Charles got a raise from $5.00 an hour to $7.00 an hour. this was a rise of what percent?

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