Recall from the previous quiz the random variables X and Y with joint probability distributionY=0 Y = 1X = 0 0.4 0.1X = 1 0.3 0.2The correlation between X and Y is (3 decimal places)You AnsweredCorrect Answers0.218 (with margin: 0.002) Question 50 / 1 ptsIn the previous question, var(X + Y) isYou AnsweredCorrect Answers0.56 (with margin: 0)PreviousNextLast Attempt Details:Time: 12 minutesCurrent Score: 2 out of 5Kept Score: 2 out of 5Unlimited AttemptsTake the Quiz Again(Will keep the highest of all your scores)

To calculate the correlation between X and Y, we first need to calculate the standard deviations of X and Y, and then use the formula for correlation:

Corr(X,Y) = Cov(X,Y) / (std_dev(X) * std_dev(Y))

Step 1: Calculate std_dev(X) and std_dev(Y)

Var(X) = E(X^2) - [E(X)]^2 = [(0^2 * 0.5) + (1^2 * 0.5)] - (0.5)^2 = [0 + 0.5] - 0.25 = 0.5 - 0.25 = 0.25 std_dev(X) = sqrt(Var(X)) = sqrt(0.25) = 0.5

Var(Y) = E(Y^2) - [E(Y)]^2 = [(0^2 * 0.7) + (1^2 * 0.3)] - (0.3)^2 = [0 + 0.3] - 0.09 = 0.3 - 0.09 = 0.21 std_dev(Y) = sqrt(Var(Y)) = sqrt(0.21) = 0.458

Step 2: Substitute Cov(X,Y), std_dev(X), and std_dev(Y) into the formula for Corr(X,Y)

Corr(X,Y) = Cov(X,Y) / (std_dev(X) * std_dev(Y)) = 0.05 / (0.5 * 0.458) = 0.05 / 0.229 = 0.218

So, the correlation between X and Y is 0.218 (rounded to three decimal places).

For the second question, var(X + Y) is calculated as follows:

Var(X + Y) = Var(X) + Var(Y) + 2Cov(X,Y) = 0.25 + 0.21 + 20.05 = 0.56

So, var(X + Y) is 0.56.