exercise:88d9acf85f: 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> A number is chosen at random from the integers 1, 2, 3, \dots, <math>n</math>. Let <math>X</math> be the number chosen. Show that <math>E(X) = (n + 1)/2</math> and <math>V(X) = (n - 1)(n + 1)/12</math>. '' Hint'': The following identity may b...")
 
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<div class="d-none"><math>
A number is chosen at random from the integers <math>1, 2, 3,\ldots, n</math>.  Let
\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> A number is chosen at random from the integers 1, 2, 3,
\dots, <math>n</math>.  Let
<math>X</math> be the number chosen.  Show that <math>E(X) = (n + 1)/2</math> and <math>V(X) = (n - 1)(n +
<math>X</math> be the number chosen.  Show that <math>E(X) = (n + 1)/2</math> and <math>V(X) = (n - 1)(n +
1)/12</math>.   '' Hint'':  The following identity may be useful:
1)/12</math>. '' Hint'':  The following identity may be useful:


<math display="block">
<math display="block">
1^2 + 2^2 + \cdots + n^2 = \frac{(n)(n+1)(2n+1)}{6}\ .
1^2 + 2^2 + \cdots + n^2 = \frac{(n)(n+1)(2n+1)}{6}\ .
</math>
</math>

Latest revision as of 22:03, 14 June 2024

A number is chosen at random from the integers [math]1, 2, 3,\ldots, n[/math]. Let [math]X[/math] be the number chosen. Show that [math]E(X) = (n + 1)/2[/math] and [math]V(X) = (n - 1)(n + 1)/12[/math]. Hint: The following identity may be useful:

[[math]] 1^2 + 2^2 + \cdots + n^2 = \frac{(n)(n+1)(2n+1)}{6}\ . [[/math]]