Value x of X P(x = x) 1 0.10 0.20 0.28 2. 4.
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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![**Title: Understanding Probability Distributions of Discrete Random Variables**
**Objective:** Learn how to assign probabilities to a discrete random variable to form a legitimate probability distribution.
**Task:** Fill in the \( P(X = x) \) values to create a valid probability distribution for the discrete random variable \( X \). The possible values for \( X \) are 1, 2, 4, 5, and 6.
| Value \( x \) of \( X \) | \( P(X = x) \) |
|----------------------|----------------|
| 1 | 0.10 |
| 2 | 0.20 |
| 4 | [ ] |
| 5 | [ ] |
| 6 | 0.28 |
**Description:**
To create a legitimate probability distribution, the sum of all probability values, \( P(X = x) \), must equal 1.
The given probabilities are:
- \( P(X = 1) = 0.10 \)
- \( P(X = 2) = 0.20 \)
- \( P(X = 6) = 0.28 \)
The probabilities for \( X = 4 \) and \( X = 5 \) are currently missing and need to be determined so that the sum of all probabilities equals 1.
**Approach:**
1. Add the provided probabilities:
\[
0.10 + 0.20 + 0.28 = 0.58
\]
2. The remaining probability is:
\[
1 - 0.58 = 0.42
\]
3. Assign the remaining probability of 0.42 to the unknown probabilities for \( X = 4 \) and \( X = 5 \), ensuring the sum across all values totals to 1.
**Conclusion:**
Understanding the principles of probability distributions ensures that all probabilities add up to one, maintaining the integrity of statistical analysis in various applications.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F33c37e08-4ae0-47a0-8176-1da90542d3cc%2Ff54aabfc-7184-47ee-8303-fb26a63c6ac2%2Fj0gg0t_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Understanding Probability Distributions of Discrete Random Variables**
**Objective:** Learn how to assign probabilities to a discrete random variable to form a legitimate probability distribution.
**Task:** Fill in the \( P(X = x) \) values to create a valid probability distribution for the discrete random variable \( X \). The possible values for \( X \) are 1, 2, 4, 5, and 6.
| Value \( x \) of \( X \) | \( P(X = x) \) |
|----------------------|----------------|
| 1 | 0.10 |
| 2 | 0.20 |
| 4 | [ ] |
| 5 | [ ] |
| 6 | 0.28 |
**Description:**
To create a legitimate probability distribution, the sum of all probability values, \( P(X = x) \), must equal 1.
The given probabilities are:
- \( P(X = 1) = 0.10 \)
- \( P(X = 2) = 0.20 \)
- \( P(X = 6) = 0.28 \)
The probabilities for \( X = 4 \) and \( X = 5 \) are currently missing and need to be determined so that the sum of all probabilities equals 1.
**Approach:**
1. Add the provided probabilities:
\[
0.10 + 0.20 + 0.28 = 0.58
\]
2. The remaining probability is:
\[
1 - 0.58 = 0.42
\]
3. Assign the remaining probability of 0.42 to the unknown probabilities for \( X = 4 \) and \( X = 5 \), ensuring the sum across all values totals to 1.
**Conclusion:**
Understanding the principles of probability distributions ensures that all probabilities add up to one, maintaining the integrity of statistical analysis in various applications.
Expert Solution
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Our aim is to find the probability of 4 and 5.
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