# Normal Approximations to the Binomial Distribution

## Normal Approximations to the Binomial Distribution Revision

**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.

**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:

**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

**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

**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

## Normal Approximations to the Binomial Distribution Example Questions

**Question 1:Â **For which of these binomial distributions could we use a normal approximation?

i) X\sim B(135,0.5)

ii) X\sim B(6,0.3)

iii) X\sim B(21,0.25)

iv) X\sim B(2000,0.001)

**[5 marks]**

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.

**Question 2:Â **X\sim B(150,0.4). Find:

i) \mathbb{P}(X<60)

ii) \mathbb{P}(X\leq 66)

iii) \mathbb{P}(40\leq X\leq 75)

**[4 marks]**

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

**Question 3:** Every day, the probability that John buys a chocolate bar is \dfrac{12}{25}. What is the probability that he buys more than 200 chocolate bars in a (non-leap) year?

**[3 marks]**

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

**Question 4: **(Harder) Find the largest value of x such that \mathbb{P}(X<x)<0.1 where X\sim B(1000,0.6)

(Hint: you will need to use the standard normal distribution.)

**[4 marks]**

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

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