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

4.4004%

Explanation:

Data provided in the question:

Mean values : 20, 30, and 15 min

Standard deviations : 1, 2, and 1.9 min

X = 1 hour = 60 minutes

Now,

New Mean will be = 20 + 30 + 15

Variance = 1² + 2² + 1.9²

or

⇒ Variance = 1 + 4 + 3.61 = 8.61

standard deviation = √8.61 = 2.93

Therefore,

Z-value =
`((60 - 65))/(2.93)`

or

Z-value = - 1.706

from standard z table

P(- 1.706) = 0.044004

Hence, % below = 0.044004 × 100%

= 4.4004%