Fill in the P(x=x) values to give a legitimate probability distribution for th discrete random variable x, whose possible values are o, 2, 3, 5, and 6. P(x = x) Value x of X 0.18 3 0.25 5 0.22 6.
Fill in the P(x=x) values to give a legitimate probability distribution for th discrete random variable x, whose possible values are o, 2, 3, 5, and 6. P(x = x) Value x of X 0.18 3 0.25 5 0.22 6.
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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![**Title: Constructing a Legitimate Probability Distribution for a Discrete Random Variable**
**Introduction:**
Understanding probability distributions is essential in statistics and probability theory. This exercise focuses on filling in the missing probabilities for a discrete random variable, ensuring it forms a legitimate probability distribution.
**Problem Statement:**
Fill in the \( P(X = x) \) values to give a legitimate probability distribution for the discrete random variable \( X \), whose possible values are 0, 2, 3, 5, and 6.
**Table of Values and Probabilities:**
| Value \( x \) of \( X \) | \( P(X = x) \) |
|---------------------|-------------|
| 0 | 0.18 |
| 2 | □ |
| 3 | 0.25 |
| 5 | 0.22 |
| 6 | □ |
**Objective:**
The goal is to determine the missing probabilities □ so that the distribution satisfies the following conditions:
1. **Non-negativity:** Each probability \( P(X = x) \) must be non-negative.
2. **Sum of Probabilities:** The sum of all probabilities must equal 1, ensuring the distribution accounts for all possible outcomes.
**Solution Steps:**
1. **Calculate the Total Known Probability:**
Add the known probabilities:
\[
0.18 + 0.25 + 0.22 = 0.65
\]
2. **Determine the Remaining Probability:**
The sum of the probabilities must equal 1. Therefore, the missing probabilities must sum to:
\[
1 - 0.65 = 0.35
\]
3. **Distribute the Remaining Probability:**
You are given two missing values (for \( X = 2 \) and \( X = 6 \)). An example solution might equally distribute the remaining probability:
\[
P(X = 2) = 0.175 \quad \text{and} \quad P(X = 6) = 0.175
\]
However, different distributions might be valid as long as both conditions are met.
**Conclusion:**
By filling in the probability values, you create a valid probability distribution for the discrete random variable \( X \). This exercise reinforces key concepts about probability distributions, including non-negativity and completeness](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F73803aea-729d-4f3c-b586-f9207ffa1dae%2F11557d34-93e2-4662-b435-c8dce1274a80%2Faasuqi9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Constructing a Legitimate Probability Distribution for a Discrete Random Variable**
**Introduction:**
Understanding probability distributions is essential in statistics and probability theory. This exercise focuses on filling in the missing probabilities for a discrete random variable, ensuring it forms a legitimate probability distribution.
**Problem Statement:**
Fill in the \( P(X = x) \) values to give a legitimate probability distribution for the discrete random variable \( X \), whose possible values are 0, 2, 3, 5, and 6.
**Table of Values and Probabilities:**
| Value \( x \) of \( X \) | \( P(X = x) \) |
|---------------------|-------------|
| 0 | 0.18 |
| 2 | □ |
| 3 | 0.25 |
| 5 | 0.22 |
| 6 | □ |
**Objective:**
The goal is to determine the missing probabilities □ so that the distribution satisfies the following conditions:
1. **Non-negativity:** Each probability \( P(X = x) \) must be non-negative.
2. **Sum of Probabilities:** The sum of all probabilities must equal 1, ensuring the distribution accounts for all possible outcomes.
**Solution Steps:**
1. **Calculate the Total Known Probability:**
Add the known probabilities:
\[
0.18 + 0.25 + 0.22 = 0.65
\]
2. **Determine the Remaining Probability:**
The sum of the probabilities must equal 1. Therefore, the missing probabilities must sum to:
\[
1 - 0.65 = 0.35
\]
3. **Distribute the Remaining Probability:**
You are given two missing values (for \( X = 2 \) and \( X = 6 \)). An example solution might equally distribute the remaining probability:
\[
P(X = 2) = 0.175 \quad \text{and} \quad P(X = 6) = 0.175
\]
However, different distributions might be valid as long as both conditions are met.
**Conclusion:**
By filling in the probability values, you create a valid probability distribution for the discrete random variable \( X \). This exercise reinforces key concepts about probability distributions, including non-negativity and completeness
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