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, 15, and 30 min, respectively, and the standard deviations are 1, 2, and 1.5 min, respectively. What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component

Answer 1

0.0314 = 3.14% probability that it takes at most 1 hour of machining time to produce a randomly selected component

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Sum of normal variables:

When normal variables are added, the mean is the sum of the means while the standard deviation is the square root of the sum of the variances.

The mean values are 20, 15, and 30 min, respectively:

This means that
`\mu = 20 + 15 + 30 = 65`

The standard deviations are 1, 2, and 1.5 min, respectively.

This means that
`\sigma = √(1^2 + 2^2 + 1.5^2) = 2.6926`

What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component?

Mean and standard deviation are in minutes, so this is the pvalue of Z when X = 60.

`Z = (X - \mu)/(\sigma)`

`Z = (60 - 65)/(2.6926)`

`Z = -1.86`

`Z = -1.86` has a pvalue of 0.0314

0.0314 = 3.14% probability that it takes at most 1 hour of machining time to produce a randomly selected component