Problem 72 - Calculus and its applications

Question

Median. Let \(x\) be a continuous random variable over \([a, b]\) with probability density function \(f .\) Then the median of the \(x\) -values is that number \(m\) for which $$ \int_{a}^{m} f(x) d x=\frac{1}{2} $$ Find each median. $$ f(x)=\frac{3}{2} x^{2},[-1,1] $$

Step-by-Step Solution

Answer

The median is 0.

Step 1: Understand the Problem

We need to find the median \(m\) of a continuous random variable with the given probability density function \(f(x) = \frac{3}{2} x^2\) over the interval \([-1, 1]\). The median \(m\) satisfies the equation \(\int_{-1}^{m} f(x) \, dx = \frac{1}{2}\).

Step 2: Set Up the Integral

To find \(m\), set up the integral of \(f(x)\) from \(a = -1\) to \(m\) such that it equals \(\frac{1}{2}\). Write the equation as \(\int_{-1}^{m} \frac{3}{2} x^2 \, dx = \frac{1}{2}\).

Step 3: Solve the Integral

First, find the indefinite integral of \(\frac{3}{2}x^2\). The antiderivative is \(\frac{3}{2} \cdot \frac{x^3}{3} = \frac{x^3}{2}\). Now use the definite integral formula to find \(\int_{-1}^{m} \frac{3}{2} x^2 \, dx = \left. \frac{x^3}{2} \right|_{-1}^m = \frac{m^3}{2} - \frac{(-1)^3}{2}\).

Step 4: Set Equation to Solve for m

The equation from the integral is \(\frac{m^3}{2} - \left(-\frac{1}{2}\right) = \frac{1}{2}\). Simplify to get \(\frac{m^3}{2} + \frac{1}{2} = \frac{1}{2}\).

Step 5: Find the Value of m

Subtract \(\frac{1}{2}\) from both sides to get \(\frac{m^3}{2} = 0\). Multiply by 2 to solve for \(m^3\), resulting in \(m^3 = 0\). Solve for \(m\) to find that \(m = 0\).

Key Concepts

Probability Density Function

The concept of a Probability Density Function (PDF) is fundamental in the realm of continuous probability distributions. A PDF is a function, denoted as \( f(x) \), that describes the likelihood of a random variable taking on a particular value within its continuous range. Unlike discrete probabilities, the probability of a continuous random variable taking on any single value is zero. This is because the number of possible values is infinite. Instead, probabilities in continuous distributions are calculated over intervals.

• The PDF must satisfy two key conditions: it must be non-negative for all values of \(x\), which means \( f(x) \geq 0 \) for all \(x\).
• The integral of the PDF over the entire space is equal to 1, ensuring that the total probability is accounted for. This is expressed as \( \int_{-\infty}^{\infty} f(x) \, dx = 1 \).

When you have the PDF of a random variable, it enables you to calculate probabilities over specific intervals by integrating the PDF over that interval.

Continuous Random Variable

A Continuous Random Variable is a type of variable that can take on an uncountable number of values within a given range. This range could be bounded or unbounded, depending on the context. For example, the time needed for a chemical reaction or the precise height of individuals in a population are continuous variables as they can take on virtually any real number between certain limits.

• Continuous random variables, unlike discrete ones, do not assume distinct values. Instead, they cover an entire range of possibilities, which is why we use PDFs to map their behavior.
• The value of a continuous random variable within a certain interval reflects its likelihood, and this is where integration plays a crucial role. We use integration because the area under the PDF curve over an interval represents the probability of the variable falling within that interval.

Understanding the nature of continuous random variables helps in appreciating these distributions' real-life applications and phenomena models.

Definite Integral

The Definite Integral is a powerful mathematical tool for calculating the area under a curve described by a function, within a specific interval. In the context of probability distributions, it allows us to derive probabilities from a PDF over a continuous range.

• The Definite Integral, represented as \( \int_{a}^{b} f(x) \, dx \), computes the signed area between the function \(f(x)\) and the x-axis from \(x = a\) to \(x = b\).
• This is crucial in a probability context, as it provides the probability that a continuous random variable falls within the interval \([a, b]\).

To determine a median, for example, we set the integral from one boundary to the median equal to 0.5 (since the median splits the distribution into two equal halves). This equation \( \int_{a}^{m} f(x) \, dx = 0.5 \) is solved to find the median. Understanding integrals thus helps interpret, manipulate, and utilize continuous probability distributions effectively.