Math Problem Statement

The p.d.f of a continuous random variable X is given by f(x) = {

x+1

k

, 2 ≤ x ≤ 4

0 otherwise

.

(a) Find k. (b) Obtain the distribution function F(x) (c) Using F(x) compute P( X ≥ 3)

(d) Find Mean and Variance of X (e) Using F(x) find P (2.5 < X < 3).

Solution

The given probability density function (p.d.f) f(x)f(x) is described as follows:

\begin{cases} \frac{x+1}{k}, & 2 \leq x \leq 4 \\ 0, & \text{otherwise} \end{cases}$$ We'll go step by step to solve each part: ### (a) Find $$k$$ For $$f(x)$$ to be a valid probability density function, the total area under the curve must be 1. This means: $$\int_{-\infty}^{\infty} f(x) \, dx = 1$$ Given that $$f(x) = 0$$ outside the interval $$2 \leq x \leq 4$$, the equation reduces to: $$\int_{2}^{4} \frac{x+1}{k} \, dx = 1$$ Let's solve for $$k$$: $$\frac{1}{k} \int_{2}^{4} (x+1) \, dx = 1$$ First, compute the integral: $$\int_{2}^{4} (x+1) \, dx = \left[ \frac{x^2}{2} + x \right]_{2}^{4} = \left( \frac{4^2}{2} + 4 \right) - \left( \frac{2^2}{2} + 2 \right)$$ $$= \left( \frac{16}{2} + 4 \right) - \left( \frac{4}{2} + 2 \right) = (8 + 4) - (2 + 2) = 12 - 4 = 8$$ Thus, we have: $$\frac{1}{k} \times 8 = 1 \quad \Rightarrow \quad k = 8$$ ### (b) Obtain the distribution function $$F(x)$$ The cumulative distribution function (C.D.F) $$F(x)$$ is found by integrating the p.d.f. $$f(x)$$ from the lower bound of the distribution (here, 2) to $$x$$. $$F(x) = \int_{2}^{x} f(t) \, dt = \int_{2}^{x} \frac{t+1}{8} \, dt$$ For $$2 \leq x \leq 4$$, we can compute the integral: $$F(x) = \frac{1}{8} \int_{2}^{x} (t+1) \, dt = \frac{1}{8} \left[ \frac{t^2}{2} + t \right]_{2}^{x}$$ Substitute the limits: $$F(x) = \frac{1}{8} \left( \left( \frac{x^2}{2} + x \right) - \left( \frac{2^2}{2} + 2 \right) \right)$$ $$= \frac{1}{8} \left( \frac{x^2}{2} + x - (2 + 2) \right) = \frac{1}{8} \left( \frac{x^2}{2} + x - 4 \right)$$ Thus, the C.D.F. for $$2 \leq x \leq 4$$ is: $$F(x) = \frac{1}{8} \left( \frac{x^2}{2} + x - 4 \right)$$ For $$x < 2$$, $$F(x) = 0$$, and for $$x > 4$$, $$F(x) = 1$$. ### (c) Using $$F(x)$$, compute $$P(X \geq 3)$$ Using the C.D.F., the probability $$P(X \geq 3)$$ is: $$P(X \geq 3) = 1 - F(3)$$ Let's compute $$F(3)$$ using the expression for $$F(x)$$: $$F(3) = \frac{1}{8} \left( \frac{3^2}{2} + 3 - 4 \right) = \frac{1}{8} \left( \frac{9}{2} + 3 - 4 \right)$$ $$= \frac{1}{8} \left( 4.5 + 3 - 4 \right) = \frac{1}{8} \times 3.5 = \frac{3.5}{8}$$ Thus: $$F(3) = 0.4375$$ Now, calculate $$P(X \geq 3)$$: $$P(X \geq 3) = 1 - 0.4375 = 0.5625$$ ### (d) Find Mean and Variance of $$X$$ The mean $$\mu$$ of a continuous random variable is given by: \[

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Math Problem Analysis

Mathematical Concepts

Probability Density Functions (PDF)
Cumulative Distribution Functions (CDF)
Continuous Random Variables
Integral Calculus
Mean and Variance of Random Variables

Formulas

k = 1 / Integral of f(x)
F(x) = Integral of f(t) from 2 to x
P(X ≥ 3) = 1 - F(3)
Mean (μ) = Integral of x * f(x)
Variance (Var(X)) = Integral of (x - μ)^2 * f(x)

Theorems

Fundamental Theorem of Calculus
Properties of Continuous Probability Distributions

Suitable Grade Level

Undergraduate (Calculus and Probability Courses)