Manufacture of a certain component requires three different

Manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal distribution, and the three times are independent of one another. The mean values are 20, 15, and 30 min, respectively, and the standard deviations are 1, 2, and 1.5 min, respectively. What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component? (Round your answer to four decimal places.)

Solution

1st we will take the sum of all he mean time = 15+30+20 = 65 min

now we have to find the s.d for whole-

Total S.D = sqrt (1^2+2^2+1.5^2)
= sqrt (1+4+2.25)
= 2.69
z = (X-Mean) / S.D

now for calculating the z score

z value corresponding to X=60 is
z = (60-65) / 2.69 = - 1.86
The area under the standard normal curve corresponding to z = - 1.86 is 0.4686
The area under the standard normal curve left to z = - 1.86 indicates the required probability.
= 0.5000 - 0.4686
= 0.0314 OR 3.14%