Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The median is 2.

Step 1: Understand the Problem

We need to find the median of a probability density function (PDF) given by \( f(x) = \frac{15}{512} x^{2}(4-x)^{2} \) for \( 0 \leq x \leq 4 \) without performing any integration.

Step 2: Identify the PDF Characteristics

The function \( f(x) = \frac{15}{512} x^{2}(4-x)^{2} \) describes a symmetrical distribution. As a quadratic function, it has reflectional symmetry around the line \( x = 2 \). This suggests that the median will also lie at \( x = 2 \), the line of symmetry.

Step 3: Use Symmetry to Determine the Median

Since the PDF is symmetric about \( x = 2 \), the median, which is the value dividing the distribution into two equal halves, must also lie at this point of symmetry. Therefore, \( x = 2 \) is the median of the distribution.

Key Concepts

Symmetrical Distributions

When we talk about symmetrical distributions, we are referring to a situation where the shape of the distribution on one side of the central point is a mirror image of the shape on the other side. Such distributions are very predictable. The central point of a symmetrical distribution, which is often its mean, is where you'd expect most of the data to cluster.

In the exercise, the probability density function (PDF) is given by

• \( f(x) = \frac{15}{512} x^{2}(4-x)^{2} \)
• for \( 0 \leq x \leq 4 \)

This function is symmetric around the line \( x = 2 \). You can visualize it like an arch centered on that point. This symmetry is crucial for simplifying problems, such as finding the median, as we'll see later on.

Probability Density Function

A probability density function (PDF) describes the likelihood of a random variable taking on a particular value. For continuous random variables, the PDF provides a function, where the integral over a specific interval, gives the probability that the variable falls within that interval.

The PDF must satisfy two basic properties:

• The function is non-negative everwyhere: \( f(x) \geq 0 \).
• The integral across its entire range equals 1: \( \int_{-\infty}^{\infty} f(x) \, dx = 1 \).

In our specific case, the function \( f(x) = \frac{15}{512} x^{2}(4-x)^{2} \) represents how the probabilities are distributed over the interval \( 0 \leq x \leq 4 \). Visualize this as a curve, where the area under the curve equals 1. Each point on this curve indicates the density of probabilities at that very point, but not the probability per se, since this is a continuous variable.

Median Calculation

To calculate the median of a probability distribution, you need to find the value where the distribution is divided into two equal halves. This value is the point at which there is a 50% chance that the random variable takes a value less than the median, and a 50% chance that it's greater.

For symmetrical distributions like the one in the exercise, you can find the median easily. Since symmetry suggests the distribution is mirrored around its center, the median must lie on the line of symmetry. Here, we have the PDF:

• \( f(x) = \frac{15}{512} x^{2}(4-x)^{2} \)
• with symmetry about \( x = 2 \).

Without any heavy mathematical lifting or integrations, you can determine that the median is simply \( x = 2 \). This conclusion stems from the characteristic that in a symmetric distribution, the median equals the mean, and both lie on the axis of symmetry.