Question

tion: Assuming k in a constant, deternine the covariance of X and Y 1) -0933 (22) Ass ume that losses from policyholders A, B, and C are i nentially distributed. The expected loss amount for policyholders A, B and IN 20o, 450, and 700, respectively, what is the probability that the,naximum loss from the three policyholders will ex A) 0.001 B) 0.381 C) 0.450 D 0.619 E) 1.000 (23) Let X and Y represent the number of hours that a person spends cooking and doing the dishes, respectively, during a 30-day period. Assume that during a 30-day period, the expected number of hours that a person spends in cooking is 20, with a variance of 40. Also assume that during a 30-day period, the expected number of hours that a a variance of 25. selected at random and observed person spends doing the dishes is 10, with 100 people are for a 30-day period, what is the probability Suppose the covariance of X and Y is 10. IF that the total number of hours spent by the group doing the dishes and cooking 900? A) 0.140 B) 0.505 C) 0.860 D) p.869 E0.875
I know these three questions are some how challenging because this the fourth time I am posting them, the circled answers are wrong, show steps to arrive to correct answers and will definitely rate more points.

Answer 1

21)

Covariance = E(XY) - E(X)E(Y)

here pdf is symmetric for x and y and so is support

first we calculate value of k

$\int _0^2\int _0^2k\left(x+y\right)dxdy=8k = 1$

k = 1/8

$E(XY) = \int _0^2\int _0^2\frac{xy}{8}\left(x+y\right)dxdy=\frac{4}{3}\quad \left(\mathrm{Decimal:\quad }\:1.33333\dots \right)$

$E(X) =\int _0^2\int _0^2\frac{x}{8}\left(x+y\right)dxdy=\frac{7}{6}\quad \left(\mathrm{Decimal:\quad }\:1.16667\dots \right)$

E(Y) = E(X) = 7/6

hence

Please post rest questions again

You are allowed to post one question at a time

Covariance = 4/3 * (7/6)^2 = 1.814