Find k such that the function is a probability density function over the given interval. Then write the probability density function. f(x)=kx, 1≤x≤8 What is the value of k? k=1 (Simplify your answer.) What is the probability density function? f(x) = 0
Find k such that the function is a probability density function over the given interval. Then write the probability density function. f(x)=kx, 1≤x≤8 What is the value of k? k=1 (Simplify your answer.) What is the probability density function? f(x) = 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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![### Probability Density Function Problem
#### Problem Statement:
Find \( k \) such that the function is a probability density function over the given interval. Then write the probability density function.
\[ f(x) = kx, \quad 1 \leq x \leq 8 \]
---
**What is the value of \( k \)?**
\[ k = \boxed{1} \] (Simplify your answer.)
**What is the probability density function?**
\[ f(x) = \boxed{} \]
---
**Explanation:**
To be a probability density function over the interval \([1, 8]\), the function \( f(x) = kx \) must satisfy two conditions:
1. The function must be non-negative over the entire interval.
2. The total area under the function over the interval must equal 1.
Let's find \( k \) by setting up the integral of \( kx \) from 1 to 8 and setting it equal to 1:
\[ \int_{1}^{8} kx \, dx = 1 \]
\[ k \int_{1}^{8} x \, dx = 1 \]
Evaluating the integral:
\[ k \left[ \frac{x^2}{2} \right]_{1}^{8} = 1 \]
\[ k \left( \frac{8^2}{2} - \frac{1^2}{2} \right) = 1 \]
\[ k \left( \frac{64}{2} - \frac{1}{2} \right) = 1 \]
\[ k \left( 32 - 0.5 \right) = 1 \]
\[ k \left( 31.5 \right) = 1 \]
\[ k = \frac{1}{31.5} \]
Therefore, the value of \( k \) simplifies to \( \frac{1}{31.5} \).
The probability density function \( f(x) \) is then:
\[ f(x) = \frac{x}{31.5} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc56b3bff-26b7-4f28-9b8d-5c65e8bb1b58%2Ff8dc6c04-9dbd-4f3f-89fb-ad28175cc245%2Fdztpwh_processed.png&w=3840&q=75)
Transcribed Image Text:### Probability Density Function Problem
#### Problem Statement:
Find \( k \) such that the function is a probability density function over the given interval. Then write the probability density function.
\[ f(x) = kx, \quad 1 \leq x \leq 8 \]
---
**What is the value of \( k \)?**
\[ k = \boxed{1} \] (Simplify your answer.)
**What is the probability density function?**
\[ f(x) = \boxed{} \]
---
**Explanation:**
To be a probability density function over the interval \([1, 8]\), the function \( f(x) = kx \) must satisfy two conditions:
1. The function must be non-negative over the entire interval.
2. The total area under the function over the interval must equal 1.
Let's find \( k \) by setting up the integral of \( kx \) from 1 to 8 and setting it equal to 1:
\[ \int_{1}^{8} kx \, dx = 1 \]
\[ k \int_{1}^{8} x \, dx = 1 \]
Evaluating the integral:
\[ k \left[ \frac{x^2}{2} \right]_{1}^{8} = 1 \]
\[ k \left( \frac{8^2}{2} - \frac{1^2}{2} \right) = 1 \]
\[ k \left( \frac{64}{2} - \frac{1}{2} \right) = 1 \]
\[ k \left( 32 - 0.5 \right) = 1 \]
\[ k \left( 31.5 \right) = 1 \]
\[ k = \frac{1}{31.5} \]
Therefore, the value of \( k \) simplifies to \( \frac{1}{31.5} \).
The probability density function \( f(x) \) is then:
\[ f(x) = \frac{x}{31.5} \]
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