Question

(3 markah/marks) d) Bilangan kecacatan dalam kabel gentian optik mempunyai taburan Poisson. Purata bilangan kelemahan dalam 50m kabel adalah 1.2. The number of flaws in a fibre optic cable follows a Poisson distribution. The average number of flaws in 50m of cable is 1.2 ) Apakah kebarangkalian hanya ada tiga kecacatan dalam kabel 150m? What is the probability of exactly three flaws in 150m of cable? (1 markah/mark) i) Apakah kebarangkalian sekurang-kurangnya dua kecacatan dalam kabel 100m? What is the probability of at least two fiaws in 100m of cable? (2 markah/marks)

Answer 1

(i)

Mean number of flaws in 50 m cable = 1.2

So,

Mean number of flaws in 150 m cable = 1.2 X 3 = 3.6

So,

Probability of exactly 3 flaws in 150 m cable is given by:

(ii)

Mean number of flaws in 100 m cable = 1.2 X 2 = 2.4

P(X2) = 1 - [P(X=0) + P(X=1)] (1)

(2)

e-2.42.40 0,091×1- 0.091 P(X = 0) =-O!

(3)

Substituting (2) & (3), equation (1)becomes:

P(X2) = 1 - (0.091 + 0.218)

= 1 - 0.309

= 0.691

So,

Answer is:

0.691