An instructor has given a short quiz consisting of two parts. For a randomly sel

Question

An instructor has given a short quiz consisting of two parts. For a randomly selected student, let X = the number of points earned on the first part and Y = the number of points earned on the second part. Suppose that the joint pmf of X and Y is given in the accompanying table. 0.03 0.04 0.01 0.06 0.15 0.15 0.02 0.20 0.13 15 0.10 0.10 0.01 10 (a) Compute the covariance for X and Y. (Round your answer to two decimal places.) Coy(X, Y) = (19025 ) x (b) Compute p for X and Y. (Round your answer to two decimal places.)

Explanation / Answer

(a) Covaiance (X,Y) = E(XY) - E(X)E(Y)

We will calculate now these three values

E(X) = xf(x) = 0 * (0.03 + 0.06 + 0.02 + 0.10) + 5 * (0.04 + 0.15 + 0.20 + 0.10) + 10 * (0.01 + 0.15 +0.13 + 0.01) = 5.45

E(Y) = yf(y) = 0 * (0.03 + 0.04 + 0.01) + 5 * (0.06 + 0.15 + 0.15) + 10 * (0.02 + 0.20 +0.13) + 15 * (0.1 + 0.10 + 0.01) = 8.45

E(XY) = xyf(x,y) = 0 * 0 * 0.03 + 0 * 5 * 0.06 + 0 * 10 * 0.02 + 0 * 15 * 0.10 + 5 * 0 * 0.04 + 5 * 5 * 0.15 + 5 * 10 * 0.20 + 5 * 15 * 0.10 + 0 * 10 * 0.01 + 5 * 10 * 0.15 + 10 * 10 * 0.13 + 15 * 10 * 0.01 = 43.25

Cov(X,Y) = 43.25 - 5.45 * 8.45 = -2.8025

(b) Correlation coefficient

= Cov(x,y) / sqrt(x y  )

x = E(X2) - E(X) 2

E(X2) = 0 * 0.21 + 52 * 0.49 + 102 * 0.30 = 42.25

x  = 42.25 - 5.452 = 12.5475

y = E(Y2) - E(Y) 2

E(Y2 ) = 02 * 0.08 + 52 * 0.36 + 102 * 0.35 + 152 * 0.21 = 91.25

y = 91.25 - 8.452 = 19.8475

= Cov(x,y) / sqrt(x y  ) = -2.8025 / sqrt(19.8475 * 12.5475)

= -2.8025/ 15.78 = -0.1776 or -0.18

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