Answer to Question #101853 in Statistics and Probability for Nidhi Shuka

Question

In a certain factory producing cycle tires, there is a small chance of 1 in 500 tires to be defective. The tires are supplied in lots of 20. Using Poisson distribution, calculate the approximate number of lots containing no defective, no defective and two defective tiers respectively in a consignment of 20,000 tiers.

Answer 1

Binomial distribution with parameters n = 20 and p = 0.002 can be approximated using Poisson with parameter m = np = 0.04

$𝑝0=𝑃(𝑋=0)=0.04^0/0!*exp(−0.04)=0.9607894$

$𝑛0=20000∗𝑝0≈19216$

The approximate number of packets containing blades with no defective blades is 19216

$𝑝2=𝑃(𝑋=2)=0.04^2/2! * exp(−0.04)=0.0007686$

$𝑛2=20000∗𝑝2≈15$

The approximate number of packets containing blades with two defective blades is 15

Answer: 19216, 15.