Question

9. (a) 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 = ]5.0

 

Answer 1

i)

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

$\sigma=\sqrt{np(1-p)}=\sqrt{100\cdot0.1(1-0.1)}=3$

$P(10 \leq X \leq12)\approx P(9.5 \leq Y \leq12.5)=$

$=P(\frac{9.5-10}{3} \leq Z\leq\frac{12.5-10}{3})=P(-0.17\leq Z\leq0.83)=$

$=0.7967-0.4325=0.3642$

ii)

$\lambda=np=10$

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

$P(10 \leq X \leq12)=P(X=10)+P(X=11)+P(X=12)$

$P(X=10)=\frac{10^{10}e^{-10}}{10!}=0.1251$

$P(X=11)=\frac{10^{11}e^{-10}}{11!}=0.1137$

$P(X=12)=\frac{10^{12}e^{-10}}{12!}=0.0948$

$P(10 \leq X \leq12)=0.1251+0.1137+0.0948=0.3336$