The current in a certain circuit as measured by an ammeter is a continuous random variable X with the following density function. 0.025x + 0.4 3 ≤x≤5 f(x) = ) = otherwise (a) Graph the pdf. f(x) f(x) f(x) 0.6 0.5 0.4 0.3 0.2 0.1 1 2 3 4 3 O O Verify that the total area under the density curve is indeed 1. [²³ 0.025x+ 0.4 dx= X = 2.3125- ]x ) x (b) Calculate P(X ≤ 4). How does this probability compare to P(X < 4)? P(X ≤ 4) = P(X < 4) O P(X ≤ 4) > P(X < 4) O P(X ≤ 4) < P(X < 4) (c) Calculate P(3.5 ≤ x ≤ 4.5). 0.5 Calculate P(4.5 < X). X 0.6 0.5 0.4 0.3 0.2 0.1 1 2 4 5 X O 0.6 0.5 0.4 0.3 0.2 0.1 1 2 3 4 f(x) 0.6 0.5 0.4 0.3 0.2 0.1 1 2 3 4 5 X
The current in a certain circuit as measured by an ammeter is a continuous random variable X with the following density function. 0.025x + 0.4 3 ≤x≤5 f(x) = ) = otherwise (a) Graph the pdf. f(x) f(x) f(x) 0.6 0.5 0.4 0.3 0.2 0.1 1 2 3 4 3 O O Verify that the total area under the density curve is indeed 1. [²³ 0.025x+ 0.4 dx= X = 2.3125- ]x ) x (b) Calculate P(X ≤ 4). How does this probability compare to P(X < 4)? P(X ≤ 4) = P(X < 4) O P(X ≤ 4) > P(X < 4) O P(X ≤ 4) < P(X < 4) (c) Calculate P(3.5 ≤ x ≤ 4.5). 0.5 Calculate P(4.5 < X). X 0.6 0.5 0.4 0.3 0.2 0.1 1 2 4 5 X O 0.6 0.5 0.4 0.3 0.2 0.1 1 2 3 4 f(x) 0.6 0.5 0.4 0.3 0.2 0.1 1 2 3 4 5 X
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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![### Understanding Continuous Random Variables and Density Functions
The current in a certain circuit measured by an ammeter is a continuous random variable \( X \) with the following density function:
\[
f(x) = \begin{cases}
0.025x + 0.4 & \text{for } 3 \le x \le 5 \\
0 & \text{otherwise}
\end{cases}
\]
#### (a) Graph the Probability Density Function (pdf)
The correct graph is shown in the fourth plot. Here's an explanation of the graph:
- The function \( f(x) \) is defined for \( x \) in the interval \([3, 5]\).
- For \( x \) outside this interval, the density is 0.
- The pdf starts at \( (3, 0.475) \), increases linearly, reaching \( (5, 0.525) \).
The graph of the pdf is a straight line starting at (3, 0.475) and ending at (5, 0.525) within the interval \([3, 5]\). Outside this interval, the graph is 0.
#### Verify the Area Under the Density Curve
To verify that the total area under the density curve is indeed 1, compute the integral:
\[
\int_{3}^{5} (0.025x + 0.4) \, dx
\]
1. Compute the indefinite integral:
\[
\int (0.025x + 0.4) \, dx = 0.0125x^2 + 0.4x + C
\]
2. Evaluate the definite integral from 3 to 5:
\[
\left[ 0.0125x^2 + 0.4x \right]_{3}^{5}
= (0.0125(5)^2 + 0.4(5)) - (0.0125(3)^2 + 0.4(3))
= (0.3125 + 2) - (0.1125 + 1.2)
= 2.3125 - 1.3125
= 1
\]
Thus, the area under the probability density function is indeed 1.
#### (b) Calculate \( P(X \leq 4) \)
To calculate \( P(X \leq 4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fda5a8041-aeed-45a6-a7fc-c4165c2d9a1f%2Fdc85a094-1f8a-4b3d-95d3-39c9515eede0%2Fszmdb9s_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Continuous Random Variables and Density Functions
The current in a certain circuit measured by an ammeter is a continuous random variable \( X \) with the following density function:
\[
f(x) = \begin{cases}
0.025x + 0.4 & \text{for } 3 \le x \le 5 \\
0 & \text{otherwise}
\end{cases}
\]
#### (a) Graph the Probability Density Function (pdf)
The correct graph is shown in the fourth plot. Here's an explanation of the graph:
- The function \( f(x) \) is defined for \( x \) in the interval \([3, 5]\).
- For \( x \) outside this interval, the density is 0.
- The pdf starts at \( (3, 0.475) \), increases linearly, reaching \( (5, 0.525) \).
The graph of the pdf is a straight line starting at (3, 0.475) and ending at (5, 0.525) within the interval \([3, 5]\). Outside this interval, the graph is 0.
#### Verify the Area Under the Density Curve
To verify that the total area under the density curve is indeed 1, compute the integral:
\[
\int_{3}^{5} (0.025x + 0.4) \, dx
\]
1. Compute the indefinite integral:
\[
\int (0.025x + 0.4) \, dx = 0.0125x^2 + 0.4x + C
\]
2. Evaluate the definite integral from 3 to 5:
\[
\left[ 0.0125x^2 + 0.4x \right]_{3}^{5}
= (0.0125(5)^2 + 0.4(5)) - (0.0125(3)^2 + 0.4(3))
= (0.3125 + 2) - (0.1125 + 1.2)
= 2.3125 - 1.3125
= 1
\]
Thus, the area under the probability density function is indeed 1.
#### (b) Calculate \( P(X \leq 4) \)
To calculate \( P(X \leq 4
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