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Chapter 4: Q45E (page 168)

A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is \({\rm{.500}}\)in. A bearing is acceptable if its diameter is within \({\rm{.004}}\)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 \({\rm{.499}}\)in. and standard deviation \({\rm{.002}}\)in. What percentage of the bearings produced will not be acceptable?

Short Answer

Expert verified

The\({\rm{7}}{\rm{.3\% }}\)of total bearings are not acceptable.

Step by step solution

01

Introduction

The standard deviation is a statistic that calculates the square root of the variance and measures the dispersion of a dataset compared to its mean.

02

Explanation

Let the diameter of the bearings be denoted by \({\rm{rv}}\) \({\rm{X}}\), then it's mean and standard deviation are given to us as:

\(\begin{array}{l}{\rm{\mu = 0}}{\rm{.499 inches }}\\{\rm{\sigma = 0}}{\rm{.002 inches }}\end{array}\)

The acceptable diameter of bearing is given to us as \({\rm{0}}{\rm{.496}}\)to \({\rm{0}}{\rm{.504}}\)inches. Hence \({\rm{P(0}}{\rm{.496 < X < 0}}{\rm{.504)}}\)denotes the fraction of the acceptable bearing.

Standardizing gives:

\({\rm{0}}{\rm{.496 < X < 0}}{\rm{.504}}\)if and only if

\(\frac{{{\rm{0}}{\rm{.496 - 0}}{\rm{.499}}}}{{{\rm{0}}{\rm{.002}}}}{\rm{ < }}\frac{{{\rm{X - 0}}{\rm{.499}}}}{{{\rm{0}}{\rm{.002}}}}{\rm{ < }}\frac{{{\rm{0}}{\rm{.504 - 0}}{\rm{.499}}}}{{{\rm{0}}{\rm{.002}}}}\frac{{{\rm{ - 0}}{\rm{.003}}}}{{{\rm{0}}{\rm{.002}}}}{\rm{ < }}\frac{{{\rm{X - 0}}{\rm{.499}}}}{{{\rm{0}}{\rm{.002}}}}{\rm{ < }}\frac{{{\rm{0}}{\rm{.005}}}}{{{\rm{0}}{\rm{.002}}}}{\rm{ - 1}}{\rm{.5 < Z < 2}}{\rm{.5}}\)

Thus

\({\rm{P(0}}{\rm{.496 < X < 0}}{\rm{.504) = P( - 1}}{\rm{.5 < Z < 2}}{\rm{.5)}}\)Here \({\rm{Z}}\)is a standard normal distribution \({\rm{rv}}\) with \(cdf\phi (z)\). Hence

\(P(0.496 < X < 0.504) = \phi (2.5) - \phi ( - 1.5)\)To get \(\phi (2.5)\)and\(\phi ( - 1.5)\), we check Appendix Table \({\rm{A}}{\rm{.3,}}\)from there

\(\phi ( - 1.5) = 0.0668{\rm{ and }}\phi (2.5) = 0.9938\)

Hence

\(\begin{array}{l}{\rm{P(0}}{\rm{.496 < X < 0}}{\rm{.504) = 0}}{\rm{.9938 - 0}}{\rm{.0668}}\\{\rm{P(0}}{\rm{.496 < X < 0}}{\rm{.504) = 0}}{\rm{.927}}\end{array}\)

Since this probability denotes the fraction of acceptable bearings, Hence

\(\begin{array}{c}{\rm{P( bearing not acceptable ) = 1 - P(0}}{\rm{.496 < X < 0}}{\rm{.504)}}\\{\rm{ = 1 - 0}}{\rm{.927P( bearing not acceptable )}}\\{\rm{ = 0}}{\rm{.073}}\end{array}\)

Hence \({\rm{7}}{\rm{.3\% }}\)of total bearings are not acceptable.

Proposition: Let Z be a continuous \({\rm{rv}}\) with \(cdf\phi (z)\). Then for any \({\rm{a}}\) and \({\rm{b}}\)with\({\rm{a < b}}\),

\(P(a < Z < b) = \phi (b) - \phi (a)\)

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