Question

44. Manufacturing Ball bearings are manufactured with a mean diameter of 5 millimeters (mm). Because of variability in the manufacturing process, the diameters of the ball bearings are approximately normally distributed, with a standard deviation of 0.02 mm. (a) What proportion of ball bearings has a diameter more than 5.03 mm

Answer 1

0.0668 = 6.68% of ball bearings has a diameter more than 5.03 mm

Explanation:

When the distribution is normal, we use 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.

Manufacturing Ball bearings are manufactured with a mean diameter of 5 millimeters (mm).

This means that
`\mu = 5`

With a standard deviation of 0.02 mm.

This means that
`\sigma = 0.02`

(a) What proportion of ball bearings has a diameter more than 5.03 mm

This is 1 subtracted by the pvalue of Z when X = 5.03. So

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

`Z = (5.03 - 5)/(0.02)`

`Z = 1.5`

`Z = 1.5` has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% of ball bearings has a diameter more than 5.03 mm