exercise:6bdf658d45: Difference between revisions
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(Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> Let <math>X</math> be a random variable with distribution function <math>m_X(x)</math> defined by <math display="block"> m_X(-1) = 1/5,\ \ m_X(0) = 1/5,\ \ m_X(1) = 2/5,\ \ m_X(2) = 1/5\ . </math> <ul><li> Let <math>Y</math> be the random varia...") |
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Let <math>X</math> be a random variable with distribution function | |||
<math>m_X(x)</math> defined by | <math>m_X(x)</math> defined by | ||
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m_X(-1) = 1/5,\ \ m_X(0) = 1/5,\ \ m_X(1) = 2/5,\ \ m_X(2) = 1/5\ . | m_X(-1) = 1/5,\ \ m_X(0) = 1/5,\ \ m_X(1) = 2/5,\ \ m_X(2) = 1/5\ . | ||
</math> | </math> | ||
<ul><li> Let <math>Y</math> be the random variable defined by the equation <math>Y = X + 3</math>. Find the | <ul style="list-style-type:lower-alpha"><li> Let <math>Y</math> be the random variable defined by the equation <math>Y = X + 3</math>. Find the | ||
distribution function <math>m_Y(y)</math> of <math>Y</math>. | distribution function <math>m_Y(y)</math> of <math>Y</math>. | ||
</li> | </li> | ||
Latest revision as of 20:59, 12 June 2024
Let [math]X[/math] be a random variable with distribution function [math]m_X(x)[/math] defined by
[[math]]
m_X(-1) = 1/5,\ \ m_X(0) = 1/5,\ \ m_X(1) = 2/5,\ \ m_X(2) = 1/5\ .
[[/math]]
- Let [math]Y[/math] be the random variable defined by the equation [math]Y = X + 3[/math]. Find the distribution function [math]m_Y(y)[/math] of [math]Y[/math].
- Let [math]Z[/math] be the random variable defined by the equation [math]Z = X^2[/math]. Find the distribution function [math]m_Z(z)[/math] of [math]Z[/math].