Math Problem Statement

Let X be a continuous random variable with PDF f_X(x)= \begin{cases} kx^2, & 0\leq x \leq 1\ 0, & \text{otherwise.} \end{cases}fX​(x)={kx2,0,​0≤x≤1otherwise.​ Find the median of X. Enter the answer correct to four decimal places.

Hint: The median m is the value such that P(X \leq m) = P(X > m) = 0.5P(X≤m)=P(X>m)=0.5

Solution

We are given a probability density function (PDF) for a continuous random variable XX as:

\begin{cases} kx^2, & 0 \leq x \leq 1, \\ 0, & \text{otherwise.} \end{cases}$$ ### Step 1: Determine the constant $$k$$ To find $$k$$, we use the fact that the total area under the PDF must equal 1, since the total probability must sum to 1. This means we need to solve the equation: $$\int_0^1 f_X(x) \, dx = 1.$$ Substituting $$f_X(x) = kx^2$$: $$\int_0^1 kx^2 \, dx = 1.$$ Now, calculate the integral: $$k \int_0^1 x^2 \, dx = k \left[ \frac{x^3}{3} \right]_0^1 = k \times \frac{1^3}{3} = \frac{k}{3}.$$ So we have: $$\frac{k}{3} = 1.$$ Solving for $$k$$: $$k = 3.$$ Thus, the PDF becomes: $$f_X(x) = \begin{cases} 3x^2, & 0 \leq x \leq 1, \\ 0, & \text{otherwise.} \end{cases}$$ ### Step 2: Find the median $$m$$ The median $$m$$ is defined as the value such that: $$P(X \leq m) = 0.5.$$ This is equivalent to finding $$m$$ such that the cumulative distribution function (CDF) up to $$m$$ equals 0.5: $$\int_0^m 3x^2 \, dx = 0.5.$$ First, calculate the integral: $$3 \int_0^m x^2 \, dx = 3 \left[ \frac{x^3}{3} \right]_0^m = m^3.$$ Now, set this equal to 0.5: $$m^3 = 0.5.$$ Solving for $$m$$: $$m = \sqrt[3]{0.5} \approx 0.7937.$$ ### Final Answer: The median of $$X$$ is approximately $$\boxed{0.7937}$$. Would you like further details or have any questions? Here are 5 related questions: 1. How do you compute the expectation $$E[X]$$ for this distribution? 2. What is the variance of $$X$$ for the given PDF? 3. How do you find the mode of this distribution? 4. Can you derive the CDF $$F_X(x)$$ from the PDF? 5. What is the probability that $$X$$ is greater than the median? **Tip:** Always ensure that the area under a probability density function equals 1, as this is a key property of PDFs.

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Math Problem Analysis

Mathematical Concepts

Probability Density Function (PDF)
Continuous Random Variables
Cumulative Distribution Function (CDF)

Formulas

∫[0 to 1] f_X(x) dx = 1 (PDF normalization)
f_X(x) = kx^2
∫[0 to m] 3x^2 dx = 0.5 (Median calculation)

Theorems

Fundamental Theorem of Calculus
Property of Continuous Random Variable Median: P(X ≤ m) = 0.5

Suitable Grade Level

Undergraduate Probability and Statistics

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