91Ó°ÊÓ

Two different testsAandBare to be given to a student chosen at random from a certain population. Suppose also that the mean score on testAis 85, and the standard deviation is 10; the mean score on testBis 90,and the standard deviation is 16; the scores on the two tests have a bivariate normal distribution, and the correlation of the two scores is 0.8.

Let A denote the first test scores, and let B denote second test scores.

Then,

\(\begin{array}{*{20}{l}}{{\mu _A} = 85\;}\\{{\sigma _A} = 10}\\{{\mu _B} = 90}\\{{\sigma _B} = 16}\\{p = 0.8}\end{array}\)

\(\)\(\)

For a \(BVN{\rm{ }}\left( {85,10,90,16,0.8} \right)\),

The probability that the sum of her scores on the two tests will be greater than 200 is

\(P\left( {A + B > 200} \right)\).

Let,\(C = A + B\)be a new variable.

The expectation of C is:

\(\begin{aligned}{}E\left( C \right) &= E\left( A \right) + E\left( B \right)\\ &= {\mu _A} + {\mu _B}\\ &= 85 + 90\\ &= 175\end{aligned}\)

The standard deviation of C is:

\(\begin{aligned}{}\sqrt {Var\left( C \right)} &= \sqrt {Var\left( A \right) + Var\left( B \right) + 2 \times p \times sd\left( A \right) \times sd\left( B \right)} \\ &= \sqrt {{\sigma _A}^2 + {\sigma _B}^2 + 2 \times p \times {\sigma _A} \times {\sigma _B}} \\ &= \sqrt {100 + 256 + 2 \times 0.8 \times 10 \times 16} \\& = \sqrt {612} \\ &= 24.73\end{aligned}\)

Therefore, C follows normal distribution, that is, \(C \sim N\left( {175,24.73} \right)\)

\(\begin{aligned}{}P\left( {C > 200} \right) &= P\left( {Z > \frac{{200 - 175}}{{24.73}}} \right)\\ &= P\left( {Z > 1.01} \right)\\ &= 0.1562\end{aligned}\)

Therefore the answer is 0.1562.