Question

) Let X be a binomial variate with n =100, p = .1.0 Find the approximate value of

P 10( ≤ X ≤12) using: (5) 

 (i) normal distribution 

 (ii)poisson distribution 

 [You may like to use the following values. 

 P(Z ≤ 67.0 ) = .0 7486, P(Z ≤ 33.0 ) = .0 6293, P(Z ≤ )0 = ]

Answer 1

(i) normal distribution

 Previously find mean and standard deviation

$\bold{E}[X]=np=100\cdot0.1=10$

$\sigma[X]=\sqrt{np(1-p)}=\sqrt{100\cdot0.1\cdot0.9}=\sqrt{9}=3$

$P (10\leq X\leq12)=$

$F\left(\dfrac{12-\bold{E}[X]}{\sigma[X]}\right)-F\left(\dfrac{10-\bold{E}[X]}{\sigma[X]}\right)=$

$F\left(\dfrac{12-10}{3}\right)-F\left(\dfrac{10-10}{3}\right)=$

$F\left(\dfrac{2}{3}\right)-F(0)\approx0.7486-0.5000=0.2486$

(ii) Poisson distribution 

Previously find parameter of Poisson distribution

$\lambda=np=100\cdot0.1=10$

Formula for Poisson probability is

$P(k)=\dfrac{\lambda^ke^{-\lambda}}{k!}$

$P (10\leq X\leq12)=$

$P(0)+P(1)+\ldots+P(10)+P(11)+P(12)-\bigg(P(0)+P(1)+\ldots+P(10)\bigg)=$

$P(11)+P(12)=$

$\dfrac{10^{11}\cdot e^{-10}}{11!}+\dfrac{10^{12}\cdot e^{-10}}{12!}=$

$\dfrac{10^{11}\cdot e^{-10}}{11!}+\dfrac{10^{11}\cdot 10\cdot e^{-10}}{11!\cdot12}=$

$\dfrac{10^{11}\cdot e^{-10}}{11!}\left(1+\dfrac{10}{12}\right)=$

$\dfrac{10^{11}\cdot e^{-10}}{11!}\left(1+\dfrac{5}{6}\right)=$

$\dfrac{10^{11}\cdot e^{-10}}{11!}\cdot\dfrac{11}{6}\approx0.2085$