5. A continuous random variable, X, has the following density function. Sx/6, 2 ≤ x ≤ 4 f(x)=0 elsewhere a. Find P(X=3). b. Find the CDF of X. c. Find P(0
5. A continuous random variable, X, has the following density function. Sx/6, 2 ≤ x ≤ 4 f(x)=0 elsewhere a. Find P(X=3). b. Find the CDF of X. c. Find P(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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![**Problem 5: Analysis of a Continuous Random Variable**
A continuous random variable, \( X \), has the following probability density function (PDF):
\[
f(x) =
\begin{cases}
x/6, & 2 \leq x \leq 4 \\
0, & \text{elsewhere}
\end{cases}
\]
**Tasks:**
a. **Find \( P(X=3) \).**
Since \( X \) is a continuous random variable, the probability of it taking any single exact value is zero. Therefore, \( P(X=3) = 0 \).
b. **Find the cumulative distribution function (CDF) of \( X \).**
To find the CDF, \( F(x) \), we integrate the PDF from the lower bound up to \( x \):
\[
F(x) = \int_{2}^{x} \frac{t}{6} \, dt = \left[ \frac{t^2}{12} \right]_2^x = \frac{x^2}{12} - \frac{4}{12}
\]
For \( x \) in the range \( 2 \leq x \leq 4 \), the CDF is:
\[
F(x) = \frac{x^2}{12} - \frac{1}{3}
\]
For other values of \( x \):
- If \( x < 2 \), \( F(x) = 0 \).
- If \( x > 4 \), \( F(x) = 1 \).
c. **Find \( P(0 < X \leq 3) \).**
To find this probability, evaluate the CDF from 2 to 3:
\[
P(0 < X \leq 3) = F(3) - F(2)
\]
Calculate:
\[
F(3) = \frac{3^2}{12} - \frac{1}{3} = \frac{9}{12} - \frac{4}{12} = \frac{5}{12}
\]
\[
F(2) = \frac{2^2}{12} - \frac{1}{3} = \frac{4}{12} - \frac{4}{12} = 0
\]
Therefore:
\[
P(0 < X \leq](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1d63ef4c-a273-47ae-b0f8-84a52d6ae66e%2F5e134b34-04c6-4aa4-a246-9ef8464fdbea%2F8fi54oo_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 5: Analysis of a Continuous Random Variable**
A continuous random variable, \( X \), has the following probability density function (PDF):
\[
f(x) =
\begin{cases}
x/6, & 2 \leq x \leq 4 \\
0, & \text{elsewhere}
\end{cases}
\]
**Tasks:**
a. **Find \( P(X=3) \).**
Since \( X \) is a continuous random variable, the probability of it taking any single exact value is zero. Therefore, \( P(X=3) = 0 \).
b. **Find the cumulative distribution function (CDF) of \( X \).**
To find the CDF, \( F(x) \), we integrate the PDF from the lower bound up to \( x \):
\[
F(x) = \int_{2}^{x} \frac{t}{6} \, dt = \left[ \frac{t^2}{12} \right]_2^x = \frac{x^2}{12} - \frac{4}{12}
\]
For \( x \) in the range \( 2 \leq x \leq 4 \), the CDF is:
\[
F(x) = \frac{x^2}{12} - \frac{1}{3}
\]
For other values of \( x \):
- If \( x < 2 \), \( F(x) = 0 \).
- If \( x > 4 \), \( F(x) = 1 \).
c. **Find \( P(0 < X \leq 3) \).**
To find this probability, evaluate the CDF from 2 to 3:
\[
P(0 < X \leq 3) = F(3) - F(2)
\]
Calculate:
\[
F(3) = \frac{3^2}{12} - \frac{1}{3} = \frac{9}{12} - \frac{4}{12} = \frac{5}{12}
\]
\[
F(2) = \frac{2^2}{12} - \frac{1}{3} = \frac{4}{12} - \frac{4}{12} = 0
\]
Therefore:
\[
P(0 < X \leq
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