Final answer:
a) The probability P(5 < X < 6 & Y < 6) is 0.171. b) The expected value of X - Y is -1. c) The variance of X - Y is 5. d) The probability P(X < Y + 1) is 0.420. e) When Cov[X,Y] = -1, the probability P(X < Y + 1) remains 0.420.
Step-by-step explanation:
(a) To find P(5 < X < 6 & Y < 6), we need to find the common area under the curves of X and Y.
Since X ~ N(5,1), we can convert the inequalities to standard normal form:
P(0 < Z < 1) = P(Z < 1) - P(Z < 0) = 0.841 - 0.5 = 0.341
Similarly, since Y ~ N(6,4), we can convert the inequality to standard normal form:
P((Y-6)/2 < 0) = P(Z < 0) = 0.500
Therefore, P(5 < X < 6 & Y < 6) = P(0 < Z < 1) * P(Z < 0) = 0.341 * 0.500 = 0.171
(b) The expected value of X - Y is the difference of the expected values: E(X-Y) = E(X) - E(Y) = 5 - 6 = -1
(c) The variance of X - Y is the sum of the variances: Var(X-Y) = Var(X) + Var(Y) = 1 + 4 = 5
(d) To find P(X < Y + 1), we need to convert the inequality to standard normal form:
P((X - (Y+1))/sqrt(Var(X)+Var(Y))) < 0 = P(((X-Y)-1)/sqrt(5)) < 0
Since X and Y are independent, the difference (X-Y) follows a normal distribution with mean (Mean(X) - Mean(Y)) and variance (Var(X) + Var(Y)). Therefore, we can convert the inequality to standard normal form:
P((Z-1)/sqrt(5)) < 0 = P(Z < 1/sqrt(5))
= 0.420
(e) When the covariance between X and Y is -1, it affects the variance of the difference (X-Y). The variance is given by Var(X-Y) = Var(X) + Var(Y) - 2*Cov(X,Y), where Cov(X,Y) is the covariance between X and Y.
Therefore, the variance is 1 + 4 - 2*(-1) = 5. Since the variance remains the same, the probability P(X < Y + 1) will also remain the same: 0.420