Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 5: Q16E (page 212)
a. For \({\rm{f}}\)(\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}\)) as given in Example \({\rm{5}}{\rm{.10}}\), compute the joint marginal density function of \({{\rm{X}}_{\rm{1}}}{\rm{and}}{{\rm{X}}_{\rm{3}}}\)alone (by integrating over x2).
b. What is the probability that rocks of types \({\rm{1 and 3}}\)together make up at most \({\rm{50\% }}\)of the sample? (Hint: Use the result of part (a).)
c. Compute the marginal pdf of \({{\rm{X}}_{\rm{1}}}\) alone. (Hint: Use the result of part (a).)
Short Answer
\(\text{f}\left({{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{3}}}\right)\text{=}\left\{\begin{matrix}\text{72}\!\!\times\!\!\text{}{{\text{x}}_{\text{1}}}\left(\text{1}{{\text{x}}_{\text{3}}}\right){{\left(\text{1}{{\text{x}}_{\text{1}}}\text{}{{\text{x}}_{\text{3}}}\right)}^{\text{2}}}\text{,}{{\text{x}}_{\text{1}}}\text{}\!\!{}^\text{3}\!\!\text{0,}{{\text{x}}_{\text{3}}}\text{}\!\!{}^\text{3}\!\!\text{0,}{{\text{x}}_{\text{1}}}\text{+}{{\text{x}}_{\text{3}}}\text{拢1}\\\text{,}\!\!~\!\!\text{otherwise}\!\!~\!\!\text{}\\\end{matrix} \right.\)..
2.\({\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{3}}}{\rm{拢0}} {\rm{.5}}} \right){\rm{ = 0}}{\rm{.531;}}\)
3. \({\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}} \right){\rm{ = }}\left\{ {\begin{aligned}{*{20}{c}}{{\rm{18}}{{\rm{x}}_{\rm{1}}}{\rm{ - 48x}}_{\rm{1}}^{\rm{2}}{\rm{ + 36x}}_{\rm{1}}^{\rm{3}}{\rm{ - 6x}}_{\rm{1}}^{\rm{5}}{\rm{,0拢}}{ {\rm{x}}_{\rm{1}}}{\rm{拢1}}}\\{{\rm{,\;otherwise\;}}}\end{aligned}} \right.\)
Step by step solution
Definition of probability
the proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.
Calculating the joint marginal density function of \({{\rm{X}}_{\rm{1}}}{\rm{and}}{{\rm{X}}_{\rm{3}}}\)alone
We have pdf from the example, thus the combined marginal density function of \({{\rm{X}}_{\rm{1}}}{\rm{and}}{{\rm{X}}_{\rm{3}}}\)is
\({{\rm{X}}_{\rm{1}}}{\rm{and}}{{\rm{X}}_{\rm{3}}}\)have a joint marginal density function.
Calculating the probability that rocks of types \({{\rm{X}}_{\rm{1}}}{\rm{and}}{{\rm{X}}_{\rm{3}}}\)together make up at most \({\rm{50\% }}\)of the sample
Using marginal pdf obtained in, we need to find the probability of an occurrence where the sum of the random variables $X 1$ and $X 2$ is less than or equal to\(0.5\) (a). Therefore
(1): all integrals are straightforward. We will not go over all of the algebra (there is a lot of algebra) that goes into calculations.
Calculating the marginal pdf of \({{\rm{X}}_{\rm{1}}}\) alone.
For the continuous random variable \({\rm{X}}\), the marginal probability density function is
\({{\rm{f}}_{\rm{X}}}{\rm{(x) = \`o }}_{{\rm{ - 楼}}}^{\rm{楼}}{\rm{nf(x,y)dy,}}\;\;\;{\rm{\;for\; - 楼 < x < 楼}}\)
The marginal pdf obtained in (a) will be used. The following statement is correct:
To conclude, the \({{\rm{X}}_{\rm{1}}}\)marginal pdf is
\({\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}} \right){\rm{ = }}\left\{ {\begin{aligned}{*{20}{c}}{{\rm{18}}{{\rm{x}}_{\rm{1}}}{\rm{ - 48x}}_{\rm{1}}^{\rm{2}}{\rm{ + 36x}}_{\rm{1}}^{\rm{3}}{\rm{ - 6x}}_{\rm{1}}^{\rm{5}}{\rm{,0拢}}{{\rm{x}}_{\rm{1}}}{\rm{拢1}}}\\{{\rm{ ,\;otherwise\;}}}\end{aligned}} \right.\)
Over 30 million students worldwide already upgrade their learning with 91影视!
