Answer to Question #202286 in Statistics and Probability for RAGHAV SOOD

Question

Let X be a binomial variate with n=100, p=0.1. Find the approximate value of

P(10 ≤X≤ 12) using:

 (i) normal distribution 

 (ii)poisson distribution 

 [You may like to use the following values. 

 P(Z ≤ 0.67)=0.7486, P(Z ≤ 0.33)=0.6293, P(Z ≤0)=0.5]

Answer 1

We have that

n=100, p=0.1

P(10 ≤ X ≤ 12) - ?

i) Normal distribution

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

Standard deviation $\sigma = \sqrt{np(1-p)}=\sqrt{100\cdot0.1(1-0.1)}=\sqrt{10\cdot0.9}=\sqrt9=3$

$P(10 ≤ X ≤ 12)=F(\frac{12-\mu}{\sigma})-F(\frac{10-\mu}{\sigma})=F(\frac{12-10}{3})-F(\frac{10-10}{3})=F(\frac{2}{3})-F(0)\approx 0.7486 - 0.5 = 0.2486$

ii) Poisson distribution

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

The poisson probability is calculated by the formula

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

$P(10 ≤ X ≤ 12)=P(10)+P(11)+P(12)=\frac{10^{10}e^{-10}}{10!}+\frac{10^{11}e^{-10}}{11!}+\frac{10^{12}e^{-10}}{12!}=0.12511+0.11374+0.09478=0.33363$