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2. A random variable X = {0, 1, 2, 3} has probability mass function: 1 1 2 Px(0) = Px(1) = Px(2) = (a) Compute E(X). (b) Compute (X). (c) Compute E(X³). Px(3) = 1 3

2. A random variable X = {0, 1, 2, 3} has probability mass function: 1 1 2 Px(0) = Px(1) = Px(2) = (a) Compute E(X). (b) Compute (X). (c) Compute E(X³). Px(3) = 1 3

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Please helpt to solve this. Can you handwrite the solution please? It helps to follow the steps properly. Thank you! 

**Problem 2: Discrete Random Variable Analysis** A random variable \(X\) that takes values in the set \(\{0, 1, 2, 3\}\) is described by the following probability mass function: - \(P_X(0) = \frac{1}{3}\) - \(P_X(1) = \frac{1}{9}\) - \(P_X(2) = \frac{2}{9}\) - \(P_X(3) = \frac{1}{3}\) Tasks: (a) Compute the expected value \(E(X)\). (b) Compute the standard deviation \(\sigma(X)\). (c) Compute the expected value of the cube of \(X\), denoted as \(E(X^3)\).
Transcribed Image Text:**Problem 2: Discrete Random Variable Analysis** A random variable \(X\) that takes values in the set \(\{0, 1, 2, 3\}\) is described by the following probability mass function: - \(P_X(0) = \frac{1}{3}\) - \(P_X(1) = \frac{1}{9}\) - \(P_X(2) = \frac{2}{9}\) - \(P_X(3) = \frac{1}{3}\) Tasks: (a) Compute the expected value \(E(X)\). (b) Compute the standard deviation \(\sigma(X)\). (c) Compute the expected value of the cube of \(X\), denoted as \(E(X^3)\).
Expert Solution
Step 1: Given

X~{0,1,2,3}

P left parenthesis X equals 0 right parenthesis equals 1 third P left parenthesis X equals 1 right parenthesis equals 1 over 9 P left parenthesis X equals 2 right parenthesis equals 2 over 9 P left parenthesis X equals 3 right parenthesis equals 1 third

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