Math Problem Statement
A continuous RV X has density function f(x)={(1/(1+x)^2,x≥0@0,otherwise) Calculate P(2<X<5)
Solution
We are given the probability density function (PDF) of a continuous random variable , defined as:
\begin{cases} \frac{1}{(1+x)^2}, & x \geq 0 \\ 0, & \text{otherwise} \end{cases}$$ We are asked to find $$P(2 < X < 5)$$, which is the probability that $$X$$ lies between 2 and 5. This probability can be computed by integrating the PDF over the interval $$(2, 5)$$: $$P(2 < X < 5) = \int_{2}^{5} f(x) \, dx$$ Substituting the PDF: $$P(2 < X < 5) = \int_{2}^{5} \frac{1}{(1+x)^2} \, dx$$ To solve this, we perform the following integration. ### Step-by-step solution: 1. Let’s use a substitution to simplify the integral: - Let $$u = 1 + x$$, so $$du = dx$$. - When $$x = 2$$, $$u = 3$$, and when $$x = 5$$, $$u = 6$$. Now, the integral becomes: $$P(2 < X < 5) = \int_{3}^{6} \frac{1}{u^2} \, du$$ 2. The integral of $$\frac{1}{u^2}$$ is: $$\int \frac{1}{u^2} \, du = -\frac{1}{u}$$ Thus, the integral evaluates to: $$P(2 < X < 5) = \left[ -\frac{1}{u} \right]_{3}^{6}$$ 3. Substituting the limits: $$P(2 < X < 5) = -\frac{1}{6} + \frac{1}{3}$$ 4. Simplifying: $$P(2 < X < 5) = \frac{1}{3} - \frac{1}{6} = \frac{1}{6}$$ Thus, the probability that $$X$$ lies between 2 and 5 is: $$P(2 < X < 5) = \frac{1}{6}$$ --- Would you like further details or have any questions? Here are 5 related questions: 1. How would we compute $$P(X > 5)$$ for this random variable? 2. Can you generalize the process for finding $$P(a < X < b)$$ for any $$a$$ and $$b$$? 3. What is the cumulative distribution function (CDF) of this random variable? 4. How would you compute the mean of this random variable? 5. Can you find the variance of this random variable? **Tip**: When solving integrals involving rational functions, substitution is often a very useful technique for simplification!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability Theory
Continuous Random Variables
Probability Density Function (PDF)
Integration
Formulas
f(x) = 1/(1+x)^2 for x ≥ 0
P(2 < X < 5) = ∫[2,5] f(x) dx
Substitution: u = 1 + x, du = dx
Theorems
Fundamental Theorem of Calculus
Suitable Grade Level
College/University Level
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