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x P(X = x) -2 2 10 0.39 0.69 0.12

x P(X = x) -2 2 10 0.39 0.69 0.12

MATLAB: An Introduction with Applications
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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**Determining Discrete Probability Distribution** In this exercise, we aim to evaluate whether the given distribution qualifies as a discrete probability distribution. The provided data is structured in a table format, showcasing values of a discrete random variable \( X \) alongside their associated probabilities: | \( x \) | -2 | 2 | 10 | |------------|------|------|------| | \( P(X = x) \) | 0.39 | 0.69 | 0.12 | To determine if this is a discrete probability distribution, two fundamental criteria must be met: 1. **Non-Negativity**: The probability of each outcome must be greater than or equal to zero. 2. **Normalization**: The sum of all probabilities must equal 1. Evaluating the table: - **Non-Negativity Check**: \[ \begin{align*} 0.39 \geq 0 \\ 0.69 \geq 0 \\ 0.12 \geq 0 \end{align*} \] All probabilities are non-negative. - **Normalization Check**: \[ 0.39 + 0.69 + 0.12 = 1.20 \] The sum of probabilities is 1.20, which does not equal 1. Since the sum of the probabilities is greater than 1, the given distribution does not meet the requirements for a discrete probability distribution. Therefore, it is not a valid discrete probability distribution.
Transcribed Image Text:**Determining Discrete Probability Distribution** In this exercise, we aim to evaluate whether the given distribution qualifies as a discrete probability distribution. The provided data is structured in a table format, showcasing values of a discrete random variable \( X \) alongside their associated probabilities: | \( x \) | -2 | 2 | 10 | |------------|------|------|------| | \( P(X = x) \) | 0.39 | 0.69 | 0.12 | To determine if this is a discrete probability distribution, two fundamental criteria must be met: 1. **Non-Negativity**: The probability of each outcome must be greater than or equal to zero. 2. **Normalization**: The sum of all probabilities must equal 1. Evaluating the table: - **Non-Negativity Check**: \[ \begin{align*} 0.39 \geq 0 \\ 0.69 \geq 0 \\ 0.12 \geq 0 \end{align*} \] All probabilities are non-negative. - **Normalization Check**: \[ 0.39 + 0.69 + 0.12 = 1.20 \] The sum of probabilities is 1.20, which does not equal 1. Since the sum of the probabilities is greater than 1, the given distribution does not meet the requirements for a discrete probability distribution. Therefore, it is not a valid discrete probability distribution.
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