Answered: Which component has the highest…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Question

Two components are being considered for use in building a system in your manufacturing plant. The lifetime of component 1 is modeled by a normal random variable with mean 20,000 hours and standard deviation of 5000 hours (ignore the fact the this model can technically give negative values for lifetime - that probability is practically zero). The lifetime of component 2 is also a normal random variable but with mean 22,000 hours and standard deviation 1000 hours. Which component has the highest probability of functioning for the life of the system if the target lifetime is 20,000 hours? What about a target lifetime of 24,000 hours?

Expert Solution

Step 1: find which component has highest probability if the target lifetime is 20,000 hours

Given,

$bold italic C bold italic o bold italic m bold italic p bold italic o bold italic n bold italic e bold italic n bold italic t bold space bold 1 bold space bold colon$ is modeled by a normal random variable with $bold italic m bold italic e bold italic a bold italic n bold left parenthesis bold italic mu bold right parenthesis bold equals bold space bold 20 bold comma bold 000$ hours and $bold italic s bold italic t bold italic a bold italic n bold italic d bold italic a bold italic r bold italic d bold space bold italic d bold italic e bold italic v bold italic i bold italic a bold italic t bold italic i bold italic o bold italic n bold left parenthesis bold italic sigma bold right parenthesis bold equals bold space bold 5000$ hours

$bold italic c bold italic o bold italic m bold italic p bold italic o bold italic n bold italic e bold italic n bold italic t bold space bold 2 bold colon$ is also a normal random variable but with $bold italic m bold italic e bold italic a bold italic n bold left parenthesis bold italic mu bold right parenthesis bold equals bold 22 bold comma bold 000$ hours and $bold italic s bold italic t bold italic a bold italic n bold italic d bold italic a bold italic r bold italic d bold space bold italic d bold italic e bold italic v bold italic i bold italic a bold italic t bold italic i bold italic o bold italic n bold left parenthesis bold italic sigma bold right parenthesis bold equals bold 1000$ hours.

Hence the $z subscript s c o r e end subscript equals fraction numerator x minus mu over denominator sigma end fraction$

For the target lifetime is $bold space bold 20 bold comma bold 000 bold space$ hours

$bold italic C bold italic o bold italic m bold italic p bold italic o bold italic n bold italic e bold italic n bold italic t bold space bold 1 bold space$ ,

$P left parenthesis x greater than 20000 right parenthesis space b u t space t h e space z subscript s c o r e end subscript space a t space x equals 20000 z subscript s c o r e end subscript equals fraction numerator x minus mu over denominator sigma end fraction space space space space space space space space space space equals fraction numerator 20000 minus 20000 over denominator 5000 end fraction space space space space space space space space space space equals 0 over 5000 z subscript s c o r e end subscript equals 0 rightwards double arrow P left parenthesis x greater than 20000 right parenthesis space equals P left parenthesis z greater than 0 right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 minus 0.5 space space space space space space space space space space........ space space space f r o m space t a b l e bold space bold space bold space bold space bold italic P bold left parenthesis bold italic x bold greater than bold 20000 bold right parenthesis bold space bold equals bold 0 bold. bold 5$

For target lifetime is $bold space bold 20 bold comma bold 000 bold space$ hours

$bold italic c bold italic o bold italic m bold italic p bold italic o bold italic n bold italic e bold italic n bold italic t bold space bold 2$

$P left parenthesis x greater than 20000 right parenthesis space b u t space t h e space z subscript s c o r e end subscript space a t space x equals 20000 z subscript s c o r e end subscript equals fraction numerator x minus mu over denominator sigma end fraction space space space space space space space space space space equals fraction numerator 20000 minus 22000 over denominator 1000 end fraction space space space space space space space space space space equals fraction numerator negative 2000 over denominator 1000 end fraction z subscript s c o r e end subscript equals negative 2 rightwards double arrow P left parenthesis x greater than 20000 right parenthesis space equals P left parenthesis z greater than negative 2 right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 minus 0.02275 space space space space space space space space space space........ space space space f r o m space t a b l e bold space bold space bold space bold space bold italic P bold left parenthesis bold italic x bold greater than bold 20000 bold right parenthesis bold space bold equals bold 0 bold. bold 97725$

Hence $bold italic c bold italic o bold italic m bold italic p bold italic o bold italic n bold italic e bold italic n bold italic t bold space bold 2$ has the highest probability of functioning for the life of the system if the target lifetime is $bold 20 bold comma bold 000$ hourse

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