Question

18. The servicing of a machine requires two separate steps, with the time needed for the

first step being an exponential random variable with mean 0.2 hour and the time for the
second step being an independent exponential random variable with mean 0.3 hour. If a
repair person has 20 machines to service, what is approximately the probability that all the
work can be completed in 8 hours?

Answer 1

Explanation:

Let X denote the first step

Let Y denote the second step

Then

E(X) = 0.2

E (Y) = 0.3

V (X) = 0.04

V (Y) = 0.09

Now,

E(X,Y) = E[X] + E{Y}

0.2 + 0.3 = 0.5

And since X and Y are independent

Therefore,

V(X , Y) = V(X) + V(Y)

= 0.04 + 0.09

= 0.13

Now required probability is

`P\{ \sum X_i+\sum Y_i<8 \}=P\{ (\sum X_i + \sum Y_i-nE[X+Y])/(√(Var(X+Y)n) ) <(8-20*0.5)/(√(0.13*20) ) \}\\\\=P\{Z_n<(8-10)/(√(2.6) ) \}\\\\=P\{Z_n<-1.24\}`

= Φ(-1.24)

= 1 - Φ (1.24)

= 1 - 0.8925

= 0.1075