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olya-2409
1 month ago
5

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.)
Mathematics
1 answer:
PIT_PIT [12.4K]1 month ago
6 0

Answer:

0.0359

Step-by-step explanation:

Provided values:

Mean durations of three independent processes are 15, 30, and 20 minutes.

The associated standard deviations are 2, 1, and 1.6 minutes, respectively.

Thus,

New Mean = 15 + 30 + 25 = 65

Variance = (standard deviation)²

or

Variance = 2² + 1² + 1.6² = 7.56

<phence>

Standard deviation = √variance

or

Standard deviation = 2.75

<pas a="" result="">

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

or

Z-value = - 1.81

Consulting the Z-table, the Probability of Z ≤ -1.81 is equal to 0.0359.

</pas></phence>
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