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Tomtit [17]
3 years ago
13

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 , , and min, respectively, and the standard deviations are , , and min, respectively. What is the probability that it takes at most hour of machining time to produce a randomly selected component
Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
5 0

Answer:

Z = (60 - x + y + z) / √a + b + c

Step-by-step explanation:

Since it is a normal distribution, we must calculate the mean and standard deviation, since we do not have data, what we will do is leave them based on these:

Thus Total Mean time = M1 + M2 + M3

given:

M1 = x

M2 = y

M3 = z

Total Mean Time M = x + y + z

Now to calculate the standard deviation we first calculate the variance.

The total Variance V = V1 + V2 + V3

Given:

V1 = a

V2 = b

V3 = c

V = a + b + c

Thus Standard deviation SD of the complete operation is

SD = √ V

SD = √a + b + c

we need to find the probability that the mean time is less than or equal to 60 minutes, the first thing is to find the value of Z.

Formula of Z is:

Z = (X - M) / SD

In this case X = 60.

On plugging the values we get

Z = (60 - x + y + z) / √a + b + c

refer to the Z table and find the Probability of Z ≤ (60 - x + y + z) / √a + b + c

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Bob          19                                       </span>Bob          23<span>
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Maria        13                                       </span>Maria        17<span>

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Week 2: (23 + 14 + 19 + 21 + 14 + 20 + 17)/7 = 128/7 = 18.28
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Week 1: 10, 10, 13, 15, 16, 17, 19            - The median is 15.
Week 2: 14, 14, 17, 19, 20, 21, 23          - The median is 19.
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Week 2: 14, 14, 17, 19, 20, 21, 23          - The mode is 14.
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Week 2: 14, 14, 17, 19, 20, 21, 23          - The median is 9 (23 - 14 = 9).
The difference in the ranges of times between Week 2 and Week 1 is 0 (9 - 9 = 0)
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