3. Bivariate distributions Aa Aa A qualifying exam for a graduate school program

Question

3. Bivariate distributions Aa Aa A qualifying exam for a graduate school program has a math section and a verbal section. Students receive a score of 1, 2, or 3 on each section. Define X as a student's score on the math section and Y as a student's score on the verbal section. Test scores vary according to the following bivariate probability distribution 2 1 0.05 0.22 0.02 x 20.10 0.10 0.30 3 0.10 0.06 0.05 Ex = , and Pv = The covariance of X and Y is . The coefficient of correlation is . The variables X and Y independent. The expected value of X + Y is , and the variance of X + Y is To be accepted to a particular graduate school program, a student must have a combined score of 4 on the qualifying exam What is the probability that a randomly selected exam taker qualifies for the program? O 0.60 O 0.56 O 0.61 O 0.63 Chebysheffs Theorem states that the proportion of observations in any population that lie within k standard deviations of the mean is at least 1 1/ k2 (for k> 1) According to Chebysheff's Theorem, there is at least a 0.75 probability that a randomly selected exam taker has a combined score between and

Explanation / Answer

fX(X) = 0.29 ; x = 1

= 0.50 ; x = 2

= 0.21 ; x = 3

fy(Y) = 0.25; Y = 1

= 0.38 ; Y = 2

= 0.37; Y = 3

x = 1 * 0.29 + 2 * 0.50 + 3 * 0.21 = 1.92

y = 1 * 0.25 + 2 * 0.38 + 3 * 0.37 = 2.12

x = sqrt [(1 - 1.92)2 * 0.29 + (2 - 1.92)2 * 0.50 + (3 - 1.92)2 * 0.21] = 0.7026

x = sqrt [(1 - 2.12)2 * 0.25 + (2 - 2.12)2 * 0.38 + (3 - 2.12)2 * 0.37] = 0.7782

Cov (X,Y) = E(XY) - E(X)E(Y)

E(XY) = 1 * 1 * 0.05 + 1 * 2 * 0.22 + 1 * 3 * 0.02 + 2 * 1 * 0.10 + 2 * 2 * 0.10 + 2 * 3 * 0.30 + 3 * 1 * 0.10 + 3 * 2 * 0.06 + 3 * 3 * 0.05 = 4.06

Cov(X,Y) = 4.06 - 1.92 * 2.12 = -0.01

Corr (X,Y) = Cov(X,Y) / x y = -0.01/ (0.7026 * 0.7782) = - 0.0182

so the events are not independet.

E(X + Y) = E(X) + E(Y) = 1.92 + 2.12 = 4.04

Var(X +Y) = Var(X)+Var(Y ) + 2 Cov(X, Y ) = 0.7026 + 0.7782 + 2 * (-0.01) = 1.46

Pr( X +Y >=3 4) = 1- Pr(1,1) - Pr(1,2) - Pr(2,1) = 1 - (0.05 + 0.10 + 0.22) = 0.63

So as per given mean and standard deviation

there is at least 0.75 proabability so it would be +- 2

which are = 4.04 +- 2 * 1.46 = (1.12, 6.96)

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