poisson distribution

Question

a manufacturing process produces on the average 1.2 defects per 100 units of output and the number of defects follows a poisson distribution. What is the probability that the next 500 units show less than three defects?

Answer 1

We will let x denote the number of defects in 500 units to begin.

Our lambda value, for which I will write here as L, is the average number of changes we can expect over the given interval. If we get 1.2 defects every 100 units, we can expect 2.4 defects for 200, 3.6 defects for 300, 4.8 defects for 400, and 6 defects for 500, just using the expected value from what we have given to us. Thus, here, L = 6.

So, if we can expect 6 defects, what do we think the probability is of getting less than 3 defects for 500 units made? This is where the function of the poisson distribution comes in. Recall that the poisson distribution looks something like f(x)=(L^x * e^-L)/x!

The probability that x < 3 is equal to P(x=0) + P(x=1) + P(x=2).

This is then equal to (6^0 * e^-6)/0! + (6^1 * e^-6)/1! + (6^2 * e^-6)/2! = .06197

I hope that this makes sense! Good luck