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Chapter 12: Problem 36

A random variable \(X\) has a cumulative distribution function \(F(x)=(x-1)^{2}\) on \(1 \leq x \leq 2 .\) Find \(b\) such that \(\operatorname{Pr}(X \leq b)=\frac{1}{4}.\)

Short Answer

Expert verified
b = 1.5

Step by step solution

01

Understand the Cumulative Distribution Function (CDF)

The CDF is given as: \[ F(x) = (x - 1)^2 \] This function is valid for the interval \(1 \leq x \leq 2\).
02

Set Up the Probability Condition

We need to find \(b\) such that \[\operatorname{Pr}(X \leq b) = \frac{1}{4}\] Since the CDF represents \(\operatorname{Pr}(X \leq x)\), we set \[ F(b) = \frac{1}{4} \]
03

Solve the Equation

Substitute the CDF function into the equation: \[ (b - 1)^2 = \frac{1}{4} \] Solve for \(b\) by taking the square root of both sides: \[ b - 1 = \pm \frac{1}{2} \] Thus, \[ b = 1 + \frac{1}{2} = 1.5 \] or \[ b = 1 - \frac{1}{2} = 0.5 \]
04

Verify the Valid Interval

Check if the solutions for \(b\) fall within the interval \(1 \leq x \leq 2\). Since \(0.5\) is outside this range, we discard it. Therefore, \[ b = 1.5 \] is the correct and valid answer.

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Key Concepts

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

Probability
Probability measures the likelihood of an event occurring, expressed as a value between 0 and 1. An event with a probability of 1 is certain to happen, while an event with a probability of 0 will not occur. In our exercise, we are asked to find a value for the random variable \(b\) such that the probability of \(X \text{ being less than or equal to } b \text{ is } \frac{1}{4}\). This means we seek to find the point where the cumulative probability reaches 0.25.

Key points you'd want to note about probability include:
  • It is always between 0 and 1, inclusive.
  • It provides a way to predict the likelihood of various outcomes.
  • The sum of probabilities for all possible outcomes of an experiment is 1.
Understanding these basics helps us when we engage with more complex concepts such as cumulative distribution functions (CDF) and their applications.
Random Variables
Random variables are quantities whose values result from the outcomes of a random phenomenon. These variables can be either discrete (taking on specific values) or continuous (taking any value within a range). In this exercise, \(X\) is a continuous random variable with a specified range from 1 to 2.
Understanding random variables involves:
  • Knowing they can represent various types of data (e.g., temperature, time, etc.).
  • Distinguishing between discrete and continuous random variables.
  • Recognizing that they have associated probability distributions that describe how probabilities are assigned to different values or ranges.

    In our scenario, the value of \(X\) follows a specific rule given by the cumulative distribution function (CDF) \(F(x) = (x - 1)^2\).
Cumulative Distribution Function (CDF)
The Cumulative Distribution Function (CDF) represents the probability that a random variable \(X\) is less than or equal to a certain value \(x\). For continuous random variables, it is a non-decreasing function that ranges from 0 to 1. In our problem, the CDF is given as \(F(x) = (x - 1)^2\) within the interval \(1 \text{ and }2\).
To solve the problem:
  • We set up the probability condition: \(F(b) = \frac{1}{4}\), where \(b\) is the value we seek.
  • This setup results from the understanding that \(F(b)\) gives the probability \(X \text{ is less than or equal to } b\).
Then we:
  • Substituted the value \(F(b) = (b - 1)^2\) into the probability condition, leading us to the equation: \((b - 1)^2 = \frac{1}{4}\).
  • Solved this equation to find \(b\), yielding \(b - 1 = \frac{1}{2} \text{ or } b - 1 = -\frac{1}{2}\), hence \(b = 1.5 \text{ or } b = 0.5\).
  • Checked the valid interval, finding \(b = 0.5\) invalid as it lies outside \(1 \text{ to } 2\), leaving us with \(b = 1.5\) as the correct answer.
Understanding how to interpret and use the CDF is crucial for solving problems involving probabilities of continuous random variables.

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Most popular questions from this chapter

Two Competing Companies In a certain town, there are two competing taxicab companies, Red Cab and Blue Cab. The taxis mix with downtown traffic in a random manner. There are three times as many Red taxis as there are Blue taxis. Suppose that you stand on a downtown street corner and count the number \(X\) of Red taxis that appear before the first Blue taxi appears. (a) Determine the formula for \(\operatorname{Pr}(X=n).\) (b) What is the likelihood of observing at least three Red taxis before the first Blue taxi? (c) What is the average number of consecutive Red taxis prior to the appearance of a Blue taxi?

A random variable \(X\) has a cumulative distribution function \(F(x)=\frac{1}{4} x^{2}\) on \(0 \leq x \leq 2 .\) Find \(b\) such that \(\operatorname{Pr}(X \leq b)=.09.\)

The time (in minutes) required to complete a certain subassembly is a random variable \(X\) with the density function \(f(x)=\frac{1}{21} x^{2}, 1 \leq x \leq 4.\) (a) Use \(f(x)\) to compute \(\operatorname{Pr}(2 \leq X \leq 3).\) (b) Find the corresponding cumulative distribution function \(F(x).\) (c) Use \(F(x)\) to compute \(\operatorname{Pr}(2 \leq X \leq 3).\)

At a certain bus stop the time between buses is a random variable \(X\) with the density function \(f(x)=6 x(10-x) / 1000,0 \leq x \leq 10 .\) Find the average time between buses.

Consider a circle with radius 1. (a) What percentage of the points lies within \(\frac{1}{2}\) unit of the center? (b) Let \(c\) be a constant with \(0
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