Math Problem Statement

Given a random variable X with the probability density function f(x) defined as: f(x) = {kx, 0 ≤ x ≤ 5; k(x-3), 5 ≤ x ≤ 10; 0, otherwise}, find the following: a) The value of k. b) The expected value E(X). c) The variance Var(X). d) P(4 ≤ X ≤ 6). e) Sketch the graph.

Solution

The image shows a probability density function (PDF) f(x)f(x) of a random variable XX. The function is defined as:

\begin{cases} kx, & 0 \leq x \leq 5 \\ k(x-3), & 5 \leq x \leq 10 \\ 0, & \text{otherwise} \end{cases}$$ The questions associated with this function are: 1. (a) Find the value of $$k$$. 2. (b) Find the expected value $$E(X)$$. 3. (c) Find the variance $$\text{Var}(X)$$. 4. (d) Find $$P(4 \leq X \leq 6)$$. 5. (e) Sketch the graph of $$f(x)$$. Let's solve each part step-by-step: ### (a) Find the value of $$k$$ To find $$k$$, we use the fact that the total area under the PDF must be 1: $$\int_{-\infty}^{\infty} f(x) \, dx = 1$$ Since the function is only nonzero from $$0$$ to $$10$$, we split the integral into two parts: $$\int_0^5 kx \, dx + \int_5^{10} k(x - 3) \, dx = 1$$ Calculating these integrals: 1. For the first part: $$\int_0^5 kx \, dx = k \left[\frac{x^2}{2}\right]_0^5 = k \cdot \frac{25}{2} = \frac{25k}{2}$$ 2. For the second part: $$\int_5^{10} k(x - 3) \, dx = k \left[\frac{(x - 3)^2}{2}\right]_5^{10} = k \cdot \frac{49}{2} = \frac{49k}{2}$$ Adding both integrals: $$\frac{25k}{2} + \frac{49k}{2} = \frac{74k}{2} = 37k$$ Setting this equal to 1: $$37k = 1 \implies k = \frac{1}{37}$$ ### (b) Find the expected value $$E(X)$$ The expected value $$E(X)$$ is given by: $$E(X) = \int_{-\infty}^{\infty} x f(x) \, dx = \int_0^5 x(kx) \, dx + \int_5^{10} x k(x - 3) \, dx$$ Substituting $$k = \frac{1}{37}$$ and evaluating: 1. For the first part: $$\int_0^5 x \cdot \frac{x}{37} \, dx = \frac{1}{37} \int_0^5 x^2 \, dx = \frac{1}{37} \cdot \frac{x^3}{3} \Bigg|_0^5 = \frac{125}{111}$$ 2. For the second part: $$\int_5^{10} x \cdot k(x - 3) \, dx = \frac{1}{37}$$

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability Density Function
Expected Value
Variance
Continuous Random Variables

Formulas

Total area under the PDF must equal 1: ∫f(x)dx = 1
Expected value: E(X) = ∫x f(x) dx
Variance: Var(X) = E(X^2) - (E(X))^2
Probability for continuous random variables: P(a ≤ X ≤ b) = ∫[a, b] f(x) dx

Theorems

Basic properties of probability density functions

Suitable Grade Level

Undergraduate Level (College)