Homework SourseAn instructor has given a short quiz consisting of two parts
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 |
Solution
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 | |||
