Let Y1 and Y2 denote the proportions of two different

Chapter 5, Problem 12E

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QUESTION:

Let \(Y_{1} \text { and } Y_{2}\)  denote the proportions of two different types of components in a sample from a mixture of chemicals used as an insecticide. Suppose that \(Y_{1} \text { and } Y_{2}\) have the joint density function given by

                          \(f\left(y_{1}, y_{2}\right)=\left\{\begin{array}{ll}

2, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1,0 \leq y_{1}+y_{2} \leq 1, \\

0, & \text { elsewhere. }

\end{array}\right.

\)

(Notice that \(Y_{1}+Y_{2} \leq 1\) because the random variables denote proportions within the same sample.) Find

a \(\mathrm{P}\left(Y_1\le3/4,\ Y_2\le3/4\right)\).

b \(\mathrm{P}\left(Y_1\le1/2,\ Y_2\le1/2\right)\).

Equation Transcription:

Text Transcription:

Y_1 and Y_2

Y_1 and Y_2

f(y_1,y_2)=

2, 0</=y_11,0</=y_2</=1,0</=y_1+y_2</=1,

0, elsewhere.

Y_1+Y_2</=1

P(Y_1</=3/4,Y_2</=3/4)

P(Y_1</=1/2,Y_2</=1/2)

Questions & Answers

QUESTION:

Let \(Y_{1} \text { and } Y_{2}\)  denote the proportions of two different types of components in a sample from a mixture of chemicals used as an insecticide. Suppose that \(Y_{1} \text { and } Y_{2}\) have the joint density function given by

                          \(f\left(y_{1}, y_{2}\right)=\left\{\begin{array}{ll}

2, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1,0 \leq y_{1}+y_{2} \leq 1, \\

0, & \text { elsewhere. }

\end{array}\right.

\)

(Notice that \(Y_{1}+Y_{2} \leq 1\) because the random variables denote proportions within the same sample.) Find

a \(\mathrm{P}\left(Y_1\le3/4,\ Y_2\le3/4\right)\).

b \(\mathrm{P}\left(Y_1\le1/2,\ Y_2\le1/2\right)\).

Equation Transcription:

Text Transcription:

Y_1 and Y_2

Y_1 and Y_2

f(y_1,y_2)=

2, 0</=y_11,0</=y_2</=1,0</=y_1+y_2</=1,

0, elsewhere.

Y_1+Y_2</=1

P(Y_1</=3/4,Y_2</=3/4)

P(Y_1</=1/2,Y_2</=1/2)

ANSWER:

Solution

Step 1 of 2

a) We have to find

Given that

                                   =0,                Otherwise

The joint distribution function is

                 

Then

                                                   =

                                                   =2(3/4 )(1/4)+(1/2)

                                                   =(3/8)+(1/2)

                                                   =7/8

Hence =7/8


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