Question

The diameters of ball bearings are distributed normally. The mean diameter is 123 millimeters and the standard deviation is 4 millimeters. Find the probability that the diameter of a selected bearing is greater than 129 millimeters. Round your answer to four decimal places.

Answer 1

0.0668

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\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 problem, we have that:

`\mu = 123, \sigma = 4`

Find the probability that the diameter of a selected bearing is greater than 129 millimeters.

This is 1 subtracted by the pvalue of Z when X = 129. So

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

`Z = (129 - 123)/(4)`

`Z = 1.5`

`Z = 1.5` has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% probability that the diameter of a selected bearing is greater than 129 millimeters.