Normal Approximations to the Binomial Distribution

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