Question

Assume the inspection time of a machine follows an exponential distribution with parameter equal to 1.

(a) Calculate the probability that the inspection time lasts more than 2 hours.

(b) Calculate the probability that the total inspection time lasts more than 4 hours if we have already spent 2 hours inspecting the machine.

(c) Every time we need to spend more that 2 hours inspecting a machine we get paid a bonus $1000. Calculate the probability that for the next 20 machines we will be paid at least $5000 extra.

Answer 1

a)

β =

1.000

probability =

P(X>2)=

1-P(X<2)=

1-(1-exp(-2/1))=

0.1353

b)

probability that the total inspection time lasts more than 4 hours if we have already spent 2 hours inspecting the machine

=P(X>4|X>2) =P(X>4)/P(X>2) =e^-4/1/e^-2/1 =e^-2 =0.1353

c)

e probability that for the next 20 machines we will be paid at least $5000 extra

=P(at least 5 machines took more than 2 hours to repair)

=

P(X>=5)=

1-P(X<=4)=

1-∑x=04     (nCx)px(1−p)(n-x)    =

0.1233