Question

Consider again the two tests A and B described in Exercise2. If a student is chosen at random, what is the probability that the sum of her scores on the two tests will be greater than 200?

Short answer

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

0.1562.

Step-by-step solution

Given information

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.

Denote the random variables

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

Then,

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

\(\)\(\)

Define a new variable

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, \(C \sim N\left( {175,24.73} \right)\)

Calculate the probability

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

Therefore the answer is 0.1562.