Normal Approximations to the Binomial Distribution

Normal Approximations to the Binomial Distribution

A LevelAQAEdexcelOCRAQA 2022

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 AQA Edexcel OCR

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<Y<a+0.5)

For example, \color{red}\mathbb{P}(X=1)\color{grey}=\color{blue}\mathbb{P}(0.5<Y<1.5)\color{grey},\color{red}\mathbb{P}(X=2)\color{grey}=\color{blue}\mathbb{P}(1.5<Y<2.5) and so on.

This table shows continuity correction in practice:

A LevelAQAEdexcelOCR

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 LevelAQAEdexcelOCR
A Level AQA Edexcel OCR

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 LevelAQAEdexcelOCR

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 LevelAQAEdexcelOCR

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<Y<75.5)=0.9947

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<x)<0.1

 

\mathbb{P}(Y<x-0.5)=0.1

 

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<x)<0.1 is 580

Related Topics

MME

The Binomial Distribution

A Level
MME

The Normal Distribution

A Level

Additional Resources

MME

Exam Tips Cheat Sheet

A Level
MME

Formula Booklet

A Level

You May Also Like...

A Level Maths Revision Cards

The best A level maths revision cards for AQA, Edexcel, OCR, MEI and WJEC. Maths Made Easy is here to help you prepare effectively for your A Level maths exams.

£14.99
View Product

A Level Maths – Cards & Paper Bundle

A level maths revision cards and exam papers for Edexcel. Includes 2022 predicted papers based on the advance information released in February 2022! MME is here to help you study from home with our revision cards and practise papers.

From: £22.99
View Product

Transition Maths Cards

The transition maths cards are a perfect way to cover the higher level topics from GCSE whilst being introduced to new A level maths topics to help you prepare for year 12. Your ideal guide to getting started with A level maths!

£8.99
View Product