Question

A.J. has 20 jobs that she must do in sequence, with the times required to do each of these jobs being independent random variables with mean 50 minutes and standard deviation 10 minutes. M.J. has 20 jobs that he must do in sequence, with the times required to do each of these jobs being independent independent random variables with mean 52 minutes and standard deviation 15 minutes.
(a) Find the probability that A.J. finishes in less than 900 minutes.
(b) Find the probability that M.J. finishes in less than 900 minutes.
(c) Find the probability that A.J. finishes before M.J.

Answer 1

Answer:a) the probability that A.J. finishes in less than 900 minutes is 0.0125

b)the probability that M.J. finishes in less than 900 minutes is 0.0183

C) Find the probability that A.J. finishes before M.J. is 0.6915

Step-by-step explanation:

Let TA.j be the total time of A.j finishes all 20 jobs

Let Tm.j be the total time m.j finishes all 20 jobs .

Then by central limit theorem, TA.j and Tm.j has approximately a normal distribution.

E(TA.j) = 20×50=1000

Var(TA.j)= 20(10×10)=2000

E(Tm.j) = 20×52= 1040

Var(Tm.j)= 20(15×15)= 4500

a) P[TA.j <900]= [(TA.j -E(TA.j))/√(var(TA.j)]

Z< (900-1000)/√2000

Z<-2.24

Checking through a standardized normal table

The prob is 0.0125

b)p[Tm.j<900] = [Z< (900- 1040)/√4500]

Z < -2.08

The probability is 0.0183

C) p[( TA.j- Tm.j)- E( TA.j- Tm.j)]/√ (var(TA.j-Tm.j))

=[ 0 - E( TA.j- Tm.j)] / (var(TA.j-Tm.j))

= 0-(-40)/√6500

Z< 40/√6500

Z< 0.49613

Z< 0.6915