Manufacture of a certain component requires three different machining operations

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 20, 15, and 30 min, respectively, and the standard deviations are 2, 1, and 1.3 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.) You may need to use the appropriate table in the Appendix of Tables to answer this question.

Explanation / Answer

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and

variance 2, then

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)

If X1, X2, X3 are independent and X1 ~ N(µ1, 12), X2 ~ N(µ2, 22), X3 ~ N(µ3, 32), then

T = (X1 + X2 + X3) ~ N(µ, 2), where µ = µ1 + µ2 + µ3, and 2 = 12 + 22 + 32 [note that the result for variance is true only if the variables are independent.] ………………(3)

Now, to work out the solution,

Let X1, X2, X3 represent the time taken for machining operations 1, 2 and 3 respectively.

Then, we are given X1 ~ N(20, 22), X2 ~ N(15, 12), X3 ~ N(30, 1.32) and X1, X2, X3 are independent.

Then, total time, T, for completing all 3 machining operations = (X1 + X2 + X3).

[vide (3) under Back-up Theory], T ~ N(µ, 2), where µ = 20 + 15 + 30 = 65 minutes, and 2 = 22 + 12 + 1.32 = 6.69 or = 2.5865.

We want P(T 60) [1 hour = 60 minutes]

[vide (2) under Back-up Theory],

P(T 60) = P[Z {(60 - 65)/2.5865}]

= P(Z - 1.9331) = 0.0266 ANSWER

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