The concept of determining medians and percentiles from Probability Density Functions (PDFs) is crucial for a comprehensive understanding of continuous random variables. This aspect of statistics enables us to quantify and interpret the behavior of a dataset or distribution, providing a deeper insight into its characteristics.

Continuous Random Variables and PDFs

• Continuous Random Variables: Can take infinite values within a range.
• Probability Density Function (PDF): A curve where the area under it between two points represents the probability of the variable falling within that range. Characteristics include:
  • Non-Negativity: Always ≥ 0.
  • Total Area = 1: Integral over its range equals 1.

Medians and Percentiles

• Median: The point dividing the distribution so 50% is to its left.
  • Calculation: Find the value where the area under the PDF to the left is 0.5.
• Percentiles: Points below which a certain percentage of data falls.
  • Calculation: For the $p^{th}$ percentile, find the value where the area under the PDF to the left equals $\frac{p}{100}$.

• Image courtesy of Online Stat

Example Problems

Example 1: Finding the Median

Calculate the median of a uniform distribution where ( X ) is distributed evenly between 0 and 10.

Solution:

• PDF: $f(x) = 0.1$ for $0 \leq x \leq 10$.
• Median: For a uniform distribution, the median is the midpoint of the range.
• Computation: Median $= \frac{0 + 10}{2} = 5$.

Conclusion: The median of the distribution is 5.

Example 2: Calculating the 25th Percentile

Determine the 25th percentile for a continuous random variable $Y$ with PDF $f(y) = 2y$ for $0 \leq y \leq 1$.

Solution:

• 25th Percentile: The point where 25% of the values lie, or the CDF is 0.25.
• Integral Setup: Integrate $f(y) = 2y$ from 0 to $y$ to find the CDF.
• Calculation: Solve $\int_{0}^{y} 2y \, dy = 0.25$.
• Solve for $y$: Find $y$ when $y^2 = 0.25$, which gives $y = \sqrt{0.25} = 0.5$.

Conclusion: The 25th percentile of $Y$ is 0.5.