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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}
      $x$&1&2&3&4&5&6\\
      $p(x)$&$\frac{1}{6}$&$\frac{1}{6}$&$\frac{1}{6}$&$\frac{1}{6}$&$\frac{1}{6}$&$\frac{1}{6}$
     \end{tabular}\\

     $\mu =3.5$

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

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

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

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

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

\medskip

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

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

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

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

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

   \end{description}

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