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 , , 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

Answer 1

Z = (60 - x + y + z) / √a + b + c

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