargers. Let $p$ be the probability that a cell phone charger is defective.

Given that $n=100$ (large) and $p=0.05$ (small). $X\sim B(100, 0.05)$.

a. Using Binomial Distribution: The probability that 3 of the 100 cell phone chargers are defective is

 \begin{aligned} P(X=3) &= \binom{100}{3}(0.05)^{3}(0.95)^{100 - 3}\\ & = 0.1396 \end{aligned}

b. Using Poisson Approximation: If $n$ is sufficiently large and $p$ is sufficiently large such that that $\lambda = n*p$ is finite, then we use Poisson approximation to binomial distribution.

Here $\lambda=n*p = 100*0.05= 5$ (finite). Thus $X\sim P(5)$ distribution.

The probability mass function of $X$ is

 \begin{aligned} P(X=x) &= \frac{e^{-5}5^x}{x!}; x=0,1,2,\cdots \end{aligned}

The probability that 3 of 100 cell phone chargers are defective screw is

 \begin{aligned} P(X = 3) &= \frac{e^{-5}5^{3}}{3!}\\ &= 0.1404 \end{aligned}

## Poisson approximation to binomial Example 5

Assume that one in 200 people carry the defective gene that causes inherited colon cancer. A sample of 800 individuals is selected at random. Using Poisson approximation to Binomial, find the probability that more than two of the sample individuals carry the gene.

#### Solution

Let $X$ be the number of people carry defective gene that causes inherited colon cancer out of $800$ selected individuals. Let $p=0.005$ be the probability that an individual carry defective gene that causes inherited colon cancer. Thus $X\sim B(800, 0.005)$.

Here $n=800$ (sufficiently large) and $p=0.005$ (sufficiently small) such that $\lambda =n*p =800*0.005= 4$ is finite. Thus we use Poisson approximation to Binomial distribution.

That is $X\sim P(4)$ distribution.

The probability mass function of $X$ is

 \begin{aligned} P(X=x) &= \frac{e^{-4}4^x}{x!}; x=0,1,2,\cdots \end{aligned}

probability that more than two of the sample individuals carry the gene is

 \begin{aligned} P(X > 2) &=1- P(X \leq 2)\\ &= 1- \big[P(X=0)+P(X=1)+P(X=2) \big]\\ &= 1-0.2381\\ & \quad \quad (\because \text{Using Poisson Table})\\ &= 0.7619 \end{aligned}

## Conclusion

In this tutorial, you learned about how to use Poisson approximation to binomial distribution for solving numerical examples. To read about theoretical proof of Poisson approximation to binomial distribution refer the link Poisson Distribution