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 15, 30, and 20 min, respectively, and the standard deviations are 2, 1, and 1.6 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

Answer:

0.0359

Step-by-step explanation:

Data provided:

mean values of three independent times are 15, 30, and 20 minutes

the standard deviations are 2, 1, and 1.6 minutes

Now,

New Mean = 15 + 30 + 25 = 65

Variance = ( standard deviation )²

or

Variance = 2² + 1² + 1.6² = 7.56

therefore,

Standard deviation = √variance

or

Standard deviation =  2.75

Thus,

Z-value = $\frac{\textup{60 - 65}}{\textup{2.75}}$

or

Z-value = - 1.81

from the Z-table

the Probability of Z ≤ -1.81 = 0.0359