Question

A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is .500 in. A bearing is acceptable if its diameter is within .004 in. of this target value. Suppose, however, that the setting has changed during the course of production, so that the bearings have normally distributed diameters with mean value .499 in. and standard deviation .002 in. What percentage of the bearings produced will not be acceptable

Answer 1

7.3% of the bearings produced will not be acceptable

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 = 0.499, \sigma = 0.002`

Target value of .500 in. A bearing is acceptable if its diameter is within .004 in. of this target value.

So bearing larger than 0.504 in or smaller than 0.496 in are not acceptable.

Larger than 0.504

1 subtracted by the pvalue of Z when X = 0.504.

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

`Z = (0.504 - 0.494)/(0.002)`

`Z = 2.5`

`Z = 2.5` has a pvalue of 0.9938

1 - 0.9938= 0.0062

Smaller than 0.496

pvalue of Z when X = -1.5

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

`Z = (0.496 - 0.494)/(0.002)`

`Z = -1.5`

`Z = -1.5` has a pvalue of 0.0668

0.0668 + 0.0062 = 0.073

7.3% of the bearings produced will not be acceptable