91Ó°ÊÓ

In statistics, the **probability density function (PDF)** is a function that describes the likelihood of a random variable to take on a particular value. For a continuous random variable, the PDF, denoted as \( f(x) \), satisfies the property that the area under the curve of \( f(x) \) over an interval represents the probability that the random variable falls within that interval.
For example, if we have \( f(x) = \frac{1}{18} x \) defined on \( 0 \leq x \leq 6 \), the area under this curve from 0 to 6 is equal to 1, ensuring it is a valid PDF. This property ensures that the total probability over all possible values is 1. Understanding the PDF is crucial for calculations involving probabilities, expectations, and other properties of random variables.

A **random variable** is a numerical outcome of a random phenomenon. It assigns a value to each outcome in a sample space. There are two types of random variables: discrete and continuous.

• **Discrete random variables** take on countable values, like the roll of a die (1, 2, 3, 4, 5, 6).
• **Continuous random variables** take on an infinite number of values within a given range, such as the height of individuals or the time it takes to run a race.

In our exercise, \( X \) is a continuous random variable defined by the PDF \( f(x) = \frac{1}{18} x \) over the interval \( 0 \leq x \leq 6 \). The task is to find its median, which involves integrating the PDF up to a point where the cumulative probability equals 0.5.

**Integration** is a fundamental concept in calculus and is extensively used in statistics. It helps in finding areas under curves, which is crucial for working with continuous random variables and their PDFs.
In the context of our exercise, integration is used to calculate the cumulative distribution function (CDF), which represents the probability that the random variable \( X \) is less than or equal to a certain value.
The integral of the PDF \( f(x) = \frac{1}{18} x \) from 0 to \( M \) is computed as follows: \ \( \int_{0}^{M} \frac{1}{18} x \, dx = \frac{1}{18} \int_{0}^{M} x \, dx = \frac{1}{18} \left[ \frac{x^2}{2} \right]_{0}^{M} = \frac{M^2}{36} \). This integration step is necessary to solve for the median.

• Use of integration helps convert PDF to CDF.
• Integration helps calculate expectations and variances of random variables.

**Median** of a continuous random variable \( X \) with a given PDF is the value \( M \) where the probability that \( X \) is less than or equal to \( M \) (the CDF evaluated at \( M \)) equals 0.5. It divides the distribution into two equal halves.
In the exercise, we solved for the median by setting up the integral of \( f(x) \) from \( 0 \) to \( M \) equal to 0.5. We had: \ \( \int_{0}^{M} \frac{1}{18} x \, dx = \frac{1}{2} \).
Performing the integration: \ \( \frac{M^2}{36} = \frac{1}{2} \).
Solving for \( M \): \ \( M^2 = 18 \), then \( M = \sqrt{18} = 3\sqrt{2} \).

• This median value indicates that 50% of the probability lies to the left of \( M \).
• Median provides a measure of central tendency especially useful in skewed distributions.

In summary, finding \ the median involves integrating the PDF and solving for the point where the cumulative probability equals 0.5.