Question

The diameters of ball bearing are distributed normally. The mean diameter is 138 millimeters and the variance is 9. Find the probability that the diameter of a selected bearing is greater than 132 millimeters

Answer 1

97.72% probability that the diameter of a selected bearing is greater than 132 millimeters

Explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation(which is the square root of the variance)
`\sigma`, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question:

`\mu = 138, \sigma = √(9) = 3`

Find the probability that the diameter of a selected bearing is greater than 132 millimeters

This is 1 subtracted by the pvalue of Z when X = 132. So

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

`Z = (132 - 138)/(3)`

`Z = -2`

`Z = -2` has a pvalue of 0.0228

1 - 0.0228 = 0.9772

97.72% probability that the diameter of a selected bearing is greater than 132 millimeters