Question

Suppose that two different testsAand B are to be givento a student chosen at random from a certain population.Suppose also that the mean score on test A is 85, and thestandard deviation is 10; the mean score on test B is 90,and the standard deviation is 16; the scores on the two testshave a bivariate normal distribution, and the correlationof the two scores is 0.8. If the student’s score on test A is 80, what is the probability that her score on test B will behigher than 90?

Short answer

The required probability is 0.472336.

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 correlationof the two scores is 0.8.

Denote the random variables

Let X denote the test scores on A, and let Y denotetest scores on B.

Then,

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

\(\)\(\)

Denote the conditional pdf

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

\(\begin{aligned}{}Y|X &= x \sim N\left( {{\mu _y} + p\frac{{{\sigma _y}}}{{{\sigma _x}}}\left( {x - {\mu _x}} \right),{\sigma _y}^2\left( {1 - {p^2}} \right)} \right)\\ &= N\left( {90 + 0.8\frac{{16}}{{10}}\left( {x - 85} \right),\,\,{{16}^2}\left( {1 - {{0.8}^2}} \right)} \right)\\ &= N\left( {90 + 1.28\left( {x - 85} \right),\,\,92.16} \right)\end{aligned}\)

Calculate the probability

It is given that the score on test A 80. Therefore X=80.the above equation reduces to

\(\begin{array}{l} = N\left( {90 + 1.28\left( {x - 85} \right),\,\,92.16} \right)\\ = N\left( {83.6\,\,92.16} \right)\end{array}\)

Therefore, the pdf is,

\(f\left( {y,\mu ,\sigma } \right) = \frac{1}{{\sqrt {2\pi } \times 92.16}}{e^{ - \frac{1}{2}}}{\left( {\frac{{y - 83.6}}{{92.16}}} \right)^2}\)

Therefore, the probability that her score on test B will be higher than 90

\(P\left( {Y > 90|X = 80} \right)\)

\(\int\limits_{90}^\infty {\frac{1}{{\sqrt {2\pi } \times 92.16}}{e^{ - \frac{1}{2}}}{{\left( {\frac{{y - 83.6}}{{92.16}}} \right)}^2}dy} = 0.472336\)

From the normal tables the value is 0.472336.