Question

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 30, 20, and 15 min, respectively, and the standard deviations are 2, 1, and 1.9 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.)

Answer 1

Explanation:

Let P(X >60) be probability of randomly selected component

U = Mean value and α = Standard deviation

Using P(X>60) = P(X - U/α > 60 - U/α)

i) For U = 30 and α =2

P(X>60) = P(X1 - U/α > 60 - 30/2)

= P(Z >15)

P(o ≤ Z1 ≤ 15) = 4.332

ii) For U = 20 and α = 1

P(X>60) = P(X2 - U/α > 60 - 20/1)

=P(Z >40)

P(0 ≤ Z2 ≤ 40) = 5.000

iii) For U =15 and α = 1.9

P(X> 60) = P(X3 - U/α > 60 - 15/1.9)

= P(Z3 > 23.68)

P(0 ≤ Z3 ≤ 23.68) = 4.911