91Ó°ÊÓ

\(f\left( {x,y} \right)\left\{ \begin{aligned}\frac{2}{5}\left( {2x + 3y} \right)for0 \le x \le 1 and 0 \le y \le 1\\0 otherwise\end{aligned} \right.\)

a)

We have to find the continues marginal density for s, as we are testing for any value of y

\(\begin{aligned}{f_x}\left( x \right) = \int\limits_{ - \infty }^\infty {f\left( {x,y} \right)} dy\\ = \int\limits_0^1 {\frac{2}{5}\left( {2x + 3y} \right)dy} \\ = \frac{4}{5}x + \frac{3}{5}\\{f_x}\left( {0.8} \right) = \frac{{37}}{{25}}\end{aligned}\)

Hence, we can say that proportion of college students obtain a score greater than 0.8 on the mathematics test is \(\frac{{37}}{{25}}\)

b)

The score in music = 0.3

\(\begin{aligned}{\rm P}\left( {{\rm A}|{\rm B}} \right) = \frac{{{\rm P}\left( {{\rm A} \cap {\rm B}} \right)}}{{{\rm P}\left( {\rm B} \right)}}\\ = {\rm P}\left( {X > 0.8|Y = 0.3} \right)\\ = \frac{{{\rm P}\left\{ {\left( {X > 0.8} \right) \cap \left( {Y = 0.3} \right)} \right\}}}{{{\rm P}\left( {Y = 0.3} \right)}}\end{aligned}\)

\(\begin{aligned}{\rm P}\left( {{\rm A}|{\rm B}} \right) = \frac{{\int\limits_{0.8}^1 {\frac{2}{5}\left( {2x + 3\left( 1 \right)} \right)dx} }}{{0.3}}\\\frac{{0.384}}{{0.3}}\\ = 1.28\end{aligned}\)

Therefore, the probability of score on the mathematics test is. 1.28. This value is greater than 0.8.

c)

\(\begin{aligned}{\rm P}\left( {{\rm A}|{\rm B}} \right) = \frac{{{\rm P}\left( {{\rm A} \cap {\rm B}} \right)}}{{{\rm P}\left( {\rm B} \right)}}\\ = {\rm P}\left( {X > 0.8|Y = 0.3} \right)\\ = \frac{{{\rm P}\left( {\left( {X > 0.8} \right) \cap \left( {Y = 0.3} \right)} \right)}}{{{\rm P}\left( {Y = 0.3} \right)}}\end{aligned}\)

\(\begin{aligned}{\rm P}\left( {{\rm A}|{\rm B}} \right) = \frac{{\int\limits_{0.8}^1 {\frac{2}{5}\left( {2x + 3\left( 1 \right)} \right)dx} }}{{0.3}}\\\frac{{0.384}}{{0.3}}\\ = 1.28\end{aligned}\)

Therefore, the probability of score on the music test is. 1.28. This value is greater than 0.8.