Normal approximation to the binomial distribution

​​In a nutshell

Given certain specific conditions, it is possible to approximate a binomial distribution with a normal distribution.

Required conditions

Given $$X \sim B(n,p)$$​ and $$Y \sim N(\mu, \sigma^2)$$​, the random variable $$X$$ can be approximated with $$Y$$ if:

• $$n$$​ is large.
• ​$$p$$​ is close to $$0.5$$​.

If $$X \approx Y$$, then $$\mu$$ and $$\sigma$$​ can be calculated using the following formulae:

​$$\boxed{\begin{aligned} \mu &= np\\ \sigma &= \sqrt{np(1-p)} \end{aligned}}$$​​

Example 1

Given the random variable $$X \sim B(250, 0.55)$$, explain why it is possible to approximate $$X$$ with a normal distribution, and calculate the values of $$\mu$$ and $$\sigma^2$$ if $$X \approx Y \sim N(\mu, \sigma^2)$$.

​$$n$$​ is large and $$p$$ is close to $$0.5$$. Therefore, it is possible to estimate $$X$$ with a normal distribution.

To calculate $$\mu$$ and $$\sigma^2$$, use the formulae:

$$\begin{aligned} \mu &= np\\&= 250 \times 0.55\\&=\underline{137.5}\\ \sigma &= \sqrt{np(1-p)}\\ \sigma^2&= \left( \sqrt{np(1-p)}\right)^2\\&= np(1-p)\\&=250\times 0.55 \times (1 - 0.55)\\&=\underline{61.875} \end{aligned}$$​​

​​Calculating probabilities

To estimate probabilities of a binomial distribution using a normal distribution, a rounding error has to be applied due to the fact that the binomial distribution is a discrete distribution.

Example 2

The random variable $$X \sim B(250, 0.55)$$ is approximated with the random variable $$Y \sim N(137.5, 61.875)$$. Estimate the following probabilities:

i) $$P(X = 140)$$​

ii) $$P(X \leq 180)$$

iii) $$P(130 \leq X \lt 150)$$​

Part i):

Use error intervals:

$$\begin{aligned} P(X=140)& \approx P(139.5 \lt Y \lt 140.5)\\ &\approx \underline{0.0482} \end{aligned}$$​​

Part ii):

Since $$180$$ is included in the inequality, the upper bound is $$180.5$$.

$$\begin{aligned} P(X \leq 180) &\approx P(Y \lt 180.5)\\ &\approx \underline{1} \end{aligned}$$​​

Part iii):

$$\begin{aligned} P(130 \leq X \lt 150) &\approx P(129.5 \lt Y \lt 149.5)\\ &\approx \underline{0.7819} \end{aligned}$$​​​