In a process of sintering (heating) two types of copper

Chapter 6, Problem 14E

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QUESTION:

In a process of sintering (heating) two types of copper powder (see Exercise 5.152), the density function for \(Y_{1}\), the volume proportion of solid copper in a sample, was given by

\(f_{1}\left(y_{1}\right)=\left\{\begin{array}{ll}  6 y_{1}\left(1-y_{1}\right), & 0 \leq y_{1} \leq 1 \\  0, & \text { elsewhere }  \end{array}\right.\)

The density function for \(Y_{2}\), the proportion of type A crystals among the solid copper, was given as

\(f_{2}\left(y_{2}\right)=\left\{\begin{array}{ll}  3 y_{2}^{2}, & 0 \leq y_{2} \leq 1 \\

0, & \text { elsewhere }  \end{array}\right.\)

The variable \(U=Y_{1} Y_{2}\) gives the proportion of the sample volume due to type A crystals. If

 \(Y_{1} \text { and } Y_{2}\) are independent, find the probability density function for .

Equation Transcription:

 {

 {

 

Text Transcription:

Y_1

f_1(y_1)={ 6y_1(1-y_1)  0 \leq y_1 \leq 1  0,    elsewhere  

Y_2

f_2(y_2)= {3y_2^2,   0 \leq y_2 \leq 1   0,  elsewhere

U=Y_1 Y_2

Y_1 and Y_2

Questions & Answers

QUESTION:

In a process of sintering (heating) two types of copper powder (see Exercise 5.152), the density function for \(Y_{1}\), the volume proportion of solid copper in a sample, was given by

\(f_{1}\left(y_{1}\right)=\left\{\begin{array}{ll}  6 y_{1}\left(1-y_{1}\right), & 0 \leq y_{1} \leq 1 \\  0, & \text { elsewhere }  \end{array}\right.\)

The density function for \(Y_{2}\), the proportion of type A crystals among the solid copper, was given as

\(f_{2}\left(y_{2}\right)=\left\{\begin{array}{ll}  3 y_{2}^{2}, & 0 \leq y_{2} \leq 1 \\

0, & \text { elsewhere }  \end{array}\right.\)

The variable \(U=Y_{1} Y_{2}\) gives the proportion of the sample volume due to type A crystals. If

 \(Y_{1} \text { and } Y_{2}\) are independent, find the probability density function for .

Equation Transcription:

 {

 {

 

Text Transcription:

Y_1

f_1(y_1)={ 6y_1(1-y_1)  0 \leq y_1 \leq 1  0,    elsewhere  

Y_2

f_2(y_2)= {3y_2^2,   0 \leq y_2 \leq 1   0,  elsewhere

U=Y_1 Y_2

Y_1 and Y_2

ANSWER:

Solution 14E

Step1 of 2:

Let us consider the random variables denotes two types of copper powder, the density function for is given by:

The density function for , the proportion of type A crystal among the solid copper, was given as:

 

We need to find the probability density function of U.


Step2 of 2:

Let us assume that  are independent so,

 

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