Question

b,c,d please

1. Let the RV X denote your course marks in STAB52 (out of 100). From the syllabus you know that X = .15Xw + 3x +.55XF, where Xw, XM, Xp are the worksheet, midterm, and final marks (all out of 100), respectively. (a) (5 points) Would you prefer the RVs Xw, XM, Xp to be independent, positively corre- lated, or negatively correlated? Justify your answer. (b) (5 points) If E(XW) = 80, E (XM) = 70, EXF|XM) = 15 +.7 XM, find E(X) under the original grading scheme. (c) (5 points) If V(Xw)=9, V(XM) = 25, V(XP) = 20 Cov(Xw, XM) = 5, Cov(Xw,XF) = 5, Cov( XM, XF) = 10 1 , find V(X) under the original grading scheme. (d) (5 points) Assume your course marks have mean E(X) = 65 and variance V(X) = 16. Find an upper bound on the probability that you will get an A i.e. X > 80). Solution: (a) The relationship between evaluations will affect the variance of the course marks, but not their expectation. If the evaluation marks are negatively correlated, the course marks will be less variable, and if they are positively correlated, the course marks will be more variable. If one is happy with their average mark, they would typically prefer lower variance and hence negatively correlated marks. ibe Log in

Answer 1

Soll Given x = 0.15 xw + 0.3%m + 0.55 XF ElXw) = b0, E(Xm) = 70 , E(XFX Xm) = 15 +0.7XM Since E[Exly)] = E(X). ::E [E(XF\ xm)] = € (15 +0.7 xo) 3 Elxr) = 15 +0.7 E(X) [ EaX+h) = Q ECX) +67 15 +0.7x70 =64 - Now Ex)= E(0.15 xw t 0.38m +0.55XF) = 0.15 Elxw) + 0.3 E(XM) +0.55 EXES 0.15x80 + 0.3 x 70 +0.55864 = 68.2 And 2 V (Xw) =9 V(XM) =25, VXF) = 20 Cow (Xw, Xm) = 5 Cow (Xw, XF) = 5, Cow (XM, X() = 10 V [ax+by+CZ) = a Vax(x) + 6* Var (4) + (² Var (2) +zablov (x, y) +2bc Cor (Y, 2) +2ca Cov(x,z1 Vex) = V(0-15 Xw +0.3 xm +0.55XES = 0.15)? V(XW) + (0:3)? V(XM) + (0.55) V(XP) . +240 15x0.3 Cov (Xw, Xm) + 2x0.3x0.55 Cov(XM, XF) +2x0.55x0.15 Cov(Xw, Xf) = 0.0225 x 9 + 0.09x25 +0.3025 x 20 +0.09 x5 + 0.33x10 + 0.165 X5 =0.2025 +2.25 +6.05 + 0.45 +3.3 +0.825 = 13.0775 Ans.