Question

Suppose that a person’s score X on a mathematics aptitude test is a number between 0 and 1, and that his score Y on a music aptitude test is also a number between 0 and 1. Suppose further that in the population of all college students in the United States, the scores X and Y are distributed according to the following joint pdf:

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

a. What proportion of college students obtain a score greater than 0.8 on the mathematics test?

b. If a student’s score on the music test is 0.3, what is the probability that his score on the mathematics test will be greater than 0.8?

c. If a student’s score on the mathematics test is 0.3, what is the probability that his score on the music test will be greater than 0.8?

Short answer

1. The proportion of college students obtain a score greater than 0.8 on the mathematics test is \(\frac{{37}}{{25}}\)
2. The probability of score on the mathematics test will be greater than 0.8 is 1.28
3. The probability that students score on the music test will be greater than 0.8 is 1.28

Step-by-step solution

Given information

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

compute the probability

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 \(\frac{{37}}{{25}}\)

Calculating the probability of score on maths test.

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.

Calculating probability for the music test

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.