Math Problem Statement
If X is a continuous random variable with the following probability density function 0 , 2 2 0 2 2 x a x x , < x f(x)= (i) Find the value of “a”. (ii) Find P[X >1].
Solution
Given the probability density function (PDF) of the continuous random variable as:
\begin{cases} a(2 - x^2), & 0 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases}$$ ### Part (i) - Finding the value of $$a$$ For any probability density function, the total area under the curve must equal 1. That is, the integral of $$f(x)$$ over the range of $$X$$ must be 1: $$\int_{0}^{2} a(2 - x^2) \, dx = 1$$ Let’s compute this integral: 1. Split the integral: $$\int_{0}^{2} a(2 - x^2) \, dx = a \int_{0}^{2} (2 - x^2) \, dx$$ 2. Compute the integral: $$a \left[ 2x - \frac{x^3}{3} \right]_0^2 = a \left( 2(2) - \frac{(2)^3}{3} \right)$$ $$= a \left( 4 - \frac{8}{3} \right) = a \left( \frac{12}{3} - \frac{8}{3} \right) = a \left( \frac{4}{3} \right)$$ 3. Set this equal to 1: $$a \cdot \frac{4}{3} = 1$$ $$a = \frac{3}{4}$$ So, the value of $$a$$ is $$\frac{3}{4}$$. ### Part (ii) - Finding $$P(X > 1)$$ To find $$P(X > 1)$$, we need to compute the integral of $$f(x)$$ from 1 to 2: $$P(X > 1) = \int_{1}^{2} f(x) \, dx = \int_{1}^{2} \frac{3}{4} (2 - x^2) \, dx$$ Let’s compute this integral: 1. Split the integral: $$P(X > 1) = \frac{3}{4} \int_{1}^{2} (2 - x^2) \, dx$$ 2. Compute the integral: $$\frac{3}{4} \left[ 2x - \frac{x^3}{3} \right]_1^2 = \frac{3}{4} \left( \left( 2(2) - \frac{(2)^3}{3} \right) - \left( 2(1) - \frac{(1)^3}{3} \right) \right)$$ $$= \frac{3}{4} \left( \left( 4 - \frac{8}{3} \right) - \left( 2 - \frac{1}{3} \right) \right)$$ $$= \frac{3}{4} \left( \left( \frac{12}{3} - \frac{8}{3} \right) - \left( \frac{6}{3} - \frac{1}{3} \right) \right)$$ $$= \frac{3}{4} \left( \frac{4}{3} - \frac{5}{3} \right) = \frac{3}{4} \cdot \frac{-1}{3} = -\frac{3}{12} = -\frac{1}{4}$$ This gives $$P(X > 1) = \frac{1}{4}$$. Thus, $$P(X > 1) = 0.25$$. ### Final Answers: - (i) $$a = \frac{3}{4}$$ - (ii) $$P(X > 1) = 0.25$$ Would you like more detailed explanations on any step or further questions? ### 5 Follow-Up Questions: 1. What is the significance of normalizing a probability density function? 2. Can you calculate $$P(0.5 \leq X \leq 1.5)$$ for the given PDF? 3. How would the PDF change if the range of $$x$$ was extended? 4. How can you use this PDF to calculate the expected value of $$X$$? 5. What are the properties of a continuous random variable's PDF? ### Tip: When integrating polynomials, always double-check your limits and simplify fractions carefully to avoid small calculation errors.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability
Continuous Random Variables
Probability Density Functions
Integration
Formulas
∫(f(x)) dx = 1 for a valid PDF
P(X > 1) = ∫[1 to 2] f(x) dx
Theorems
Fundamental Theorem of Calculus
Properties of Probability Density Functions
Suitable Grade Level
College Level
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