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QUESTION Given the continuous pdf $f(x)=\frac{2}{x^2},\ 1 \leq x
\leq 2$,
  determine the mean and variance of x and find the probability
  that x exceeds 1.5. Calculate also the median and the quartile
  range for x and state the interquartile range ($Q_3-Q_1$).


ANSWER
 $f(x)=\frac{2}{x^2}\ 1 \leq x \leq 2$

  \begin{eqnarray*}
   \mu&=&\int_1^2 x\frac{2}{x^2}\,dx
   =\int_1^2 \frac{2}{x}\,dx\\
   &=&[2\ln x]_1^2=2 \ln 2
  \end{eqnarray*}

  \begin{eqnarray*}
   E(X^2)&=&\int_1^2 x^2\frac{2}{x^2}\,dx
   =\int_1^2 2\,dx\\
   &=&[2x]_1^2= 2\\
   \sigma^2&=&2-(2\ln 2)^2
  \end{eqnarray*}

  $F(x)=\int_1^x \frac{2}{u^2}\,du=[-\frac{2}{u}]_1^x=2-\frac{2}{x}\\
  P(X \leq
  1.5)=1-F(1,5)=1-(2-\frac{2}{1.5})=1-\frac{2}{3}=\frac{1}{3}$\\
  Median M:\ $ F(M)=2-\frac{2}{m}=\frac{1}{2}$ therefore
  $\frac{2}{m}=\frac{3}{2},\ M=\frac{4}{3}$\\
  Quartile $Q_1:\ F(Q_!)=2-\frac{2}{Q_1}=\frac{1}{4}$ therefore
  $\frac{2}{Q_1}=\frac{7}{2},\ Q_1=\frac{8}{7}$\\
  Quartile $Q_3:\ F(Q_3)=2-\frac{2}{Q_3}=\frac{3}{4}$
  therefore$\frac{2}{Q_3}=\frac{5}{4},\ Q_3=\frac{8}{5}$\\
  Interquartile range
  $=Q_3-Q_1=\frac{8}{5}-\frac{8}{7}=\frac{16}{35}$

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