## Normal approximation to Poisson distribution

In this tutorial we will discuss some numerical examples on Poisson distribution where normal approximation is applicable. For large value of the $\lambda$ (mean of Poisson variate), the Poisson distribution can be well approximated by a normal distribution with the same mean and variance.

Let $X$ be a Poisson distributed random variable with mean $\lambda$.

The mean of $X$ is $\mu=E(X) = \lambda$ and variance of $X$ is $\sigma^2=V(X)=\lambda$.

The general rule of thumb to use normal approximation to Poisson distribution is that $\lambda$ is sufficiently large (i.e., $\lambda \geq 5$).

For sufficiently large $\lambda$, $X\sim N(\mu, \sigma^2)$. That is $Z=\frac{X-\mu}{\sigma}=\frac{X-\lambda}{\sqrt{\lambda}} \sim N(0,1)$.

## Formula for continuity corrections

Poisson distribution is a discrete distribution, whereas normal distribution is a continuous distribution. When we are using the normal approximation to Poisson distribution we need to make correction while calculating various probabilities.

- $P(X=A)=P(A-0.5<X<A+0.5)$
- $P(X<A)=P(X<A-0.5)$
- $P(X\leq A)=P(X<A+0.5)$
- $P(A< X\leq B)=P(A-0.5<X<B+0.5)$
- $P(A\leq X< B)=P(A-0.5<X<B-0.5)$
- $P(A\leq X\leq B)=P(A-0.5<X<B+0.5)$

## Example 1

The mean number of kidney transplants performed per day in the United States in a recent year was about 45. Find the probability that on a given day,

a. exactly 50 kidney transplants will be performed,

b. at least 65 kidney transplants will be performed, and

c. no more than 40 kidney transplants will be performed.

### Solution

Let $X$ denote the number of kidney transplants per day. The mean number of kidney transplants performed per day in the United States in a recent year was about 45. $\lambda = 45$. $X$ follows Poisson distribution, i.e., $X\sim P(45)$.

Since $\lambda= 45$ is large enough, we use normal approximation to Poisson distribution. That is $Z=\dfrac{X-\lambda}{\sqrt{\lambda}}\to N(0,1)$ for large $\lambda$. (We use continuity correction)

a. The probability that on a given day, exactly 50 kidney transplants will be performed is
```
$$
\begin{aligned}
P(X=50) &= P(49.5< X< 50.5)\\
& \quad\quad (\text{Using continuity correction})\\
&= P\bigg(\frac{49.5-45}{\sqrt{45}}<\frac{X-\lambda}{\sqrt{\lambda}}<\frac{50.5-45}{\sqrt{45}}\bigg)\\
&= P(0.67< Z<0.82)\\
& = P(Z<0.82) - P(Z< 0.67)\\
&= 0.7939-0.7486\\
& \quad\quad (\text{Using normal table})\\
&= 0.0453
\end{aligned}
$$
```

b. The probability that on a given day, at least 65 kidney transplants will be performed is
```
$$
\begin{aligned}
P(X\geq 65) &= 1-P(X\leq 64)\\
&= 1-P(X\leq 64.5)\\
& \quad\quad (\text{Using continuity correction})\\
&= 1-P\bigg(\frac{X-\lambda}{\sqrt{\lambda}}<\frac{64.5-45}{\sqrt{45}}\bigg)\\
&= 1-P(Z\leq 3.06)\\
&= 1-0.9989\\
& \quad\quad (\text{Using normal table})\\
&= 0.0011
\end{aligned}
$$
```

c. The probability that on a given day, no more than 40 kidney transplants will be performed is

```
$$
\begin{aligned}
P(X< 40) &= P(X< 39.5)\\
& \quad\quad (\text{Using continuity correction})\\
&= P\bigg(\frac{X-\lambda}{\sqrt{\lambda}}<\frac{39.5-45}{\sqrt{45}}\bigg)\\
&= P(Z<-0.82)\\
& = P(Z<-0.82) \\
&= 0.2061\\
& \quad\quad (\text{Using normal table})
\end{aligned}
$$
```

## Example 2

A radioactive element disintegrates such that it follows a Poisson distribution. If the mean number of particles ($\alpha$) emitted is recorded in a 1 second interval as 69, evaluate the probability of:

a. Less than 60 particles are emitted in 1 second.

b. Between 65 and 75 particles inclusive are emitted in 1 second.

### Solution

Let $X$ denote the number of particles emitted in a 1 second interval. The mean number of $\alpha$-particles emitted per second $69$. Thus $\lambda = 69$ and given that the random variable $X$ follows Poisson distribution, i.e., $X\sim P(69)$.

Since $\lambda= 69$ is large enough, we use normal approximation to Poisson distribution. That is $Z=\dfrac{X-\lambda}{\sqrt{\lambda}}\to N(0,1)$ for large $\lambda$. (We use continuity correction)

a. The probability that less than 60 particles are emitted in 1 second is
```
$$
\begin{aligned}
P(X< 60) &= P(X< 59.5)\\
& \quad\quad (\text{Using continuity correction})\\
&= P\bigg(\frac{X-\lambda}{\sqrt{\lambda}}<\frac{59.5-69}{\sqrt{69}}\bigg)\\
&= P(Z<-1.14)\\
& = P(Z<-1.14) \\
&= 0.1271\\
& \quad\quad (\text{Using normal table})
\end{aligned}
$$
```

b. The probability that between $65$ and $75$ particles (inclusive) are emitted in 1 second is
```
$$
\begin{aligned}
P(65\leq X\leq 75) &= P(64.5< X< 75.5)\\
& \quad\quad (\text{Using continuity correction})\\
&= P\bigg(\frac{64.5-69}{\sqrt{69}}<\frac{X-\lambda}{\sqrt{\lambda}}<\frac{75.5-69}{\sqrt{69}}\bigg)\\
&= P(-0.54<Z<0.78)\\
&= P(Z<0.78)- P(Z< -0.54) \\
&= 0.7823-0.2946\\
& \quad\quad (\text{Using normal table})\\
&= 0.4877
\end{aligned}
$$
```