Problem 6 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

The random variable \(Z\) has probability density function \(f(z)=3 z^{2} / 19\) for \(2 \leq z \leq 3\) and \(f(z)=0\) elsewhere. Determine \(\mathrm{E}[Z]\). Before you do the calculation: will the answer lie closer to 2 than to 3 or the other way around?

Step-by-Step Solution

Answer

The expected value \(\mathrm{E}[Z] \approx 2.5658\) is closer to 3.

Step 1: Understanding the Problem

We need to find the expected value of a continuous random variable \(Z\) with a given probability density function (PDF) \(f(z) = \frac{3z^2}{19}\) for the interval \(2 \leq z \leq 3\). The expected value \(\mathrm{E}[Z]\) is calculated using the formula \[\mathrm{E}[Z] = \int_{a}^{b} z f(z) \, dz\] where \(a = 2\) and \(b = 3\).

Step 2: Set Up the Integral for Expected Value

To find \(\mathrm{E}[Z]\), set up the definite integral as follows: \[\mathrm{E}[Z] = \int_{2}^{3} z \left(\frac{3z^2}{19}\right) \, dz\] Simplify this to \[\mathrm{E}[Z] = \frac{3}{19} \int_{2}^{3} z^3 \, dz\] Now, we will calculate this integral.

Step 3: Evaluate the Integral

Evaluate the integral \(\int_{2}^{3} z^3 \, dz\). First, find the antiderivative of \(z^3\), which is \(\frac{z^4}{4}\). Evaluate from 2 to 3:\[\int_{2}^{3} z^3 \, dz = \left[\frac{z^4}{4}\right]_{2}^{3} = \frac{3^4}{4} - \frac{2^4}{4}\] This simplifies to \[\frac{81}{4} - \frac{16}{4} = \frac{65}{4}\].

Step 4: Calculate Expected Value

Now substitute the result of the integral back into the formula for \(\mathrm{E}[Z]\):\[\mathrm{E}[Z] = \frac{3}{19} \cdot \frac{65}{4}\] Simplify this expression to get \[\mathrm{E}[Z] = \frac{195}{76}\]. Convert to decimal form: \[\frac{195}{76} \approx 2.5658\].

Step 5: Compare to Given Range

The calculated expected value \(\mathrm{E}[Z] = 2.5658\) approximates to a value between 2 and 3. Since 2.5658 is closer to 3 than to 2, the expectation is slightly skewed towards the upper bound of the interval.

Key Concepts

Probability Density Function

The concept of a probability density function (PDF) is fundamental to understanding continuous random variables. A PDF, represented as \( f(z) \) in our equation, provides a way to describe the likelihood of a continuous random variable, like \( Z \), taking on a particular value within an interval. This means the PDF helps us calculate probabilities over a range of values rather than at any distinct point. For example, the PDF for the random variable \( Z \) is given by \( f(z) = \frac{3z^2}{19} \) within the range \( 2 \leq z \leq 3 \) and is zero outside this range.
This means that if you were to graph this PDF, the curve would only exist between \( z = 2 \) and \( z = 3 \), ensuring the total area under the curve equals 1. This area represents the entire probability, confirming the function describes a valid probability distribution. The shape of this function shows how the likelihood of \( Z \) changes as \( z \) varies from 2 to 3.

Continuous Random Variable

A continuous random variable is one that can take an infinite number of possible values. Unlike discrete variables, which have distinct, separate values (like the roll of a die), continuous variables can be any value within a given range (like measuring the exact height of a plant). The random variable \( Z \) in our example is continuous because it takes on a continuum of possible values within the interval \( 2 \leq z \leq 3 \).
This is important because, for continuous variables, we cannot point to a single probability for a specific value but rather within a range. The probability density function (PDF) helps with this by indicating probabilities over intervals. The expected value, which we are calculating, is essentially a type of average that gives us a sense of the "center" of the distribution for \( Z \) over the interval.

Definite Integral

To calculate the expected value of the continuous random variable \( Z \), we use the concept of a definite integral. This is a powerful tool in calculus that helps us find the exact total area under the curve of a function over a specific interval. In this case, we need to evaluate the integral \( \int_{2}^{3} z \cdot f(z) \, dz \), which simplifies to \( \frac{3}{19} \int_{2}^{3} z^3 \, dz \).
The definite integral yields the accumulated value of the function \( z^3 \) multiplied by the scaling factor \( \frac{3}{19} \) over the interval from 2 to 3. The result of this integration process gives us the average or expected value \( \mathrm{E}[Z] \) for our random variable \( Z \).
Overall, definite integrals are essential for calculating expectations, providing the means to incorporate and balance the contributions of all values \( Z \) might take over its range.

Antiderivative

When attempting to solve definite integrals, one crucial step is finding the antiderivative. An antiderivative is essentially the reverse of a derivative—it uncovers the original function from its rate of change. For our problem, we identified the function's antiderivative, starting from \( z^3 \), as \( \frac{z^4}{4} \).
Consider the purpose of an antiderivative: it allows us to compute the integral by evaluating the newly found function over the interval's boundaries. In our example, we calculated \( \left[\frac{z^4}{4}\right]_{2}^{3} \), providing us \( \frac{81}{4} - \frac{16}{4} = \frac{65}{4} \).
This step simplifies the process of integration, enabling us to determine the total effect or contribution of the function \( z^3 \) across the interval from 2 to 3. Mastery of finding antiderivatives is key to efficiently solving problems involving definite integrals.