The following is a joint probability mass function.x y fxr(x,y)1 1 1/41.5 2 1/81.5 3 1/42.5 4 1/43 5 1/8Determine the mean of X and the mean of Y.Round your answers to three decimal places (e.g. 98.765).Statistical Tables and CharPart 1E(X) =

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Answered: Determine the mean of X and the mean of…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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a. What value of c will make f(x) a valid probability mass function ?b. Compute P (1 < X < 6).
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The General Social Survey asked a sample of adults how many siblings (brothers and sisters) they had (X) and also how many children they had (Y). We show results for those who had no more than 4 children and no more than 4 siblings. Assume that the joint probability mass function is given in the following contingency table: y x 0 1 2 3 4 0 0.03 0.01 0.02 0.01 0.01 1 0.09 0.05 0.08 0.03 0.01 2 0.09 0.05 0.07 0.04 0.02 3 0.06 0.04 0.07 0.04 0.02 4 0.04 0.03 0.04 0.03 0.02 Find ρ(X, Y). (Round the final answer to four decimal places.) ρ(X, Y) =
The General Social Survey asked a sample of adults how many siblings (brothers and sisters) they had (X) and also how many children they had (Y). We show results for those who had no more than 4 children and no more than 4 siblings. Assume that the joint probability mass function is given in the following contingency table: y x 0 1 2 3 4 0 0.03 0.01 0.02 0.01 0.01 1 0.09 0.05 0.08 0.03 0.01 2 0.09 0.05 0.07 0.03 0.02 3 0.06 0.04 0.07 0.04 0.02 4 0.04 0.04 0.04 0.03 0.02 Find the conditional expectation E(Y|X = 4). (Round the final answer to four decimal places.)
The General Social Survey asked a sample of adults how many siblings (brothers and sisters) they had (X) and also how many children they had (Y). We show results for those who had no more than 4 children and no more than 4 siblings. Assume that the joint probability mass function is given in the following contingency table: y x 0 1 2 3 4 0 0.03 0.01 0.02 0.01 0.01 1 0.09 0.05 0.08 0.03 0.01 2 0.09 0.05 0.07 0.03 0.02 3 0.06 0.04 0.07 0.04 0.02 4 0.04 0.04 0.04 0.03 0.02 Find the conditional probability mass function pY|X(y|4). (Round the final answer to four decimal places.) pY|X(0|4) = pY|X(1|4) = pY|X(2|4) = pY|X(3|4) = pY|X(4|4) = Find the conditional probability mass function pX|Y (x|3). (Round the final answer to four decimal places.) pX|Y(0|3) = pX|Y(1|3) = pX|Y(2|3) = pX|Y(3|3) = pX|Y(4|3) = Find the conditional expectation E(X|Y = 3). (Round the final answer to four decimal places.) E(X|Y = 3) =
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Suppose that that joint probability mass function of X and Y is given in the following table. p(x,y) y 0 1 0 0.09 0.03 x 1 ? 0.17 2 0.13 0.09 Find the expected value of X.(Hint: First, find p(1,0). Then, you might want to find the (marginal) pdf of X).(Answer as a decimal number, and round to 2 decimal places).
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Determine the mean of X and the mean of Y. Round your answers to three decimal places (e.g. 98.765).