Answer to Question #251020 in Statistics and Probability for joy

Question

Digital limited is a manufacturing company that deals in the production of soft drink bottles. During a quality assurance exercise, it was noted that of every production lot, 0.2 per cent of soda bottles are found to be defective. The soda bottles are packed in crates each containing 25 bottles. A soft drinks manufacturer recently bought 1,500 crates of soda bottles from Digital Limited. Using the Poisson distribution, determine the number of crates that will contain (i) No defective soda bottle  (ii)At least two defective soda bottles                              

Answer 1

$N=1500 \\ n = 25 \\ p = 0.002 \\ λ=np=25 \times 0.002 = 0.05$

Let the random variable X denote the number of defective bottles in a crates of 25. Then by Poisson probability law, the probability of x defective bottles is a crate is

$P(X=x) = \frac{e^{-0.05}(0.05)^x}{x!}$

The frequency number of crates containing x defective bottles is

$f(x) = N \times P(X=x)$

(i) The number of crates that will contain no defective soda bottle

$= 1500 \times P(X=0) \\ = \frac{1500 \times e^{-0.05} \times (0.05)^0}{0!} \\ = 1426.8 \\ ≈1427$

(ii) The number of crates that will contain at least two defective soda bottles

$= 1500 \times P(X≥2) \\ = 1500 (1 -P(X<2)) \\ = 1500(1 -[P(X=0) +P(X=1)]) \\ = 1500(1-[\frac{e^{-0.05} \times (0.05)^0}{0!} + \frac{e^{-0.05} \times (0.05)^1}{1!}]) \\ = 1500(1-[0.9512 + 0.9512 \times 0.05]) \\ = 1.95 \\ ≈ 2$