Question

Suppose that X is a binomial random variable with n = 100 and p = 0.1

a) compute the exact probability that X is less than 4.

b) approximate the probability that X is less than 4 and compare to the result in part (a).

c) approximate the probability that 8 < X < 12

Answer 1

$a:\\P\left( X<4 \right) =\sum_{i=0}^3{P\left( X=i \right)}=\sum_{i=0}^3{C_{100}^{i}\cdot 0.1^i\cdot \left( 1-0.1 \right) ^{100-i}}=\\=0.9^{100}+100\cdot 0.1\cdot 0.9^{99}+C_{100}^{2}\cdot 0.1^2\cdot 0.9^{98}+C_{100}^{3}\cdot 0.1^3\cdot 0.9^{97}=0.00783649\\b:\\P\left( X<4 \right) =P\left( \hat{p}<0.04 \right) =P\left( \sqrt{n}\frac{\hat{p}-p}{\sqrt{p\left( 1-p \right)}}<\sqrt{n}\frac{0.04-p}{\sqrt{p\left( 1-p \right)}} \right) \approx \\\approx P\left( Z<\sqrt{100}\frac{0.04-0.1}{\sqrt{0.1\left( 1-0.1 \right)}} \right) =P\left( Z<-2 \right) =\varPhi \left( -2 \right) =0.02275\\c:\\P\left( 8<X<12 \right) =P\left( 0.08<\hat{p}<0.12 \right) =P\left( \sqrt{n}\frac{0.08-p}{\sqrt{p\left( 1-p \right)}}<\sqrt{n}\frac{\hat{p}-p}{\sqrt{p\left( 1-p \right)}}<\sqrt{n}\frac{0.12-p}{\sqrt{p\left( 1-p \right)}} \right) \approx \\\approx P\left( \sqrt{100}\frac{0.08-0.1}{\sqrt{0.1\left( 1-0.1 \right)}}<Z<\sqrt{100}\frac{0.12-0.1}{\sqrt{0.1\left( 1-0.1 \right)}} \right) =\\=P\left( -0.6667<Z<0.6667 \right) =2\varPhi \left( 0.6667 \right) -1=2\cdot 0.7475-1=0.495$