x P(X = x) -2 2 10 0.39 0.69 0.12
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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Question
![**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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d050742-a776-4688-997d-9b380febe959%2F3a5ab88f-7be1-47d6-94c2-c4dd36767d34%2Fozf0puq_processed.png&w=3840&q=75)
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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