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 $n(1-p)$. $np=2000times 0.001=2<5$. Hence, we cannot use the approximation.

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$n=150,p=0.4$

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

$Ysim N(np,np(1-p))$

$Ysim N(150times 0.4,150times 0.4times (1-0.4))$

$Ysim N(60,36)$

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

ii) $mathbb{P}(Xleq 66)=mathbb{P}(Y<66.5)=0.8606$

i) $mathbb{P}(40leq Xleq 75)=mathbb{P}(39.5<Y<75.5)=0.9947$

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We can model this with $Xsim Bleft(365,dfrac{12}{25}right)$.

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

$Ysim N(np,np(1-p))$

$Ysim N(365times 0.48,365times 0.48times 0.52)$

$Ysim N(175.2,91.104)$

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

$mathbb{P}(X>200)=0.004$

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

$Ysim N(np,np(1-p))$

$Ysim N(1000times 0.6,1000times 0.6times 0.4)$

$Ysim N(600,240)$

$mathbb{P}(X<x)<0.1$

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

Convert to standard normal $Z$:

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

Use percentage points table:

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

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

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

$x=600.5-1.2816sqrt{240}$

$x=580.65$

Hence, the highest whole number $x$ such that $mathbb{P}(X<x)<0.1$ is $580$

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