(a) Verify that the following function is a valid PMF r(x) = (")c0.3)*(0.7)5-x = 0,1,2,3,4,5 %3D (b) Give the completely defined CDF for this distribution
(a) Verify that the following function is a valid PMF r(x) = (")c0.3)*(0.7)5-x = 0,1,2,3,4,5 %3D (b) Give the completely defined CDF for this distribution
A First Course in Probability (10th Edition)
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![### Exercise on Valid Probability Mass Function (PMF) and Cumulative Distribution Function (CDF)
**Problem Statement:**
(a) **Verify that the following function is a valid PMF:**
\[ r(x) = \binom{5}{x} (0.3)^x (0.7)^{5-x}, \quad x = 0, 1, 2, 3, 4, 5 \]
(b) **Give the completely defined CDF for this distribution.**
**Solution Outline:**
(a) To verify that \( r(x) \) is a valid Probability Mass Function (PMF), we need to ensure two key properties:
1. **Non-negativity**: \( r(x) \geq 0 \) for all allowable values of \( x \).
2. **Normalization**: The sum of all probabilities must equal 1, i.e.,
\[ \sum_{x=0}^{5} r(x) = 1 \]
(b) The Cumulative Distribution Function (CDF), \( R(x) \), is the sum of the PMF up to \( x \):
\[ R(x) = \sum_{k=0}^{x} r(k) \]
Stay tuned as we delve into the calculations and explanations for each part.
### Detailed Analysis
**Part (a): Verification of PMF**
1. **Non-negativity:**
Since the binomial coefficient \(\binom{5}{x} \), and the terms \((0.3)^x\) and \((0.7)^{5-x}\) are non-negative for all \(x\), \( r(x) \) is non-negative.
2. **Normalization:**
Sum \( r(x) \) over \( x \) from 0 to 5:
\[
\sum_{x=0}^{5} \binom{5}{x} (0.3)^x (0.7)^{5-x}
\]
The above expression is the binomial expansion of \((0.3 + 0.7)^5\), which equals 1. Therefore:
\[
\sum_{x=0}^{5} r(x) = 1
\]
Since both properties hold, \( r(x) \) is a valid PMF.
**Part (b): CDF Definition**
Calculate the Cumulative Distribution Function](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0b9151a0-b51c-4e98-8bd5-faad9be5df3c%2F7c4b6341-64a9-445d-b13f-cbdc22e9ba39%2F4riim5g.png&w=3840&q=75)
Transcribed Image Text:### Exercise on Valid Probability Mass Function (PMF) and Cumulative Distribution Function (CDF)
**Problem Statement:**
(a) **Verify that the following function is a valid PMF:**
\[ r(x) = \binom{5}{x} (0.3)^x (0.7)^{5-x}, \quad x = 0, 1, 2, 3, 4, 5 \]
(b) **Give the completely defined CDF for this distribution.**
**Solution Outline:**
(a) To verify that \( r(x) \) is a valid Probability Mass Function (PMF), we need to ensure two key properties:
1. **Non-negativity**: \( r(x) \geq 0 \) for all allowable values of \( x \).
2. **Normalization**: The sum of all probabilities must equal 1, i.e.,
\[ \sum_{x=0}^{5} r(x) = 1 \]
(b) The Cumulative Distribution Function (CDF), \( R(x) \), is the sum of the PMF up to \( x \):
\[ R(x) = \sum_{k=0}^{x} r(k) \]
Stay tuned as we delve into the calculations and explanations for each part.
### Detailed Analysis
**Part (a): Verification of PMF**
1. **Non-negativity:**
Since the binomial coefficient \(\binom{5}{x} \), and the terms \((0.3)^x\) and \((0.7)^{5-x}\) are non-negative for all \(x\), \( r(x) \) is non-negative.
2. **Normalization:**
Sum \( r(x) \) over \( x \) from 0 to 5:
\[
\sum_{x=0}^{5} \binom{5}{x} (0.3)^x (0.7)^{5-x}
\]
The above expression is the binomial expansion of \((0.3 + 0.7)^5\), which equals 1. Therefore:
\[
\sum_{x=0}^{5} r(x) = 1
\]
Since both properties hold, \( r(x) \) is a valid PMF.
**Part (b): CDF Definition**
Calculate the Cumulative Distribution Function
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