Math Problem Statement

Solve this equation fully

A continuous random variable X has PDF f(x) where f(x) = cx^2 for 1 ≤ x ≤ 4 and 0 otherwise. Solve for the following: a) Find the value of c. b) Find F(x). c) Find the median of X. d) Find the mode of X.

Solution

Let's solve each part of the problem step by step.

Given:

cx^2 & \text{for } 1 \leq x \leq 4 \\ 0 & \text{otherwise} \end{cases}$$ #### **a) Find the value of $$c$$.** To find $$c$$, we need to use the fact that the total probability of a probability density function (PDF) over its entire domain must equal 1: $$\int_{-\infty}^{\infty} f(x) \, dx = 1.$$ Since $$f(x) = 0$$ outside the interval $$[1, 4]$$, we only need to integrate within this interval: $$\int_1^4 cx^2 \, dx = 1.$$ Let's calculate this: $$\int_1^4 cx^2 \, dx = c \int_1^4 x^2 \, dx.$$ The integral of $$x^2$$ is: $$\int x^2 \, dx = \frac{x^3}{3}.$$ Now apply the limits: $$c \left[\frac{x^3}{3}\right]_1^4 = c \left(\frac{4^3}{3} - \frac{1^3}{3}\right).$$ $$= c \left(\frac{64}{3} - \frac{1}{3}\right).$$ $$= c \left(\frac{63}{3}\right).$$ $$= 21c.$$ Set this equal to 1: $$21c = 1.$$ $$c = \frac{1}{21}.$$ So, $$c = \frac{1}{21}$$. --- #### **b) Find $$F(x)$$, the cumulative distribution function (CDF).** The cumulative distribution function $$F(x)$$ is defined as: $$F(x) = \int_{-\infty}^x f(t) \, dt.$$ Since $$f(x) = 0$$ for $$x < 1$$, we have: $$F(x) = 0 \text{ for } x < 1.$$ For $$1 \leq x \leq 4$$: $$F(x) = \int_1^x \frac{1}{21} t^2 \, dt.$$ The integral of $$t^2$$ is: $$\int t^2 \, dt = \frac{t^3}{3}.$$ Now apply the limits from 1 to $$x$$: $$F(x) = \frac{1}{21} \left[\frac{t^3}{3}\right]_1^x.$$ $$F(x) = \frac{1}{21} \left(\frac{x^3}{3} - \frac{1^3}{3}\right).$$ $$F(x) = \frac{1}{21} \left(\frac{x^3 - 1}{3}\right).$$ $$F(x) = \frac{x^3 - 1}{63}.$$ For $$x > 4$$, $$F(x) = 1$$. So, the CDF is: $$F(x) = \begin{cases} 0 & \text{for } x < 1, \\ \frac{x^3 - 1}{63} & \text{for } 1 \leq x \leq 4, \\ 1 & \text{for } x > 4. \end{cases}$$ --- #### **c) Find the median of $$X$$.** The median $$m$$ is the value such that $$F(m) = 0.5$$. From the CDF for $$1 \leq x \leq 4$$: $$F(m) = \frac{m^3 - 1}{63} = 0.5.$$ Solving for $$m$$: $$\frac{m^3 - 1}{63} = 0.5.$$ $$m^3 - 1 = 31.5.$$ $$m^3 = 32.5.$$ $$m = \sqrt[3]{32.5}.$$ Thus, the median is $$m = \sqrt[3]{32.5} \approx 3.17$$. --- #### **d) Find the mode of $$X$$.** The mode is the value of $$x$$ that maximizes $$f(x)$$. Since $$f(x) = cx^2$$ for $$1 \leq x \leq 4$$, and $$x^2$$ increases as $$x$$ increases, the mode is at the upper bound of the interval, which is $$x = 4$$. Thus, the mode is $$4$$. --- Would you like further details or have any questions? Here are 5 related questions for further exploration: 1. What is the expected value $$E(X)$$ for this distribution? 2. How do we calculate the variance $$\text{Var}(X)$$ of this random variable? 3. Can you find the skewness of this distribution? 4. How would you determine if $$f(x)$$ is a valid probability density function without integration? 5. What are the implications of having a mode at the boundary of the interval for $$f(x)$$? **Tip:** Remember that for a continuous random variable, the PDF can help find probabilities over intervals, but not at specific points since $$P(X = x) = 0$$ for any specific $$x$$.

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Math Problem Analysis

Mathematical Concepts

Probability Density Function (PDF)
Cumulative Distribution Function (CDF)
Median
Mode
Integration

Formulas

∫[1, 4] f(x) dx = 1
F(x) = ∫[1, x] f(t) dt
Median m satisfies F(m) = 0.5
Mode is the value of x that maximizes f(x)

Theorems

Probability Density Function Theorem (Total Probability = 1)
Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12 or Undergraduate

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