Suppose Verbal GRE scores are also distributed normally with a mean of 150 and a

Question

Suppose Verbal GRE scores are also distributed normally with a mean of 150 and a standard deviation of 10 (this is idential to the mean and standard distrubution of Quantitative scores). Suppose that the correlation between an individual’s math and verbal scores is 0.80, and that scores are “jointly” normally distributed.

(i) Calculate the covariance between Quantitative and Verbal scores.

(ii) Find the mean and standard deviation of the sum of the Verbal and Quantitative scores.

(iii) What is the probability that the sum of the two scores is above 310?

(iv) A student tells you that her Quantitative GRE score is 160. What is the conditional mean and variance of her Verbal score. What is the probability that her Verbal score is above 155?

Explanation / Answer

Here Let say X is the Quantitative scores of a random person where population mean x = 150 and x = 10

similarly, Y is the Verbal GRE score of a random person where population   mean  y = 150 and y = 10

Correlation = 0.80

(a) COV(X,Y) = Correal (X,Y) * x * y = 0.80 * 10 * 10 = 80

(b) Here if Z = X + Y

the mean of Z E(Z) = E(X +Y) = 150 + 150 = 300

Var(Z) = Var(X) + Var(Y) + 2 * COV(X,Y) = 10 * 10 + 10 * 10 + 2 * 80 = 360

Std (Z) = sqrt(360) = 18.974

(c) Pr(Z > 310) = Pr( Z > 310 ; 300 ; 18.974) = 1 - NORMAL(Z < 310 ; 300 ; 18.974)

z = (310 - 300)/18.974 = 0.527

Pr(Z > 310) = Pr( Z > 310 ; 300 ; 18.974) = 1 - NORMAL(Z < 310 ; 300 ; 18.974) = 1 - Pr(z < 0.527) = 1 - 0.700 = 0.30

(iv) Here X = 160

Here E(Y l X = 160) = y + r y/x (x -x) = 150 + 0.8 * 10/10 * (160 - 150) = 158

Var(Y l X = 160) = y2 (1 - r2) = 102 (1 - 0.82) = 100 * 0.36 = 36

STD(Y l X = 160) = 6

Pr(Y > 155 l X) = 1 - Pr(Y < 155 l X)

z = (155 - 158)/6 = -0.5

Pr(Y > 155 l X) = 1 - Pr(Y < 155 l X) = 1- Pr(z < -0.5) = 1 - 0.3085 = 0.6915

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