Normal Approximations to the Binomial Distribution

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Normal Approximations to the Binomial Distribution

In some cases, a binomial distribution can be approximated by a normal distribution. This can be useful as binomial distributions with large $n$ can be difficult to work with.

The approximation is:

$\color{red}X\sim B(n,p)\color{grey}\approx\color{blue}Y\sim N(np,np(1-p))$

Make sure you are happy with the following topics before continuing.

A Level

Continuity Correction

An obvious problem with this approximation is that the binomial distribution is discrete while the normal distribution is continuous. This means that the binomial distribution takes fixed values with certain probabilities, but the normal distribution only takes values on ranges, i.e.

• For discrete distributions, such as binomial, we can work out $\color{red}\mathbb{P}(X=0),\mathbb{P}(X=1)$ and so on.
• For continuous distributions, such as normal, $\color{blue}\mathbb{P}(Y=0)=\mathbb{P}(Y=1)=0$, and we can only work out probabilities of ranges of values.

This means that, for our approximation, we need continuity correction, which works like this:

$\color{red}\mathbb{P}(X=a)\color{grey}=\color{blue}\mathbb{P}(a-0.5

For example, $\color{red}\mathbb{P}(X=1)\color{grey}=\color{blue}\mathbb{P}(0.5 and so on.

This table shows continuity correction in practice:

A Level

Conditions for the Approximation

You can only use the approximation under some circumstances. You must make sure the conditions hold before you use the approximation. You can use the approximation when:

$p\approx 0.5$ and $n$ is large

OR

both $np>5$ and $n(1-p)>5$

A Level
A Level

Example 1: When $n$ is Large

$X\sim B(250,0.55)$. Find $\mathbb{P}(X\leq 130)$.

[2 marks]

$n=250$ which is large, and $p=0.55$ which is close to $0.5$, so we can use the approximation.

$Y\sim N(250\times 0.55,250\times 0.55\times (1-0.55))$

$Y\sim N(137.5,61.875)$

$\mathbb{P}(X\leq 130)=\mathbb{P}(Y<130.5)$ (continuity correction)

$\mathbb{P}(X\leq 130)=0.1702$

A Level

Example 2: When $np$ and $n(1-p)$ are greater than $5$

$X\sim B(20,0.7)$. Find $\mathbb{P}(X<15)$.

[2 marks]

$n=20$ and $p=0.7$

$20\times 0.7=14>5$ and $20\times (1-0.7)=6>5$, so we can use the approximation.

$Y\sim N(20\times 0.7,20\times 0.7\times (1-0.7))$

$Y\sim N(14,4.2)$

$\mathbb{P}(X<15)=\mathbb{P}(Y<14.5)$ (continuity correction)

$\mathbb{P}(X<15)=0.5964$

A Level

Example Questions

i) $n$ is large and $p=0.5$, so we can use the approximation.

ii) $n$ is not large, and $np=1.8<5$, so neither condition is met so we cannot use the approximation.

iii) $n$ is not large, but $np=5.25>5$ and $n(1-p)=15.75>5$, so we can use the approximation.

iv) $n$ is large but $p$ is not close to $0.5$ so we must check $np$ and $n(1-p)$. $np=2000\times 0.001=2<5$. Hence, we cannot use the approximation.

$n=150,p=0.4$

$n$ is large and $p$ is close to $0.5$ so we can use the approximation.

$Y\sim N(np,np(1-p))$

$Y\sim N(150\times 0.4,150\times 0.4\times (1-0.4))$

$Y\sim N(60,36)$

i) $\mathbb{P}(X<60)=\mathbb{P}(Y<59.5)=0.4668$

ii) $\mathbb{P}(X\leq 66)=\mathbb{P}(Y<66.5)=0.8606$

i) $\mathbb{P}(40\leq X\leq 75)=\mathbb{P}(39.5

We can model this with $X\sim B\left(365,\dfrac{12}{25}\right)$.

$n=365$ which is large and $p=\dfrac{12}{25}=0.48$ which is close to $0.5$, so we can use the approximation.

$Y\sim N(np,np(1-p))$

$Y\sim N(365\times 0.48,365\times 0.48\times 0.52)$

$Y\sim N(175.2,91.104)$

$\mathbb{P}(X>200)=\mathbb{P}(Y>200.5)$

$\mathbb{P}(X>200)=0.3906$

$n=1000$ which is large and $p=0.6$ which is close to $0.5$ so we can use the normal approximation.

$Y\sim N(np,np(1-p))$

$Y\sim N(1000\times 0.6,1000\times 0.6\times 0.4)$

$Y\sim N(600,240)$

$\mathbb{P}(X

$\mathbb{P}(Y

Convert to standard normal $Z$:

$\mathbb{P}\left(Z<\dfrac{x-0.5-600}{\sqrt{240}}\right)=0.1$

Use percentage points table:

$\dfrac{x-0.5-600}{\sqrt{240}}=-1.2816$

$x-0.5-600=-1.2816\sqrt{240}$

$x-600.5=-1.2816\sqrt{240}$

$x=600.5-1.2816\sqrt{240}$

$x=580.65$

Hence, the highest whole number $x$ such that $\mathbb{P}(X is $580$

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A Level

A Level

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