From cb0ac0ae922b4840096bcc5c8fc3d86d841ea74e Mon Sep 17 00:00:00 2001 From: andyeisenberg Date: Thu, 23 Apr 2026 10:40:11 -0400 Subject: [PATCH] Quiz 5 solutions --- source/main.ptx | 1 + source/quizzes/quiz-05.ptx | 204 +++++++++++++++++++++++++++++++++++++ 2 files changed, 205 insertions(+) create mode 100644 source/quizzes/quiz-05.ptx diff --git a/source/main.ptx b/source/main.ptx index 22c12b1..eaef2b8 100644 --- a/source/main.ptx +++ b/source/main.ptx @@ -42,6 +42,7 @@ + diff --git a/source/quizzes/quiz-05.ptx b/source/quizzes/quiz-05.ptx new file mode 100644 index 0000000..7f4f3ee --- /dev/null +++ b/source/quizzes/quiz-05.ptx @@ -0,0 +1,204 @@ + + + + + Quiz 5 + + + +

+ The following work should be completed individually. + Use of notes or textbooks is not allowed. + You may use a scientific calculator, not a graphing calculator or phone app. +

+ +

+ Show all work unless instructed otherwise. +

+ + + + + + + +

+ The joint and marginal distributions of X and Y are given below. + Find \Cov(X, Y) and \rho_{X, Y}. +

+ + + Joint Distribution Table + + + + + Y = 1 + Y = 2 + Y = 3 + + + + X = 0 + 0.1 + 0.15 + 0.1 + + + + X = 1 + 0.15 + 0.2 + 0.3 + + +
+ + + Distribution Table for <m>X</m> + + + + x + 0 + 1 + + + + \Pr(X = x) + 0.35 + 0.65 + + +
+ + + Distribution Table for <m>Y</m> + + + + y + 1 + 2 + 2 + + + + \Pr(Y = y) + 0.25 + 0.35 + 0.4 + + +
+ + + +

+ \Cov(X, Y) = \E(XY) - \E(X)\E(Y). We have: + + \E(X) \amp = 0(0.35) + 1(0.65) = 0.65 + \E(Y) \amp = 1(0.25) + 2(0.35) + 3(0.4) = 2.15 + \E(XY) \amp = (0)(1)(0.1) + (0)(2)(0.15) + (0)(3)(0.1) + + \amp \quad\, (1)(1)(0.15) + (1)(2)(0.2) + (1)(3)(0.3) = 1.45 \\ + \Cov(X, Y) \amp = 1.45 - (0.65)(2.15) = \boxed{0.0525} + + Since X is an indicator random variable, we have \Var(X) = (0.65)(0.35) = 0.2275. + For \Var(Y): + + \E(Y^2) \amp = 1^2(0.25) + 2^2(0.35) + 3^2(0.4) = 5.25 + \Var(Y) \amp = \E(Y^2) - (\E(Y))^2 = 5.25 - (2.15)^2 = 0.6275 + + So the correlation is: + + \rho_{X, Y} \amp = \frac{\Cov(X, Y)}{\sqrt{\Var(X)\Var(Y)}} = \frac{0.0525}{\sqrt{(0.2275)(0.6275)}} \approx \boxed{0.139}. + +

+ + + + + + + + +

+ A sample of 30 measurements are taken and a best fit line is calculated, resulting in the data below (the final row of the table shows the sums for each column). + Find the RSS, SST, and coefficient of determination. +

+ + + Sample Data + + + + x_i + y_i + best fit predicted y_i + res^2 + (y - \textrm{avg }y)^2 + + + + 1.94 + 3.31 + 10.53 + 52.13 + 886.55 + + + + 2.63 + 10.38 + 17.37 + 48.76 + 515.34 + + + + \vdots + \vdots + \vdots + \vdots + \vdots + + + + 3.94 + 19.47 + 30.31 + 117.35 + 185.24 + + + + 6.35 + 53.42 + 54.22 + 0.64 + 413.51 + + + + sum: + 992.55 + 992.55 + 953.03 + 38397.35 + + +
+ + + +

+ The RSS is the sum of res^2, 953.03, and the SST is the sum of (y - \textrm{avg})^2, 38397.35. + Then the coefficient of determination is: + + r^2 = 1 - \frac{\text{RSS}}{\text{SST}} = 1 - \frac{953.03}{38397.35} \approx \boxed{0.975} + +

+ + + + \ No newline at end of file