## Normal Approximation to Binomial Distribution

Let $X$ be a binomially distributed 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 normal approximation to binomial distribution is that the sample size $n$ is sufficiently large if $np \geq 5$ and $n(1-p)\geq 5$.

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

## Formula

Continuity correction for normal approximation to binomial distribution are as follows:

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

## Example

In a certain Binomial distribution with probability of success $p=0.20$ and number of trials $n = 30$.

Compute

a. the probability of getting 5 successes,

b. the probability of getting at least 5 successes,

c. the probability of getting between 5 and 10 (inclusive) successes.

### Solution

Let $X$ denote the number of successes in 30 trials and let $p$ be the probability of success.

Given that $n =30$ and $p=0.2$. Thus $X\sim B(30, 0.2)$.

Here `$n*p = 30\times 0.2 = 6>5$`

and `$n*(1-p) = 30\times (1-0.2) = 24>5$`

, we use Normal approximation to Binomial distribution.

Mean of $X$ is

```
$$
\begin{aligned}
\mu&= n*p \\
&= 30 \times 0.2 \\
&= 6.
\end{aligned}
$$
```

and standard deviation of $X$ is
```
$$
\begin{aligned}
\sigma &= \sqrt{n*p*(1-p)} \\
&= \sqrt{30 \times 0.2 \times (1- 0.2)}\\
&=2.1909.
\end{aligned}
$$
```

a. Using the continuity correction, $P(X=5)$ can be written as $P(5-0.5<X<5+0.5)=P(4.5<X<5.5)$.

The $Z$-scores that corresponds to $4.5$ and $5.5$ are

```
$$
\begin{aligned}
z_1=\frac{4.5-\mu}{\sigma}=\frac{4.5-6}{2.1909}\approx-0.68
\end{aligned}
$$
```

and
```
$$
\begin{aligned}
z_2=\frac{5.5-\mu}{\sigma}=\frac{5.5-6}{2.1909}\approx-0.23
\end{aligned}
$$
```

Thus the probability of getting exactly 5 successes is
```
$$
\begin{aligned}
P(X= 5) & = P(4.5<X<5.5)\\
&=P(z_1< Z< z_2)\\
&=P(-0.68<Z<-0.23)\\
&=P(Z<-0.23)-P(Z<-0.68)\\
& = 0.409-0.2483\\
& = 0.1607
\end{aligned}
$$
```

b. Using the continuity correction, the probability of getting at least 5 successes i.e., $P(X\geq 5)$ can be written as $P(X\geq5)=P(X\geq 5-0.5)=P(X\geq4.5)$.

The $Z$-score that corresponds to $4.5$ is

```
$$
\begin{aligned}
z=\frac{4.5-\mu}{\sigma}=\frac{4.5-6}{2.1909}\approx-0.68
\end{aligned}
$$
```

Thus, the probability of getting at least 5 successes is

```
$$
\begin{aligned}
P(X\geq 5) &= P(X\geq4.5)\\
&= 1-P(X<4.5)\\
&= 1-P(Z<-0.68)\\
& = 1-0.2483\\
& = 0.7517
\end{aligned}
$$
```

c. Using the continuity correction, the probability of getting between 5 and 10 (inclusive) successes is $P(5\leq X\leq 10)$ can be written as $P(5-0.5<X<10+0.5)=P(4.5<X<10.5)$.

The $Z$-scores that corresponds to $4.5$ and $10.5$ are respectively

```
$$
\begin{aligned}
z_1=\frac{4.5-\mu}{\sigma}=\frac{4.5-6}{2.1909}\approx-0.68
\end{aligned}
$$
```

and
```
$$
\begin{aligned}
z_2=\frac{10.5-\mu}{\sigma}=\frac{10.5-6}{2.1909}\approx2.05
\end{aligned}
$$
```

```
$$
\begin{aligned}
P(5\leq X\leq 10) &= P(5-0.5 < X <10+0.5)\\
&= P(4.5 < X <10.5)\\
&=P(-0.68\leq Z\leq 2.05)\\
&=P(Z\leq 2.05)-P(Z\leq -0.68)\\
&=0.9798-0.2483\\
&=0.7315
\end{aligned}
$$
```