Math Problem Statement
Solution
The uploaded image contains assignments on "Probability and Random Processes." I will proceed by analyzing the problems and providing solutions step by step.
Problem P.1
The problem describes a probability density function (pdf) as follows:
\begin{cases} Cx^2 & \text{if } 0 \leq x \leq 1 \\ C(2 - x) & \text{if } 1 < x \leq 2 \\ 0 & \text{otherwise} \end{cases}$$ **Questions:** 1. (a) Find the constant $$C$$. 2. (b) Find $$\Pr(0.75 \leq X \leq 1.5)$$. 3. (c) Find the value of $$\alpha$$ such that $$\Pr(X \leq \alpha) = 0.9$$. ### Solution to P.1 **(a) Finding $$C$$** To find the constant $$C$$, we need to use the property that the total probability must be 1: $$\int_{-\infty}^{\infty} f_X(x) dx = 1.$$ Since $$f_X(x)$$ is non-zero only from 0 to 2, we have: $$\int_0^1 Cx^2 dx + \int_1^2 C(2 - x) dx = 1.$$ Calculating each integral: $$\int_0^1 Cx^2 dx = C \int_0^1 x^2 dx = C \left[ \frac{x^3}{3} \right]_0^1 = \frac{C}{3},$$ $$\int_1^2 C(2 - x) dx = C \int_1^2 (2 - x) dx = C \left[ 2x - \frac{x^2}{2} \right]_1^2 = C \left( 4 - 2 - (2 - 0.5) \right) = \frac{C}{2}.$$ Adding them: $$\frac{C}{3} + \frac{C}{2} = 1 \Rightarrow \frac{2C}{6} + \frac{3C}{6} = 1 \Rightarrow \frac{5C}{6} = 1 \Rightarrow C = \frac{6}{5}.$$ **(b) Finding $$\Pr(0.75 \leq X \leq 1.5)$$** $$\Pr(0.75 \leq X \leq 1.5) = \int_{0.75}^1 \frac{6}{5} x^2 dx + \int_1^{1.5} \frac{6}{5} (2 - x) dx.$$ Calculate each integral separately: $$\int_{0.75}^1 \frac{6}{5} x^2 dx = \frac{6}{5} \left[ \frac{x^3}{3} \right]_{0.75}^1 = \frac{6}{15} (1 - 0.421875) = 0.23125,$$ $$\int_1^{1.5} \frac{6}{5} (2 - x) dx = \frac{6}{5} \left[ 2x - \frac{x^2}{2} \right]_1^{1.5} = 0.27.$$ Adding both: $$\Pr(0.75 \leq X \leq 1.5) = 0.23125 + 0.27 = 0.50125.$$ **(c) Finding $$\alpha$$ such that $$\Pr(X \leq \alpha) = 0.9$$** To find $$\alpha$$, we solve: $$\int_0^{\alpha} f_X(x) dx = 0.9.$$ ### Problem P.2 A Gaussian random variable $$x$$ is with mean $$\mu = 1$$ and variance $$\sigma^2 = 3$$. Find: 1. **(a)** $$\Pr(X > \sqrt{3})$$ and $$\Pr(X < 1)$$ 2. **(b)** $$\Pr(|X - 1| > 12)$$ and $$\Pr(|X - 1| < 6)$$ 3. **(c)** $$\Pr(X > 1 + \sqrt{3})$$ ### P.2 Solution Follow to find: 1. **Tip:** Use Gaussian table. **Need Clarifications?** Let me know.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Probability Density Functions
Gaussian Distribution
Integration
Conditional Probability
Formulas
f_X(x) = Cx^2 for 0 ≤ x ≤ 1
f_X(x) = C(2 - x) for 1 < x ≤ 2
Total Probability ∫ f_X(x) dx = 1
Gaussian Distribution PDF: f(x) = (1 / √(2πσ^2)) * e^(-(x - μ)^2 / 2σ^2)
Theorems
Fundamental Theorem of Calculus
Properties of Probability Density Functions
Gaussian Distribution Properties
Suitable Grade Level
Undergraduate (University Level)
Related Recommendation
Probability Problems with Normal Distribution: Calculating Probabilities for Continuous Random Variables
Normal and Exponential Distribution Problems: Cosmic Radiation and Probability Density
Conditional Probability and Expected Cost Calculation for Random Variable X
Solve Probability Questions on Bank Service Times, Signal Density, and Wind Speeds
Solve Probability and Distribution Problems for Discrete and Continuous Variables