3. Let X denote the temperature at which a certain chemical reaction takes place. Suppose that X has pdf: f(x) Find the expected value E(2X). 9 (4- x²), -1 < x < 2
3. Let X denote the temperature at which a certain chemical reaction takes place. Suppose that X has pdf: f(x) Find the expected value E(2X). 9 (4- x²), -1 < x < 2
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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Problem 1P
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Can someone please help me to answer the following question
![### Problem 3: Expected Value of a Random Variable
**Problem Statement:**
Let \( X \) denote the temperature at which a certain chemical reaction takes place. Suppose that \( X \) has the probability density function (pdf):
\[ f(x) = \frac{1}{9}(4 - x^2), \quad -1 < x < 2 \]
Find the expected value \( E(2X) \).
**Solution Process:**
1. **Understanding the Probability Density Function (pdf):**
- The given pdf is \( f(x) = \frac{1}{9}(4 - x^2) \) which is valid within the interval \( -1 < x < 2 \).
- This pdf describes the probability distribution of the random variable \( X \).
2. **Expected Value of 2X:**
- The expected value \( E(aX) \) of a random variable for a constant \( a \) is given by \( aE(X) \).
- Hence, \( E(2X) = 2E(X) \).
3. **Compute \( E(X) \):**
- The expected value \( E(X) \) for a continuous random variable with pdf \( f(x) \) is calculated by:
\[ E(X) = \int_{-\infty}^{\infty} x f(x) \, dx \]
- Given our interval, it becomes:
\[ E(X) = \int_{-1}^{2} x f(x) \, dx \]
- Substituting the pdf:
\[ E(X) = \int_{-1}^{2} x \cdot \frac{1}{9}(4 - x^2) \, dx \]
- Simplifying and solving this integral will give you \( E(X) \).
**Explanation of the Diagram:**
- If there were a graph, it would typically illustrate the pdf \( f(x) \) (a parabola opening downwards) between \( x = -1 \) and \( x = 2 \).
- The area under this curve from \( x = -1 \) to \( x = 2 \) would represent the total probability, which is 1.
4. **Final Calculation:**
- Once \( E(X) \) is computed, multiply this result by 2 to find](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F322ab9b5-8b73-42ca-a79a-5c7ee3611144%2Fc8c1e337-e437-4283-80d1-2aeb2ba79181%2Fxhxpohh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 3: Expected Value of a Random Variable
**Problem Statement:**
Let \( X \) denote the temperature at which a certain chemical reaction takes place. Suppose that \( X \) has the probability density function (pdf):
\[ f(x) = \frac{1}{9}(4 - x^2), \quad -1 < x < 2 \]
Find the expected value \( E(2X) \).
**Solution Process:**
1. **Understanding the Probability Density Function (pdf):**
- The given pdf is \( f(x) = \frac{1}{9}(4 - x^2) \) which is valid within the interval \( -1 < x < 2 \).
- This pdf describes the probability distribution of the random variable \( X \).
2. **Expected Value of 2X:**
- The expected value \( E(aX) \) of a random variable for a constant \( a \) is given by \( aE(X) \).
- Hence, \( E(2X) = 2E(X) \).
3. **Compute \( E(X) \):**
- The expected value \( E(X) \) for a continuous random variable with pdf \( f(x) \) is calculated by:
\[ E(X) = \int_{-\infty}^{\infty} x f(x) \, dx \]
- Given our interval, it becomes:
\[ E(X) = \int_{-1}^{2} x f(x) \, dx \]
- Substituting the pdf:
\[ E(X) = \int_{-1}^{2} x \cdot \frac{1}{9}(4 - x^2) \, dx \]
- Simplifying and solving this integral will give you \( E(X) \).
**Explanation of the Diagram:**
- If there were a graph, it would typically illustrate the pdf \( f(x) \) (a parabola opening downwards) between \( x = -1 \) and \( x = 2 \).
- The area under this curve from \( x = -1 \) to \( x = 2 \) would represent the total probability, which is 1.
4. **Final Calculation:**
- Once \( E(X) \) is computed, multiply this result by 2 to find
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