Fill in the P (X=x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -6, -4,-1, 0, and 2. Value x of X -6 -4 -1 0 2 X P ( X = x) 0.22 0.16 0.16 0
Fill in the P (X=x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -6, -4,-1, 0, and 2. Value x of X -6 -4 -1 0 2 X P ( X = x) 0.22 0.16 0.16 0
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: Completing a Probability Distribution Table**
**Introduction:**
When given a discrete random variable \( X \) with specific possible values, it's important to assign probabilities \( P(X=x) \) in such a way that they form a legitimate probability distribution. This involves ensuring the sum of probabilities equals 1, and each probability is between 0 and 1.
**Problem Statement:**
We need to fill in the \( P(X=x) \) values to create a valid probability distribution for the discrete random variable \( X \), whose possible values are \(-6, -4, -1, 0, \) and \( 2 \).
**Probability Distribution Table:**
The table below shows some probabilities already assigned to certain values of \( X \). Our task is to complete this table.
| Value \( x \) of \( X \) | \( P(X = x) \) |
|--------------------------|----------------|
| -6 | 0.22 |
| -4 | 0.16 |
| -1 | 0.16 |
| 0 | [ ] |
| 2 | [ ] |
**Explanation:**
- The probabilities for \( X = -6, -4, \) and \( -1 \) are given as \( 0.22, 0.16, \) and \( 0.16 \) respectively.
- To complete the table, the sum of all probabilities must equal 1, that is:
\[
0.22 + 0.16 + 0.16 + P(X=0) + P(X=2) = 1
\]
- Calculate \( P(X=0) \) and \( P(X=2) \) such that the above condition holds.
**Steps:**
1. Add the given probabilities: \( 0.22 + 0.16 + 0.16 = 0.54 \).
2. Subtract this sum from 1: \( 1 - 0.54 = 0.46 \).
3. Distribute the remaining probability between \( P(X=0) \) and \( P(X=2) \) ensuring each remains between 0 and 1.
**Conclusion:**
The values for \( P(X=0) \) and \( P(X=2) \) can be calculated to ensure the distribution is valid](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F964640eb-05b8-48b6-8066-4e5b4ee841dd%2F2c1be45e-4564-46d2-8d9c-92374480f13b%2Fiymd5hb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Completing a Probability Distribution Table**
**Introduction:**
When given a discrete random variable \( X \) with specific possible values, it's important to assign probabilities \( P(X=x) \) in such a way that they form a legitimate probability distribution. This involves ensuring the sum of probabilities equals 1, and each probability is between 0 and 1.
**Problem Statement:**
We need to fill in the \( P(X=x) \) values to create a valid probability distribution for the discrete random variable \( X \), whose possible values are \(-6, -4, -1, 0, \) and \( 2 \).
**Probability Distribution Table:**
The table below shows some probabilities already assigned to certain values of \( X \). Our task is to complete this table.
| Value \( x \) of \( X \) | \( P(X = x) \) |
|--------------------------|----------------|
| -6 | 0.22 |
| -4 | 0.16 |
| -1 | 0.16 |
| 0 | [ ] |
| 2 | [ ] |
**Explanation:**
- The probabilities for \( X = -6, -4, \) and \( -1 \) are given as \( 0.22, 0.16, \) and \( 0.16 \) respectively.
- To complete the table, the sum of all probabilities must equal 1, that is:
\[
0.22 + 0.16 + 0.16 + P(X=0) + P(X=2) = 1
\]
- Calculate \( P(X=0) \) and \( P(X=2) \) such that the above condition holds.
**Steps:**
1. Add the given probabilities: \( 0.22 + 0.16 + 0.16 = 0.54 \).
2. Subtract this sum from 1: \( 1 - 0.54 = 0.46 \).
3. Distribute the remaining probability between \( P(X=0) \) and \( P(X=2) \) ensuring each remains between 0 and 1.
**Conclusion:**
The values for \( P(X=0) \) and \( P(X=2) \) can be calculated to ensure the distribution is valid
Expert Solution
Step 1
| x | p(x) |
| -6 | 0.22 |
| -4 | 0.16 |
| -1 | 0.16 |
| 0 | a |
| 2 | b |
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