Problem 10 pointsJ Suppose we survey students at a college regarding the number

Question

Problem 10 pointsJ Suppose we survey students at a college regarding the number of hours per day spent reading ebooks and (paper) books. The following table gives the joint distribution of random variables X and Y, where X is the average number of ebook reading hours and Y is the average number of book reading hours Pr Y (x, y) 0 0 0.05 0.10 1 0.21 0.22 0.05 0.24 0.13 (a) Find the marginal distribution ofX and provide it below: Px(x) (b) Find the marginal distribution of Y and provide it below: (c) Let R X Y be the total number of hours spent reading per day. Find the distribution of R PR (r)

Explanation / Answer

Solution

First Question

Part (a)

Marginal Probability of X

X takes values 1, 2, 3

P(X = 1) = P(X = 1, Y = 0) + P(X = 1, Y = 1) + P(X = 1, Y = 2) = 0.45

P(X = 2) = P(X = 2, Y = 0) + P(X = 2, Y = 1) + P(X = 2, Y = 2) = 0.40

P(X = 3) = P(X = 3, Y = 0) + P(X = 3, Y = 1) + P(X = 3, Y = 2) = 0.15

Thus, Marginal Probabilty of X is:

x 1 2 3 p(x) 0.45 0.40 0.15

Part (b)

Marginal Probability of y

Y takes values 9, 1, 2

P(Y = 0) = P(X = 1, Y = 0) + P(X = 2, Y = 0) + P(X = 3, Y = 0) = 0.15

P(Y = 1) = P(X = 1, Y = 1) + P(X = 2, Y = 1) + P(X = 3, Y = 1) = 0.48

P(X = 2) = P(X = 1, Y = 2) + P(X = 2, Y = 2) + P(X = 3, Y = 2) = 0.37

Thus, Marginal Probabilty of Y is:

y 0 1 2       p(y) 0.15 0.48 0.37

Part (c)

R = X + Y = 1, 2, 3, 4, 5

P(R = 1) = P(X = 1,Y = 0) = 0

similarly, P(R = 2) = 0.26, P(R = 3) = 0.56, P(R = 4) = 0.18, P(R = 5) = 0

Second Question

Expected value = E(X) = sum{x.p(x)} = 0.2

Variance of X = V(X) = E(X2) - {E(X)}2 where E(X2) = sum{x2.p(x)} = 1.62

So, V(X) = 1.58

Standard Deviation of X = sq.rt of V(X) = 1.257

Done      

  

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