cosulen Dhe probubily densty functren OTherwise which randen werclde Xfellews Find pobebility Tht
cosulen Dhe probubily densty functren OTherwise which randen werclde Xfellews Find pobebility Tht
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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![### Probability Density Function Example
#### Problem Statement
Consider the probability density function (PDF) described mathematically as follows:
\[
f(x) =
\begin{cases}
3(1 - x^2), & \text{if } -1 \leq x \leq 1 \\
0, & \text{otherwise}
\end{cases}
\]
Where a random variable \( X \) follows this distribution.
#### Task
Find the probability that \( 0 \leq X \leq \frac{1}{2} \).
### Explanation
The given PDF \( f(x) \) is defined piecewise. For values of \( x \) between \(-1\) and \( 1 \) inclusive, the function \( f(x) \) is \( 3(1 - x^2) \). For all other values of \( x \), the function is \( 0 \), meaning that the probability is zero outside the interval \([-1, 1]\).
To solve the problem, we need to integrate the PDF over the interval \([0, \frac{1}{2}]\):
\[
P\left(0 \leq X \leq \frac{1}{2}\right) = \int_{0}^{1/2} 3(1 - x^2) \, dx
\]
### Steps to Calculate the Probability
1. **Set up the integral:**
\[ \int_{0}^{1/2} 3(1 - x^2) \, dx \]
2. **Integrate the function:**
\[ \int_{0}^{1/2} 3(1 - x^2) \, dx = 3 \int_{0}^{1/2} (1 - x^2) \, dx \]
3. **Perform the integration:**
\[ 3 \left[ x - \frac{x^3}{3} \right]_{0}^{1/2} \]
4. **Evaluate the definite integral:**
\[
3 \left( \left( \frac{1}{2} - \frac{(1/2)^3}{3} \right) - \left( 0 - 0 \right) \right)
= 3 \left( \frac{1}{2} - \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faaa612e4-bb6a-4a7c-9bcb-a4980796bdd1%2F29e49642-03b0-451d-810e-6672b3d76197%2Fexthh48_reoriented.jpeg&w=3840&q=75)
Transcribed Image Text:### Probability Density Function Example
#### Problem Statement
Consider the probability density function (PDF) described mathematically as follows:
\[
f(x) =
\begin{cases}
3(1 - x^2), & \text{if } -1 \leq x \leq 1 \\
0, & \text{otherwise}
\end{cases}
\]
Where a random variable \( X \) follows this distribution.
#### Task
Find the probability that \( 0 \leq X \leq \frac{1}{2} \).
### Explanation
The given PDF \( f(x) \) is defined piecewise. For values of \( x \) between \(-1\) and \( 1 \) inclusive, the function \( f(x) \) is \( 3(1 - x^2) \). For all other values of \( x \), the function is \( 0 \), meaning that the probability is zero outside the interval \([-1, 1]\).
To solve the problem, we need to integrate the PDF over the interval \([0, \frac{1}{2}]\):
\[
P\left(0 \leq X \leq \frac{1}{2}\right) = \int_{0}^{1/2} 3(1 - x^2) \, dx
\]
### Steps to Calculate the Probability
1. **Set up the integral:**
\[ \int_{0}^{1/2} 3(1 - x^2) \, dx \]
2. **Integrate the function:**
\[ \int_{0}^{1/2} 3(1 - x^2) \, dx = 3 \int_{0}^{1/2} (1 - x^2) \, dx \]
3. **Perform the integration:**
\[ 3 \left[ x - \frac{x^3}{3} \right]_{0}^{1/2} \]
4. **Evaluate the definite integral:**
\[
3 \left( \left( \frac{1}{2} - \frac{(1/2)^3}{3} \right) - \left( 0 - 0 \right) \right)
= 3 \left( \frac{1}{2} - \
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