Let f(x) = 0, C:C x² +1¹ 0, if x ≤ 0 if 0
A First Course in Probability (10th Edition)
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ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
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Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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![**Probability Density Functions: Solving for the Constant \( c \)**
To determine for what value of \( c \) the function \( f(x) \) is a probability density function (PDF), consider the following piecewise function:
\[
f(x) =
\begin{cases}
0, & \text{if } x \leq 0 \\
\frac{cx}{x^2 + 1}, & \text{if } 0 < x < \sqrt{e - 1} \\
0, & \text{if } x \geq \sqrt{e - 1}
\end{cases}
\]
We seek the value of \( c \) that will ensure \( f(x) \) meets the criteria of a probability density function. Specifically, \( f(x) \) must integrate to 1 over its entire domain.
1. **Function Conditions Analysis:**
- The function \( f(x) \) is 0 when \( x \leq 0 \) and when \( x \geq \sqrt{e - 1} \).
- The function \( f(x) \) is defined as \( \frac{cx}{x^2 + 1} \) within \( 0 < x < \sqrt{e - 1} \).
2. **Integration to Solve for \( c \):**
To find \( c \), we integrate \( f(x) \) over its entire valid domain \( (0, \sqrt{e - 1}) \) and set the integral equal to 1:
\[
\int_{0}^{\sqrt{e - 1}} \frac{cx}{x^2 + 1} \, dx = 1
\]
By solving this integral, we determine the specific value of \( c \) that makes \( f(x) \) a valid probability density function. Hence, the integration process and understanding the distribution forms a crucial component of probability and statistics education.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e7c908-505f-4c35-8ac4-e541f2ab69b2%2F39a96cec-3b10-41a1-9f9b-71cfc64d12cd%2Fhseupz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Probability Density Functions: Solving for the Constant \( c \)**
To determine for what value of \( c \) the function \( f(x) \) is a probability density function (PDF), consider the following piecewise function:
\[
f(x) =
\begin{cases}
0, & \text{if } x \leq 0 \\
\frac{cx}{x^2 + 1}, & \text{if } 0 < x < \sqrt{e - 1} \\
0, & \text{if } x \geq \sqrt{e - 1}
\end{cases}
\]
We seek the value of \( c \) that will ensure \( f(x) \) meets the criteria of a probability density function. Specifically, \( f(x) \) must integrate to 1 over its entire domain.
1. **Function Conditions Analysis:**
- The function \( f(x) \) is 0 when \( x \leq 0 \) and when \( x \geq \sqrt{e - 1} \).
- The function \( f(x) \) is defined as \( \frac{cx}{x^2 + 1} \) within \( 0 < x < \sqrt{e - 1} \).
2. **Integration to Solve for \( c \):**
To find \( c \), we integrate \( f(x) \) over its entire valid domain \( (0, \sqrt{e - 1}) \) and set the integral equal to 1:
\[
\int_{0}^{\sqrt{e - 1}} \frac{cx}{x^2 + 1} \, dx = 1
\]
By solving this integral, we determine the specific value of \( c \) that makes \( f(x) \) a valid probability density function. Hence, the integration process and understanding the distribution forms a crucial component of probability and statistics education.
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