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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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Most popular questions from this chapter

In each case, determine the value of the constant\({\rm{c}}\)that makes the probability statement correct. a.\({\rm{\Phi (c) = }}{\rm{.9838}}\)b.\({\rm{P(0}} \le {\rm{Z}} \le {\rm{c) = }}{\rm{.291}}\)c.\({\rm{P(c}} \le {\rm{Z) = }}{\rm{.121}}\)d.\({\rm{P( - c}} \le {\rm{Z}} \le {\rm{c) = }}{\rm{.668}}\)e.\({\rm{P(c}} \le {\rm{|Z|) = }}{\rm{.016}}\)

The accompanying sample consisting of observations \({\rm{n = 20}}\) on dielectric breakdown voltage of a piece of epoxy resin appeared in the article "Maximum Likelihood Estimation in the \({\rm{3}}\)-Parameter Weibull Distribution (IEEE Trans. on Dielectrics and Elec. Insul., \({\rm{1996}}\): \({\rm{43 - 55)}}\). The values of \({\rm{(i - }}{\rm{.5)/n}}\) for which \({\rm{z}}\) percentiles are needed are \({\rm{(1 - }}{\rm{.5)/20 = }}{\rm{.025,(2 - }}{\rm{.5)/20 = }}\)\({\rm{.075, \ldots }}\), and\({\rm{.975}}\). Would you feel comfortable estimating population mean voltage using a method that assumed a normal population distribution?

\(\begin{array}{*{20}{c}}{{\rm{ Observation }}}&{{\rm{24}}{\rm{.46}}}&{{\rm{25}}{\rm{.61}}}&{{\rm{26}}{\rm{.25}}}&{{\rm{26}}{\rm{.42}}}&{{\rm{26}}{\rm{.66}}}\\{{\rm{ zpercentile }}}&{{\rm{ - 1}}{\rm{.96}}}&{{\rm{ - 1}}{\rm{.44}}}&{{\rm{ - 1}}{\rm{.15}}}&{{\rm{ - }}{\rm{.93}}}&{{\rm{ - }}{\rm{.76}}}\\{{\rm{ Observation }}}&{{\rm{27}}{\rm{.15}}}&{{\rm{27}}{\rm{.31}}}&{{\rm{27}}{\rm{.54}}}&{{\rm{27}}{\rm{.74}}}&{{\rm{27}}{\rm{.94}}}\\{{\rm{ zpercentile }}}&{{\rm{ - }}{\rm{.60}}}&{{\rm{ - }}{\rm{.45}}}&{{\rm{ - }}{\rm{.32}}}&{{\rm{ - }}{\rm{.19}}}&{{\rm{ - }}{\rm{.06}}}\\{{\rm{ Observation }}}&{{\rm{27}}{\rm{.98}}}&{{\rm{28}}{\rm{.04}}}&{{\rm{28}}{\rm{.28}}}&{{\rm{28}}{\rm{.49}}}&{{\rm{28}}{\rm{.50}}}\\{{\rm{ zpercentile }}}&{{\rm{.06}}}&{{\rm{.19}}}&{{\rm{.32}}}&{{\rm{.45}}}&{{\rm{.60}}}\\{{\rm{ Observation }}}&{{\rm{28}}{\rm{.87}}}&{{\rm{29}}{\rm{.11}}}&{{\rm{29}}{\rm{.13}}}&{{\rm{29}}{\rm{.50}}}&{{\rm{30}}{\rm{.88}}}\\{{\rm{ zpercentile }}}&{{\rm{.76}}}&{{\rm{.93}}}&{{\rm{1}}{\rm{.15}}}&{{\rm{1}}{\rm{.44}}}&{{\rm{1}}{\rm{.96}}}\end{array}\)

Let \({\rm{X}}\) denote the temperature at which a certain chemical reaction takes place. Suppose that \({\rm{X}}\) has pdf

\(f(x) = \left\{ {\begin{aligned}{{}{}}{\frac{1}{9}\left( {4 - {x^2}} \right)}&{ - 1拢x拢2} \\0&{{\text{ }}otherwise{\text{ }}}\end{aligned}} \right.\)

a. Sketch the graph of \({\rm{f(x)}}\).

b. Determine the cdf and sketch it.

c. Is \({\rm{0}}\) the median temperature at which the reaction takes place? If not, is the median temperature smaller or larger than\({\rm{0}}\)?

d. Suppose this reaction is independently carried out once in each of ten different labs and that the pdf of reaction time in each lab is as given. Let \({\rm{Y = }}\) the number among the ten labs at which the temperature exceeds \({\rm{1}}\). What kind of distribution does \({\rm{Y}}\) have? (Give the names and values of any parameters.)

Let \({\rm{X}}\) be a continuous \({\rm{rv}}\) with cdf

\({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{c}}{\rm{0}}&{{\rm{x}} \le {\rm{0}}}\\{\frac{{\rm{x}}}{{\rm{4}}}\left( {{\rm{1 + ln}}\left( {\frac{{\rm{4}}}{{\rm{x}}}} \right)} \right)}&{{\rm{0 < x}} \le {\rm{4}}}\\{\rm{1}}&{{\rm{x > 4}}}\end{array}} \right.\)

(This type of cdf is suggested in the article 鈥淰ariability in Measured Bedload Transport Rates鈥� (Water 91影视 Bull., \({\rm{1985:39 - 48}}\)) as a model for a certain hydrologic variable.) What is a. \({\rm{P(X}} \le {\rm{1)}}\)? b. \({\rm{P(1}} \le {\rm{X}} \le {\rm{3)}}\)? c. The pdf of \({\rm{X}}\)?

The article 鈥淪econd Moment Reliability Evaluation vs. Monte Carlo Simulations for Weld Fatigue Strength鈥� (Quality and Reliability Engr. Intl., \({\rm{2012: 887 - 896}}\)) considered the use of a uniform distribution with \({\rm{A = }}{\rm{.20}}\) and \({\rm{B = 4}}{\rm{.25}}\) for the diameter \({\rm{X}}\) of a certain type of weld (mm).

a. Determine the pdf of \({\rm{X}}\) and graph it.

b. What is the probability that diameter exceeds \({\rm{3 mm}}\)?

c. What is the probability that diameter is within \({\rm{1 mm}}\) of the mean diameter?

d. For any value \({\rm{a}}\) satisfying \({\rm{.20 < a < a + 1 < 4}}{\rm{.25}}\), what is \({\rm{P(a < X < a + 1)}}\)?
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