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\begin{document}
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QUESTION

In each of the following cases write down the probability
   function for the given variable and calculate its mean and
   variance.

    \begin{description}

     \item[(i)]the score on an unbiased die,

     \item[(ii)]the sum of the scores on two unbiased dice,

     \item[(iii)]the maximum score on one die when two dice are
      tossed,

     \item[(iv)]the score on a die which is biased so that the
     probability that it lands with any face uppermost is
     proportional to the number on that face.
    \end{description}

   In case (ii) comment on your results.

\bigskip

ANSWER

 \begin{description}

    \item[(i)]
$$$$
     \begin{tabular}{ccccccc}
      $x$&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6\\
      $p(x)$&amp;$\frac{1}{6}$&amp;$\frac{1}{6}$&amp;$\frac{1}{6}$&amp;$\frac{1}{6}$&amp;$\frac{1}{6}$&amp;$\frac{1}{6}$
     \end{tabular}\\

     $\mu =3.5$

     \begin{eqnarray*}
      \sigma ^2&amp;=&amp;\frac{1}{6}(1^2+2^2+3^2+4^2+5^2+6^2)-3.5^2\\
      &amp;=&amp;\frac{1}{6}6713-\frac{49}{4}=\frac{35}{12}
     \end{eqnarray*}

\medskip
    \item[(ii)]
$$ $$
     \begin{tabular}{cc|cccccc}
      &amp;&amp;&amp;&amp;first&amp;&amp;&amp;\\
      &amp;sum&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6\\
      \hline
      &amp;1&amp;2&amp;3&amp;4&amp;5&amp;6&amp;7\\
      &amp;2&amp;3&amp;4&amp;5&amp;6&amp;7&amp;8\\
      &amp;3&amp;4&amp;5&amp;6&amp;7&amp;8&amp;9\\
      second&amp;4&amp;5&amp;6&amp;7&amp;8&amp;9&amp;10\\
      &amp;5&amp;6&amp;7&amp;8&amp;9&amp;10&amp;11\\
      &amp;6&amp;7&amp;8&amp;9&amp;10&amp;11&amp;12\\
     \end{tabular}

     Each probability is $\frac{1}{36}$ since the two dice are
     independent.

     \begin{tabular}{ccccccccccccc}
     x&amp;2&amp;3&amp;4&amp;5&amp;6&amp;7&amp;8&amp;9&amp;10&amp;11&amp;12\\
     p(x)&amp;$\frac{1}{36}$&amp;$\frac{2}{36}$&amp;$\frac{3}{36}$&amp;$\frac{4}{36}$
     &amp;$\frac{5}{36}$&amp;$\frac{6}{36}$&amp;$\frac{5}{36}$&amp;$\frac{4}{36}$
     &amp;$\frac{3}{36}$&amp;$\frac{2}{36}$&amp;$\frac{1}{36}$
     \end{tabular}

     $\mu=7$ (calculate or symmetry)

     $\sigma ^2=\frac{1}{36}(2^2+23^2+34^2+\ldots
     +112^2)-49=\frac{1974}{36}-49=\frac{35}{6}$

     $\mu =7=2\times 3.5=2\times \mu \textrm{ from (i)}\\
     \sigma^2=\frac{35}{6}=2 \times \frac{35}{12}=2\times \sigma^2
     \textrm{ from (i)}$

    \item[(iii)]
$$$$
     \begin{tabular}{cc|cccccc}
      &amp;&amp;&amp;&amp;first&amp;&amp;&amp;\\
      &amp;max&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6\\
      \hline
      &amp;1&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6\\
      &amp;2&amp;2&amp;2&amp;3&amp;4&amp;5&amp;6\\
      &amp;3&amp;3&amp;3&amp;3&amp;4&amp;5&amp;6\\
      second&amp;4&amp;4&amp;4&amp;4&amp;4&amp;5&amp;6\\
      &amp;5&amp;5&amp;5&amp;5&amp;5&amp;5&amp;6\\
      &amp;6&amp;6&amp;6&amp;6&amp;6&amp;6&amp;6
     \end{tabular}

     Each probability is $\frac{1}{36}$ since the dice are
     independent.

\medskip

     \begin{tabular}{ccccccc}
      $x$&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6\\
      $p(x)$&amp;$\frac{1}{36}$&amp;$\frac{3}{36}$&amp;$\frac{5}{36}$&amp;$\frac{7}{36}$
      &amp;$\frac{9}{36}$&amp;$\frac{11}{36}$
     \end{tabular}

     $\mu = \frac{1}{36}(1\times 1+2\times 3+3\times 5+4\times 7+5
     \times 9+11 \times 36)=\frac{161}{36}=4.472\\
     \sigma ^2=\frac{1}{36}(1^2\times 1+2^2\times 3+3^2\times 5+4^2\times
     7+5^2
     \times 9+11^2 \times 36)-\mu ^2=1.971$

    \item[(iv)]
$$$$
     \begin{tabular}{ccccccc}
      $x$&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6\\
      $p(x)$&amp;$k$&amp;$2k$&amp;$3k$&amp;$4k$&amp;$5k$&amp;$6k$
     \end{tabular}

     $k+2k+3k+\ldots 6k=21k=1$ therefore $k=\frac{1}{21}$

     \begin{tabular}{ccccccc}
      $x$&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6\\
      $p(x)$&amp;$\frac{1}{21}$&amp;$\frac{2}{21}$&amp;$\frac{3}{21}$&amp;$\frac{4}{21}$&amp;$\frac{5}{21}$&amp;$\frac{6}{21}$
     \end{tabular}

     \begin{eqnarray*}
      \mu&amp;=&amp;\frac{1}{21}(1\times 1+2\times 2+\ldots +6\times 6)\\
      &amp;=&amp;\frac{1}{21} \frac{7 \times 7\times
      13}{6}=\frac{13}{3}=4.333
     \end{eqnarray*}

     \begin{eqnarray*}
      \sigma ^2&amp;=&amp;\frac{1}{21}(1^2\times 1+2^2\times 2+\ldots +6^2\times 6)-(\frac{13}{3})^2\\
      &amp;=&amp;\frac{1}{21} (\frac{7 \times 7\times
      7}{2})^2-\frac{169}{9}=\frac{20}{9}=2.222
     \end{eqnarray*}

   \end{description}

\end{document}
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