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Statistics / Applied Statistics and Probability for Engineers 6 / Chapter 5.4 / Problem 65E

Applied Statistics and Probability for Engineers | 6th Edition | ISBN: 9781118539712 | Authors: Douglas C. Montgomery, George C. Runger

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A plastic casing for a magnetic disk is composed of two

Chapter 5, Problem 65E

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QUESTION:

Problem 65E

A plastic casing for a magnetic disk is composed of two halves. The thickness of each half is normally distributed with a mean of 2 millimeters, and a standard deviation of 0.1 millimeter and the halves are independent.

(a) Determine the mean and standard deviation of the total thickness of the two halves.

(b) What is the probability that the total thickness exceeds 4.3 millimeters?

Questions & Answers

QUESTION:

Problem 65E

A plastic casing for a magnetic disk is composed of two halves. The thickness of each half is normally distributed with a mean of 2 millimeters, and a standard deviation of 0.1 millimeter and the halves are independent.

(a) Determine the mean and standard deviation of the total thickness of the two halves.

(b) What is the probability that the total thickness exceeds 4.3 millimeters?

ANSWER:

Answer

Step 1 of 2

(a)

We have given a plastic casing for a magnetic disk is composed of two halves.

The thickness of each half is normally distributed with a mean of $2\ mm,$ and a standard deviation of $0.1\ mm$ and the halves are independent.

We are asked to find the mean and standard deviation of the total thickness of the two halves.

Let $T$ denote the total thickness.

Then, $T\ =X+Y$ ……….(1)

If ${X}_{1},\ {X}_{2},........,{X}_{p}$ are independent, normal random variables with $E({X}_{i})={\mu{}}_{i}\ and\ V({X}_{i})={\sigma{}}_{i}^{2},\ for\ i=1,\ 2,....p,$

$Y\ ={c}_{1}{X}_{1}+{c}_{2}{X}_{2}+.........+{c}_{p}{X}_{p}$

Is a normal random variable with

$E(Y)\ ={c}_{1}{\mu{}}_{1}+{c}_{2}{\mu{}}_{2}+.........+{c}_{p}{\mu{}}_{p}$ …….(2)

$V(Y)={{c}^{2}}_{1}{\sigma{}}_{1}^{2}+{{c}^{2}}_{2}{\sigma{}}_{2}^{2}+.........+{{c}^{2}}_{p}{\sigma{}}_{p}^{2}$ ………(3)

Hence the mean and standard deviation of the total thickness of the two halves using equation $(2)\ and\ (3).$

$E(T)\ ={c}_{X}{\mu{}}_{X}+{c}_{Y}{\mu{}}_{Y}$ ……….(4)

Compare equation (1) and (4), we get ${C}_{X}={C}_{Y}=1\ and\ {\mu{}}_{X}=2mm,{\ \mu{}}_{Y}=2mm\ given\$

$E(T)\ =2+2=4\ mm$

$V(T)=0.{1}^{2}+0.{1}^{2}=0.02\ m{m}^{2}$

${\sigma{}}_{T}=\sqrt{V(T)}=0.1414\ mm.$

Hence the mean and standard deviation of the total thickness of the two halves is $4\ mm\ and\ 0.1414\ mm$ respectively.

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