Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

A PDF for a continuous random variable \(X\) is given. Use the \(P D F\) to find (a) \(P(X \geq 2)\), (b) \(E(X)\), and (c) the CDF: $$ f(x)= \begin{cases}\frac{\pi}{8} \sin (\pi x / 4), & \text { if } 0 \leq x \leq 4 \\ 0, & \text { otherwise }\end{cases} $$

Step-by-Step Solution

Answer

(a) \(P(X \geq 2) = \frac{1}{2}\); (b) \(E(X) = 2\); (c) \(F(x) = -\frac{1}{2} \cos \left(\frac{\pi x}{4}\right) + \frac{1}{2}\) for \(0 \leq x \leq 4\).

Step 1: Understand the PDF

We are given the probability density function (PDF) for a random variable \(X\). The PDF is defined as \(f(x) = \frac{\pi}{8} \sin \left(\frac{\pi x}{4}\right)\) for \(0 \leq x \leq 4\), and 0 otherwise. This function describes the likelihood of \(X\) taking on a particular value within this interval.

Step 2: Find \(P(X \geq 2)\)

To find \(P(X \geq 2)\), we first calculate the cumulative probability from 2 to 4 using the integral of the PDF: \[ P(X \geq 2) = \int_{2}^{4} \frac{\pi}{8} \sin \left(\frac{\pi x}{4}\right) \, dx. \]Calculate this integral to find the probability.

Step 3: Integrate \(P(X \geq 2)\)

Evaluating the integral from Step 2, we have: \[ \int \frac{\pi}{8} \sin \left(\frac{\pi x}{4}\right) \, dx = -\frac{1}{2} \cos \left(\frac{\pi x}{4}\right) + C, \]where \(C\) is the constant of integration. Applying the limits from 2 to 4 gives:\[ -\frac{1}{2} \left[ \cos \left(\pi\right) - \cos \left(\frac{\pi}{2}\right) \right] = -\frac{1}{2} (-1 - 0) = \frac{1}{2}. \]Thus, \(P(X \geq 2) = \frac{1}{2}\).

Step 4: Find \(E(X)\) (Expected Value of X)

The expected value is found using:\[ E(X) = \int_{0}^{4} x \cdot f(x) \, dx = \int_{0}^{4} x \frac{\pi}{8} \sin \left(\frac{\pi x}{4}\right) \, dx. \]The integration by parts is used here; let \(u = x\) and \(dv = \frac{\pi}{8} \sin \left(\frac{\pi x}{4}\right) \, dx\). This requires multiple steps to solve but ultimately evaluates to 2.

Step 5: Find the CDF \(F(x)\)

The cumulative distribution function (CDF) \(F(x)\) is found by integrating the PDF from \(0\) to \(x\):\[ F(x) = \int_{0}^{x} \frac{\pi}{8} \sin \left(\frac{\pi t}{4}\right) \, dt. \]This calculates to \[ F(x) = -\frac{1}{2} \cos \left(\frac{\pi x}{4}\right) + \frac{1}{2} \text{ for } 0 \leq x \leq 4. \]This function is 0 for \(x < 0\) and 1 for \(x > 4\).

Key Concepts

Continuous Random Variables

A continuous random variable is a type of random variable that can take on an infinite number of possible values. Unlike discrete random variables, which have a countable number of possible outcomes, continuous random variables vary over a continuous range. For example, the amount of rainfall in a day or the exact height of students in a class are continuous random variables.

The behavior of continuous random variables is described by a probability density function (PDF), which helps determine the likelihood of the variable assuming a specific range of values. While the PDF gives insight into the potential distribution of values, it does not give the actual probabilities because the probability at a specific point is technically zero. Instead, probabilities are calculated over intervals.

In our exercise, the variable was defined by the PDF: \( f(x) = \frac{\pi}{8} \sin \left(\frac{\pi x}{4}\right) \) for the interval \( 0 \leq x \leq 4 \).This function tells us where values of the random variable are more likely to occur within this range.

Expected Value

The expected value of a continuous random variable can be viewed as its average or mean value over its entire range. It serves as a measure of the central tendency of the probability distribution. Calculating the expected value involves integrating the product of the variable and its probability density function over the variable's range.

In mathematical terms, the expected value \( E(X) \) is given by the integral:\[ E(X) = \int_{a}^{b} x \, f(x) \, dx \]where \( f(x) \) is the PDF and \( [a, b] \) is the range of the variable.

For our specific problem, the expected value was calculated using:\[ E(X) = \int_{0}^{4} x \, \frac{\pi}{8} \sin \left(\frac{\pi x}{4}\right) \, dx \]which ultimately evaluated to 2, representing the average value the variable can be expected to have between 0 and 4.

Cumulative Distribution Function

The cumulative distribution function (CDF) is a function that describes the probability that a continuous random variable will have a value less than or equal to a specific value. It provides a cumulative probability which accumulates as we move along the range of possible values.

Mathematically, for a probability density function \( f(x) \), the CDF \( F(x) \) is calculated using:\[ F(x) = \int_{a}^{x} f(t) \, dt \]which gives you a function representing the accumulated area under the PDF curve from the beginning of the range up to \( x \).

In our exercise, the CDF was found by integrating the PDF from 0 to \( x \) and resulted in:\[ F(x) = -\frac{1}{2} \cos \left(\frac{\pi x}{4}\right) + \frac{1}{2} \]for \( 0 \leq x \leq 4 \). This CDF can be used to determine the probability of outcomes ranging from any value within the interval.