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Suppose that X and Y are random variables with E(X) = 2, E(Y) = 1, E(X²) = 5, E(Y²) = 10, E(XY) = 1. Problem 10: Find Cor(X, Y) and Var(X + Y). {HINT: The correlation of a bivariate (X, Y) is given by cor(X, Y) = \frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}} Problem 11: Assume that X + Y is normally distributed. Find P(X + Y < 5)

          Suppose that X and Y are random variables with
E(X) = 2, E(Y) = 1, E(X²) = 5, E(Y²) = 10, E(XY) = 1.
Problem 10: Find Cor(X, Y) and Var(X + Y).
{HINT: The correlation of a bivariate (X, Y) is given by cor(X, Y) = \frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}
Problem 11: Assume that X + Y is normally distributed. Find P(X + Y < 5)
        
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Suppose that X and Y are random variables with
E(X) = 2, E(Y) = 1, E(X²) = 5, E(Y²) = 10, E(XY) = 1.
Problem 10: Find Cor(X, Y) and Var(X + Y).
HINT: The correlation of a bivariate (X, Y) is given by cor(X, Y) = (Cov(X,Y))/(√(Var(X)Var(Y)))
Problem 11: Assume that X + Y is normally distributed. Find P(X + Y < 5)

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Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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Suppose that X and Y are random variables with E(X) = 2, E(Y) = 1, E(X^2) = 5, E(Y^2) = 10, E(XY) = 1. Problem 10: Find Cov(X, Y) and Var(X + Y). Cov(X, Y) {HINT: The covariance of a bivariate (X, Y) is given by Cov(X, Y) = E(XY) - E(X)E(Y)} [5] Problem 11: Assume that X + Y is normally distributed. Find P(X + Y < 5) [5]
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Transcript

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00:01 So, from question we have this relationship y is equal to the a plus b x.
00:04 So, now we will solve the a part of the question here variance y is equal to the variance of a plus b x which is equal to the variance of b x.
00:22 So, now we will use the formula that variance of k x is equal to the k square variance x and variance a plus b x is equal to the variance of b x.
00:40 So, when we put when we use this formula we will have variance of y is equal to the b square sigma square x as variance x is equal to the sigma square x.
00:57 So, this is the answer for the a part of the question.
00:59 So, now for the b part for the b part.
01:01 So, here coefficient of variance of x comma y is equal to the e of x comma y minus e of x multiply by e of y.
01:14 So, when we simplify this we have e of x multiply by a plus b x minus e of x x multiply by e of a plus b x.
01:29 So, when we solve this further we have it is equal to the e of a x plus b x square minus e of x multiply by e of a plus b multiply by e of x.
01:48 When we simplify this further we have here a e x plus b e x square minus a e x plus b e of x square whole square...
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