Quality Control - Poisson Distribution

Question

A toy manufacturing company tests the quality of the toys it manufactures. On a daily basis, 20 toys are taken at random for testing and to check that 95% meet the toy specification. If more than 2 of the 20 toys are of an unacceptable standard then the production must be temporarily stopped.

Calculate the probability if the inspectors pass the day's toy production as acceptable if 75% of the day's toys meet the specification.

I believe Poisson Distribution is required to solve this problem.

Using the Poisson Distribution formula, this is what I have attempted:

$Success(P) = 20*0.75 = 15$

$= \frac{(e^-20)*(20^15)}{15!}$

$= 0.076$

Is that correct?

Answer 1

Let $X \sim \mathsf{Binom}(20, .25)$ be the number of defective toys seen in a random sample of 20. You seek $P(X > 2) = 1 - P(X \le 2),$ which is easy to compute using the formula for the binomial PDF. [You'd need to find $P(X = 0)+P(X=1)+P(X=2)$ and subtract from $1.$]

In R statistical software (where pbinom is the binomial CDF), the answer is obtained as

 1 - pbinom(2, 20, .25)
 ## 0.9087396

---

This is not a situation for a good Poisson approximation. The Poisson rate is $\lambda = 5$ defects in 20. The figure below shows the binomial probabilities as bars. Probabilities for $\mathsf{Pois}(5)$ are shown as open circles.