Question

Let the number of aberrations in the production of a large scale lens have a Poisson distribution. We want the probability that a given lens contains at most one aberration to be greater than 0.99.
a) Find the largest value of the mean that this distribution can take.
b) Assume the mean is µ = 0.11. Determine the probability that a batch of 5 lens will have at least two lenses with at least one defect each.

Answer 1

SO here the distribution is Poisson with mean .

so here as we know about the poisson distribution

f(x) = e^-x/x!

so here

Pr(x < = 1) = Pr(x = 0) + Pr(x = 1)

= e^-/0! + e^-/1!

= e^- [1 + ]

so here as we know

e^{​​​​​​​-} [1 + ] > 0.99

so here now we have to find the value of ^​ = 0.149

so here the largest value of the mean the distribution can take is around 0.149 or 0.15.

(b) Here µ = 0.11

so here first we will find that a random lens will have at least one defect each

Pr(x >= 1) = 1 - Pr(x = 0) = 1 - e^-0.11 = 1 - 0.8958 = 0.1042

so here now we have to find  the probability that a batch of 5 lens will have at least two lenses with at least one defect each.

that would a binomial distribution with n = 5 and p = 0.1042

and if y is the nuber of lenses with atleast one defect each.

then we have to calculate

Pr(y >= 2) = 1 - Pr(y = 0) - Pr(y = 1)  

= 1- ⁵C₀ (0.1042)⁰ (1 - 0.1042)⁵ - ⁵C₁ (0.1042)¹(1 - 0.1042)⁴

= 1 - 0.57695 - 0.3354

= 0.0876