The manufacture of a certain part requires three different machine operations. T

Question

The manufacture of a certain part requires three different machine operations. The time on machine 1 has mean of 0.4 hours and standard deviation 0.1 hours; The time on machine two has mean of 0.45 hours and standard deviation 0.15 hours; while the time on machine 3 has mean 0.35 hours and standard deviation 0.125 hours. The times needed on the three machines are independent. Suppose that 65 parts are manufactured.

a) What is the distribution of the total time on machine 1?

b) What is the probability that the total time used by all three machines together is between 50 and 55 hours?

Explanation / Answer

ans=

Let X=time on machine 1 in hours, EX = 0.4, and St. Dev. X = 0.1. Y=time on machine 2,

EX= 0.45, and St. Dev. Y= 0.15. X and Y independent.

X1, X2, …, X65 are times for the 65 parts on machine 1. Y1, Y2, …, Y65 are times for the 65 parts on machine2.

Sx=X1+X2+…+X65= total time on machine 1;

Sx has approximately Normal distribution with mean ESx= 65(0.4)= 26, and VarSx =65 (0.1)2= 0.65, so the distribution of Sx is approximately N(26, 6.5). Let Sy=Y1+Y2+…+Y65= total time on machine 2;

Sy has approximately Normal distribution with mean ESy= 65(0.45)= 29.25, and VarSy =65 (0.15)2= 1.4625, so the distribution of Sy is approximately N(29.25, 1.4625)

b) Need distribution of the total time on machines 1 and 2.
Let T=total time for the 65 parts on both machines, then T= Sx+Sy.
Since both Sx and Sy are approximately normal, their sum is also approximately
normal with ET=ESx+ESy= 26+29.25=55.25 and variance VarT=VarSy + VarSx = 0.65 +
1.4625 = 2.1125. So T’s distribution is approximately N(55.25, 2.1125).
Then, P(50 < T < 55)= standardization = P(-3.61 <Z < -0.17) = 0.4325 – 0.0002 = 0.432

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