Answer to Question #259059 in Statistics and Probability for Gaurav

Question

in a certain factory producing cycle tyres there is a small chance of 1 in 500 tyres to be defective. the tyres are supplied in lots of 10 . using Poisson distribution, calculate the approximate number of lots containing no defective one defective and two defective tyres respectively in consignment of 10,000 lots

Answer 1

Binomial distribution with parameters $n = 10$ and $p =1/500= 0.002$ can be approximated using Poisson with parameter $\lambda = np=10(0.002) = 0.02.$

(a)

$$P(X=0)=\dfrac{e^{-0.02}\cdot(0.02)^0}{0!}=e^{-0.02}$$

$$\approx0.98019867$$

$$0.98019867(10000)=9802$$

(b)

$$P(X=1)=\dfrac{e^{-0.02}\cdot(0.02)^1}{1!}=0.02e^{-0.02}$$

$$\approx0.01960397$$

$$0.01960397(10000)=196$$

(c)

$$P(X=2)=\dfrac{e^{-0.02}\cdot(0.02)^2}{2!}=0.0002e^{-0.02}$$

$$\approx0.00019604$$

$$0.00019604(10000)=2$$