1. Consider the Continuous Distribution Given by its Probability Density Function f(x) = x € [10, 30] [10,30] A. Confirm that f(x) is indeed a Probability Density Function. Thus you have to show that the following two conditions are true 1. f(x) > 0 II. The Area under its graph is equal to 1. Hint: Its a graph will be a Triangle x < 10 0 B. Confirm that F(x) = (x − 10)² ÷ 400 10 ≤ x ≤ 30 is the 1 x > 30 Cumulative Distribution Function (CDF) of this probability density function. Hint: Area under a curve is gonna be a triangle, compute its area at a given point. 2-10 200 0
1. Consider the Continuous Distribution Given by its Probability Density Function f(x) = x € [10, 30] [10,30] A. Confirm that f(x) is indeed a Probability Density Function. Thus you have to show that the following two conditions are true 1. f(x) > 0 II. The Area under its graph is equal to 1. Hint: Its a graph will be a Triangle x < 10 0 B. Confirm that F(x) = (x − 10)² ÷ 400 10 ≤ x ≤ 30 is the 1 x > 30 Cumulative Distribution Function (CDF) of this probability density function. Hint: Area under a curve is gonna be a triangle, compute its area at a given point. 2-10 200 0
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![# Continuous Distribution and Probability Density Function
## 1. Consider the Continuous Distribution Given by its Probability Density Function
\[ f(x) =
\begin{cases}
\frac{x-10}{200} & x \in [10, 30] \\
0 & x \notin [10, 30]
\end{cases}
\]
### A. Confirm that \( f(x) \) is indeed a Probability Density Function.
To do this, you must demonstrate that the following two conditions are true:
1. \( f(x) \geq 0 \)
2. The area under its graph is equal to 1.
- **Hint:** Its graph will be a triangle.
### B. Confirm that
\[ F(x) =
\begin{cases}
0 & x < 10 \\
\frac{(x-10)^2}{400} & 10 \leq x \leq 30 \\
1 & x > 30
\end{cases}
\]
is the Cumulative Distribution Function (CDF) of this probability density function.
- **Hint:** The area under a curve is going to be a triangle, compute its area at a given point.
### Explanation of the Graph:
The function \( f(x) \) is defined as a piecewise function with a linear segment between \( x = 10 \) and \( x = 30 \), forming a triangle when graphed. The base of this triangle is along the x-axis from \( 10 \) to \( 30 \), while the height is determined by the slope \( \frac{1}{200} \). This ensures the total area under the curve is 1, satisfying the requirement for a probability density function.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F15077d0c-06b0-46d9-ae2e-319ac77cceb5%2Fc160c328-245f-4de6-940d-d822431628cf%2Fxhwv36e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:# Continuous Distribution and Probability Density Function
## 1. Consider the Continuous Distribution Given by its Probability Density Function
\[ f(x) =
\begin{cases}
\frac{x-10}{200} & x \in [10, 30] \\
0 & x \notin [10, 30]
\end{cases}
\]
### A. Confirm that \( f(x) \) is indeed a Probability Density Function.
To do this, you must demonstrate that the following two conditions are true:
1. \( f(x) \geq 0 \)
2. The area under its graph is equal to 1.
- **Hint:** Its graph will be a triangle.
### B. Confirm that
\[ F(x) =
\begin{cases}
0 & x < 10 \\
\frac{(x-10)^2}{400} & 10 \leq x \leq 30 \\
1 & x > 30
\end{cases}
\]
is the Cumulative Distribution Function (CDF) of this probability density function.
- **Hint:** The area under a curve is going to be a triangle, compute its area at a given point.
### Explanation of the Graph:
The function \( f(x) \) is defined as a piecewise function with a linear segment between \( x = 10 \) and \( x = 30 \), forming a triangle when graphed. The base of this triangle is along the x-axis from \( 10 \) to \( 30 \), while the height is determined by the slope \( \frac{1}{200} \). This ensures the total area under the curve is 1, satisfying the requirement for a probability density function.
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