Fill in the P ((X=x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -3, –2, -1, 2, and 4. Value x of X P(X = x) -3 -2 0.23 -1 0.16 4 0.13
Fill in the P ((X=x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -3, –2, -1, 2, and 4. Value x of X P(X = x) -3 -2 0.23 -1 0.16 4 0.13
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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![Fill in the \( P(X = x) \) values to give a legitimate probability distribution for the discrete random variable \( X \), whose possible values are \( -3, -2, -1, 2, \) and \( 4 \).
| Value \( x \) of \( X \) | \( P(X = x) \) |
|----------------|---------------|
| -3 | [ ] |
| -2 | 0.23 |
| -1 | 0.16 |
| 2 | [ ] |
| 4 | 0.13 |
**Description:**
This table provides a set of values for a discrete random variable \( X \) and requires completion to form a valid probability distribution.
- Each \( P(X = x) \) value must be filled in such that the sum of all probabilities equals 1.
- The probabilities \( P(X = -2) \), \( P(X = -1) \), and \( P(X = 4) \) are given as 0.23, 0.16, and 0.13, respectively.
- You need to calculate and fill in the missing probabilities for \( x = -3 \) and \( x = 2 \).
**Instructions:**
1. Add together the given probabilities:
0.23 (for \( x = -2 \)) + 0.16 (for \( x = -1 \)) + 0.13 (for \( x = 4 \)) = 0.52.
2. Subtract this sum from 1 to determine the total probability remaining for the missing values.
1 - 0.52 = 0.48.
3. Assign probabilities to \( P(X = -3) \) and \( P(X = 2) \) such that their sum equals 0.48. Possible distributions could be equal (0.24 each) or any other combination that sums to 0.48.
This exercise helps in understanding the creation and verification of a probability distribution for discrete random variables.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4b8881e-39ca-4374-b179-41a202733fdb%2F1a132418-9221-4b0f-aea2-389689dec30b%2Fa105i4f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Fill in the \( P(X = x) \) values to give a legitimate probability distribution for the discrete random variable \( X \), whose possible values are \( -3, -2, -1, 2, \) and \( 4 \).
| Value \( x \) of \( X \) | \( P(X = x) \) |
|----------------|---------------|
| -3 | [ ] |
| -2 | 0.23 |
| -1 | 0.16 |
| 2 | [ ] |
| 4 | 0.13 |
**Description:**
This table provides a set of values for a discrete random variable \( X \) and requires completion to form a valid probability distribution.
- Each \( P(X = x) \) value must be filled in such that the sum of all probabilities equals 1.
- The probabilities \( P(X = -2) \), \( P(X = -1) \), and \( P(X = 4) \) are given as 0.23, 0.16, and 0.13, respectively.
- You need to calculate and fill in the missing probabilities for \( x = -3 \) and \( x = 2 \).
**Instructions:**
1. Add together the given probabilities:
0.23 (for \( x = -2 \)) + 0.16 (for \( x = -1 \)) + 0.13 (for \( x = 4 \)) = 0.52.
2. Subtract this sum from 1 to determine the total probability remaining for the missing values.
1 - 0.52 = 0.48.
3. Assign probabilities to \( P(X = -3) \) and \( P(X = 2) \) such that their sum equals 0.48. Possible distributions could be equal (0.24 each) or any other combination that sums to 0.48.
This exercise helps in understanding the creation and verification of a probability distribution for discrete random variables.
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