Question

1.19 The number of sick leaves for male workers is on average three times that of female workers. The number of sick leaves of all workers in a day is Poisson distributed with mean 4.5. (a) Find the probability of the event of having no sick leave in a day. (b) Find the probability of the event of having one male sick leave and one female sick leave in a day.

Answer 1

a)

P(no sick leave)=P(no sick leave for male and no sick leave for female) =P(X=0)=e^-4.5*4.5⁰/0! =0.0111

b)

for male expected leaves (A)=3 times* female leaves (B)

therefore A=3*B

also A+B=4.5

hence A =3.375 and B=1.125

therefore P(1 male leave and 1 female sick leave)=e^-3.375*3.375¹/1!*e^-1.125*1.125¹/1!

=0.1155*0.3652=0.0422