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.

p(x, y)

(a) Compute the covariance for X and Y. (Round your answer to two decimal places.)
Cov(X, Y) =

(b) Compute ? for X and Y. (Round your answer to two decimal places.)
? =

y

p(x, y)

0 5 10 15 x 0 0.01 0.06 0.02 0.10 5 0.04 0.17 0.20 0.10 10 0.01 0.15 0.13 0.01

Explanation / Answer

E(X) = sum(X * P(x) )    = 5.55

E(Y) = sum(Y * P(Y) ) = 8.55

E(XY) = sum (X * Y * P(X,Y)) = 43.75

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

= -3.70

Var(X) = sum(X^2 * P(X)) - E(X)^2 = 42.75 - 5.55^2 = 11.9475

standard deviation(X) = sqrt(11.9475) = 3.4565

Var(Y) = sum (Y^2 * P(Y)) - E(Y)^2 = 91.75 - 8.55^2 = 18.6475

standard deviation (Y) = sqrt(18.6475) = 4.3183

correlation = covariance/std(x) * std(Y)

= -3.70/3.4565 * 4.3183

= -0.2479

0 5 10 15 P(x) x*P(X) X^2*P(X) 0 0.01 0.06 0.02 0.1 0.19 0 0 5 0.04 0.17 0.2 0.1 0.51 2.55 12.75 10 0.01 0.15 0.13 0.01 0.3 3 30 P(Y) 0.06 0.38 0.35 0.21 Y*P(Y) 0 1.9 3.5 3.15 Y^2 * P(Y) 0 9.5 35 47.25

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.