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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. $$ \begin{aligned} &f(x)= \begin{cases}\frac{1}{x^{2}} & \text { if } x \geq 1 \\ 0 & \text { if } x<1\end{cases} \\ &P(1 \leq X \leq 3) \text { and } P(X \geq 2) \end{aligned} $$

Short Answer

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

Step by step solution

01

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 \( X \).
02

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, \[ P(1 \leq X \leq 3) = \left[ -\frac{1}{x} \right]_{1}^{3} = \left(-\frac{1}{3} \right) - \left(-1\right) = 1 - \frac{1}{3} = \frac{2}{3} \]
03

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, \[ P(X \geq 2) = \left[ -\frac{1}{x} \right]_{2}^{\infty} = 0 - (-\frac{1}{2}) = \frac{1}{2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

PERSONAL HEALTH Jules decides to go on a diet for 6 weeks, with a goal of losing between 10 and 15 pounds. Based on his body configuration and metabolism, his doctor determines that the amount of weight he will lose can be modeled by a continuous random variable \(X\) with probability density function \(f(x)\) of the form \(f(x)= \begin{cases}k(x-10)^{2} & \text { for } 10 \leq x \leq 15 \\ 0 & \text { otherwise }\end{cases}\) If the doctor's model is valid, how much weight should Jules expect to lose? [Hint: First determine the value of the constant \(k\).]

In each of Exercises 3 through 6 , determine whether the given random variable \(X\) is discrete or continuous. \(X\) measures the annual distance flown by a randomly selected airplane from a particular airline.

INSURANCE SALES Let \(X\) be a random variable that measures the time (in minutes) that a person spends with an agent choosing a life insurance policy, and let \(Y\) measure the time (in minutes) the agent spends doing paperwork once the client has selected a policy. Suppose the joint probability density function for \(X\) and \(Y\) is \(f(x, y)= \begin{cases}\frac{1}{300} e^{-x / 30} e^{-y / 10} & \text { for } x \geq 0, y \geq 0 \\ 0 & \text { otherwise }\end{cases}\) a. Find the probability that choosing the policy takes more than 20 minutes. b. Find the probability that the entire transaction (policy selection and paperwork) will take more than half an hour. c. How much more time would you expect to spend selecting the policy than completing the paperwork?

TRAFFIC MANAGEMENT The distance (in feet) between successive cars on a freeway is modeled by the random variable \(X\) with probability density function $$ f(x)= \begin{cases}0.25 x e^{-x / 2} & \text { if } x \geq 0 \\ 0 & \text { otherwise }\end{cases} $$ a. Find the probability that a randomly selected pair of cars will be less than 10 feet apart. b. What is the average distance between successive cars on the freeway?

WHEEL OF FORTUNE A wheel of fortune is divided into 20 circular sectors of equal area. The wheel is spun and a payoff is made according to the color of the region on which the indicator lands. One region is gold and pays \(\$ 50\) when hit; five regions are blue and pay \(\$ 25\); and four other regions are red and pay \(\$ 10\). The remaining 10 regions are black and pay nothing. Would you be willing to pay \(\$ 10\) to play this game?

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