Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Without doing any integration, find the median of the random variable that has \(\mathrm{PDF} \quad f(x)=\frac{15}{512} x^{2}(4-x)^{2}\), \(0 \leq x \leq 4\).

Step-by-Step Solution

Answer

The median is 2.

Step 1: Understanding the Median

The median of a continuous random variable is a value \( m \) such that the probability of the random variable being less than or equal to \( m \) is 0.5. Mathematically, \( P(X \leq m) = 0.5 \). Since we're dealing with a PDF, this means integrating the PDF from 0 to \( m \) should equal 0.5.

Step 2: Recognizing Symmetry

The given PDF \( f(x) = \frac{15}{512} x^{2}(4-x)^{2} \) is symmetric about \( x = 2 \). This occurs because the PDF is a quadratic multiplied by a factor resulting in symmetry since the two terms \( x^2 \) and \((4-x)^2\) are symmetric around \( x=2 \).

Step 3: Conclusion on Median

Due to the symmetry of the PDF about \( x = 2 \), the median is equal to the point of symmetry. This is because the area under the curve from 0 to 2 equals the area from 2 to 4, thereby dividing the area under the curve into two equal halves. Thus, the median \( m \) is 2.

Key Concepts

Continuous Random Variable

A continuous random variable is a type of variable that can assume an infinite number of values within a certain range. Unlike discrete random variables, which have distinct and isolated values, continuous random variables can take on any value in a given interval. This allows them to describe outcomes that can be measured rather than counted.

An example is the random variable representing the time it takes for a student to finish an exam. This time can be any number, like 50.32 minutes, or 75.68 minutes. Continuous random variables are crucial in fields like statistics and engineering, as they help to model real-world scenarios that involve continuous data.

Since these variables can take on infinite values, their probabilities are represented using a function called a Probability Density Function (PDF). This function will be explained more in the next section.

Probability Density Function

The Probability Density Function (PDF) is a fundamental concept when dealing with continuous random variables. A PDF describes the likelihood of a random variable taking on a certain value. For continuous data, the PDF is often a function or curve, and it tells us how probabilities are distributed over the values of the random variable.

Key features of a PDF include:

• The area under the curve of a PDF over the entire range of possible values is equal to 1.
• The probability that a continuous random variable equals a specific value is technically zero since there are infinite possibilities, but we can find probabilities over intervals.

For instance, if we have a PDF that defines the probability of times taken to finish exams, the PDF would help us calculate the probability that the time taken is between 40 and 60 minutes by integrating the function over this interval.

When visualizing a PDF, it is crucial to understand that higher peaks represent a higher likelihood of the random variable being near those values.

Symmetry in PDFs

The concept of symmetry simplifies the analysis of many PDFs. When a PDF is symmetric around a particular point, this symmetry can be a powerful tool for finding central tendendies, such as the mean and the median.

Symmetry in a PDF means that for any value on the left of the symmetry point, there is a corresponding point on the right at the exact same distance with the same height in the PDF. This is analogous to a seesaw perfectly balanced around its center. In our exercise, the PDF is given as \(f(x) = \frac{15}{512} x^{2}(4-x)^{2}\).

This function is symmetric around \(x = 2\) because both components \(x^2\) and \((4-x)^2\) produce the same value at equal distances from 2. Such symmetry is very handy because it helps us determine that the median, a type of center location in the set of values, is also at \(x = 2\) without even having to perform complex calculations like integrations.

Median Calculation

The median of a continuous random variable is a value that divides the probability distribution into two equal halves. For continuous variables with a given PDF, the median \(m\) satisfies the condition that the probability of the variable being less than or equal to \(m\) is 50% (i.e., \(P(X \leq m) = 0.5\)).

In non-symmetric PDFs, finding the median can demand precise integration of the PDF from the lower bound up to \(m\) until the area equals 0.5. However, symmetry adds an advantageous shortcut: if a PDF is symmetric, as seen in previous sections, the median is located exactly at the symmetry point.

In the provided exercise, since the PDF is symmetric around \(x = 2\), we quickly conclude that the median is \(m = 2\) without performing any integral calculations. Symmetry simplifies median calculations by letting us directly use the symmetrically balanced point as our answer.