Problem 8 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 7 through \(10, f(x)\) is a probability density function for a particular continuous random variable \(X\). In each case, find the indicated probabilities. $$ \begin{aligned} &f(x)= \begin{cases}\frac{1}{x^{2}} & \text { if } x \geq 1 \\ 0 & \text { if } x<1\end{cases} \\ &P(1 \leq X \leq 3) \text { and } P(X \geq 2) \end{aligned} $$

Step-by-Step Solution

Answer

P(1 \leq X \leq 3) = \frac{2}{3} and P(X \geq 2) = \frac{1}{2}

Step 1: Understand the probability density function

The given function is a piecewise function: \[ f(x) = \begin{cases} \frac{1}{x^{2}} & \text{if } x \geq 1 \ 0 & \text{if } x < 1 \end{cases} \] This function defines the probability density function (PDF) for the random variable \( X \).

Step 2: Find the probability \(P(1 \leq X \leq 3)\)

To find \(P(1 \leq X \leq 3)\), we need to integrate the PDF from 1 to 3: \[ P(1 \leq X \leq 3) = \int_{1}^{3} \frac{1}{x^{2}} \,dx \] The antiderivative of \(\frac{1}{x^{2}}\) is \(-\frac{1}{x}\). So, \[ P(1 \leq X \leq 3) = \left[ -\frac{1}{x} \right]_{1}^{3} = \left(-\frac{1}{3} \right) - \left(-1\right) = 1 - \frac{1}{3} = \frac{2}{3} \]

Step 3: Find the probability \(P(X \geq 2)\)

To find \(P(X \geq 2)\), we need to integrate the PDF from 2 to infinity: \[ P(X \geq 2) = \int_{2}^{\infty} \frac{1}{x^{2}} \,dx \] The antiderivative of \(\frac{1}{x^{2}}\) is \(-\frac{1}{x}\). So, \[ P(X \geq 2) = \left[ -\frac{1}{x} \right]_{2}^{\infty} = 0 - (-\frac{1}{2}) = \frac{1}{2} \]

Key Concepts

Continuous Random Variable

A continuous random variable is a variable that can take on an infinite number of values within a given range. Unlike a discrete random variable, which has specific values, a continuous random variable can represent any value within an interval, making it ideal for representing real-world phenomena like time, temperature, or, in our case, probabilities.To work with continuous random variables, we often use a probability density function (PDF) which helps in determining the likelihood of the variable falling within a particular range. The area under the PDF curve within a specified range gives the probability that the continuous random variable falls within that range.

Integration

Integration is a fundamental concept in calculus that helps in computing the accumulated quantities, like areas under curves. When dealing with a probability density function (PDF) of a continuous random variable, we use integration to find the probability that the variable falls within a specific range.For example, to determine the probability that our random variable X lies between 1 and 3, we integrate the PDF from 1 to 3:\[ P(1 \leq X \leq 3) = \int_{1}^{3} \frac{1}{x^{2}} \,dx \]In this exercise, the integral gives us the exact probability by calculating the area under the curve of \( f(x) = \frac{1}{x^{2}} \) from 1 to 3.

Piecewise Function

A piecewise function is a function that is defined by different expressions based on the input value. These expressions are valid over different intervals.In the given exercise, the function \( f(x) \) is defined as:\[ f(x) = \begin{cases} \frac{1}{x^{2}} & \text{if } x \geq 1 \ 0 & \text{if } x < 1 \end{cases} \]This means that for all values of x greater than or equal to 1, the function is \( \frac{1}{x^{2}} \), and for values less than 1, the function is zero. Due to this piecewise nature, we need to carefully choose the correct portion of the function while integrating over a specified interval.

Antiderivative

An antiderivative, also known as an indefinite integral, is the reverse process of differentiation. It helps us in finding a function whose derivative is the given function.In our problem, the given PDF is \( f(x) = \frac{1}{x^{2}} \). To solve integrals like \( \int_{1}^{3} \frac{1}{x^{2}} \,dx \), we need to find the antiderivative of \( f(x) \). The antiderivative of \( \frac{1}{x^{2}} \) is \( -\frac{1}{x} \), as the derivative of \( -\frac{1}{x} \) returns \( \frac{1}{x^{2}} \).Proceeding with solving the integral using the antiderivative, we compute:\[ P(1 \leq X \leq 3) = \left[ -\frac{1}{x} \right]_{1}^{3} = \left(-\frac{1}{3} \right) - \left(-1\right) = 1 - \frac{1}{3} = \frac{2}{3} \]By finding antiderivatives, we simplify complex integrals into manageable calculations.