Question

In the manufacturing process of automotive windshield glass, it is known that the average number of air bubbles found in a 1 [cm3] is 0.12. To detect the presence of bubbles in the crystal, samples of 10 [cm3] are analyzed. Assuming that the number of bubbles per cm3 follows a Poisson distribution and that this number is independent of one cm3 to another, it is requested

a) Calculate the probability that a 10 cm3 sample does not contain air bubbles

b) Calculate the minimum number of samples of 10 cm3 that must be examined so that it can be assured, with a probability of at least 0.99, that bubbles are present in them.

Answer 1

Here  the average number of air bubbles found in a 1 [cm3] is 0.12. Here samples of 10 [cm3] are analyzed

So, the expected number of air bubble found in 10 [cm3] is 10 * 0.12 = 1.2

so here if x is the number of air bubbles in 10 [cm3] then x ~ POISSON (1.2)

P(X) = e^-1.2 1.2^x/x!

(a) Here we have to find

P(x = 0) = e^-1.2 1.2⁰/0! = 0.3012

(b) here let say there are k number of samples of 10 cm³ that must be examined so that it can be assured, with a probability of at least 0.99. So, here expected number of bubbles in k samples = k * 0.2 = 1.2 k

Here,

P(X) = e^-1.2k (1.2k)^x/x!

P(x > 0) > 0.99

P(x > 0) = 1 - P(x = 0)

1 - P(x = 0) > 0.99

P(x = 0) < 0.01

e^-0.12k < 0.01

taking logarithm both sides

0.12 k > 4.6052

k > 38.4

so here at least 39 number of samples of 10 cm3 that must be examined so that it can be assured, with a probability of at least 0.99, that bubbles are present in them.