Math Problem Statement
Let X be a random variable with distribution function F(x) where F(x) = 0 if x < 0, and F(x) = 1 - 2/3 * e^(-x/2) - 1/3 * e^(-floor(x/2)) for x ≥ 0. Find P(4 ≤ X < 6).
Solution
We are given a random variable with the cumulative distribution function defined as:
\begin{cases} 0, & \text{if } x < 0 \\ 1 - \frac{2}{3} e^{-\frac{x}{2}} - \frac{1}{3} e^{-\left\lfloor \frac{x}{2} \right\rfloor}, & \text{if } x \geq 0 \end{cases}$$ where $$\left\lfloor x \right\rfloor$$ denotes the greatest integer less than or equal to $$x$$. The problem asks for the value of the probability $$P(4 \leq X < 6)$$. ### Step 1: Understanding the distribution function Since $$F(x)$$ is the cumulative distribution function, we can compute probabilities using: $$P(a \leq X < b) = F(b) - F(a)$$ Thus, in this case: $$P(4 \leq X < 6) = F(6) - F(4)$$ ### Step 2: Computing $$F(6)$$ For $$x = 6$$, the cumulative distribution function $$F(6)$$ will be: $$F(6) = 1 - \frac{2}{3} e^{-\frac{6}{2}} - \frac{1}{3} e^{-\left\lfloor \frac{6}{2} \right\rfloor}$$ $$F(6) = 1 - \frac{2}{3} e^{-3} - \frac{1}{3} e^{-3}$$ $$F(6) = 1 - e^{-3}$$ ### Step 3: Computing $$F(4)$$ For $$x = 4$$, the cumulative distribution function $$F(4)$$ will be: $$F(4) = 1 - \frac{2}{3} e^{-\frac{4}{2}} - \frac{1}{3} e^{-\left\lfloor \frac{4}{2} \right\rfloor}$$ $$F(4) = 1 - \frac{2}{3} e^{-2} - \frac{1}{3} e^{-2}$$ $$F(4) = 1 - e^{-2}$$ ### Step 4: Final calculation Now, we can find $$P(4 \leq X < 6)$$ by subtracting: $$P(4 \leq X < 6) = F(6) - F(4)$$ $$P(4 \leq X < 6) = (1 - e^{-3}) - (1 - e^{-2})$$ $$P(4 \leq X < 6) = e^{-2} - e^{-3}$$ Thus, the correct answer is: $$\boxed{e^{-2} - e^{-3}}$$ This corresponds to option **(a)**. --- Would you like a deeper explanation of any step? Here are five related questions you might explore: 1. How do cumulative distribution functions (CDFs) relate to probability density functions (PDFs)? 2. What is the meaning of the floor function $$\left\lfloor x \right\rfloor$$ in the context of probability distributions? 3. How would you compute the mean or variance of this random variable $$X$$? 4. What is the significance of exponential terms in probability distributions? 5. How can we derive the probability mass function from a given CDF? **Tip:** Always verify the domain of the random variable when applying a CDF to compute probabilities.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability
Cumulative Distribution Function
Exponential Functions
Formulas
P(a ≤ X < b) = F(b) - F(a)
F(x) = 1 - (2/3) * e^(-x/2) - (1/3) * e^(-floor(x/2)) for x ≥ 0
Theorems
Cumulative Distribution Function Properties
Suitable Grade Level
University-level (Probability and Statistics)
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