真空题。 2.0分) function [0₂ 1/2, F(x)= 3/4 Ⓒ x <0 0

MATLAB: An Introduction with Applications
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Author:Amos Gilat
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**Problem 1: (2 points)**
Given a function \( F(x) \) defined as:

\[
F(x) = 
\begin{cases}
0, & x < 0 \\
1/2, & 0 \leq x < 0.5 \\
3/4, & 0.5 \leq x < 2 \\
1, & x \geq 2
\end{cases}
\]

This function is the cumulative distribution function (c.d.f) for some discrete random variable \( X \). Determine the probability mass function (p.m.f) for \( X \).

**Answer:**


**Explanation:**

This problem involves finding the probability mass function (p.m.f) from the given cumulative distribution function (c.d.f) of a discrete random variable \( X \). The c.d.f, \( F(x) \), provides the cumulative probabilities up to certain points, and from these, we can derive the individual probabilities associated with specific values of \( X \).
Transcribed Image Text:**Problem 1: (2 points)** Given a function \( F(x) \) defined as: \[ F(x) = \begin{cases} 0, & x < 0 \\ 1/2, & 0 \leq x < 0.5 \\ 3/4, & 0.5 \leq x < 2 \\ 1, & x \geq 2 \end{cases} \] This function is the cumulative distribution function (c.d.f) for some discrete random variable \( X \). Determine the probability mass function (p.m.f) for \( X \). **Answer:** **Explanation:** This problem involves finding the probability mass function (p.m.f) from the given cumulative distribution function (c.d.f) of a discrete random variable \( X \). The c.d.f, \( F(x) \), provides the cumulative probabilities up to certain points, and from these, we can derive the individual probabilities associated with specific values of \( X \).
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