/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Let \(X\) denote the amount of t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(X\) denote the amount of time for which a book on 2 -hour reserve at a college library is checked out by a randomly selected student and suppose that \(X\) has density function $$ f(x)= \begin{cases}.5 x & 0 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases} $$ Calculate the following probabilities: a. \(P(X \leq 1)\) b. \(P(.5 \leq X \leq 1.5)\) c. \(P(1.5

Short Answer

Expert verified
a. 0.25, b. 0.5, c. 0.4375

Step by step solution

01

Understand the Density Function

The given density function is defined as \[ f(x) = \begin{cases} 0.5x & \text{for } 0 \leq x \leq 2 \ 0 & \text{otherwise} \end{cases} \]This function describes how the variable \( X \) is distributed over the interval from 0 to 2.
02

Set Up Part a: Find Probability P(X ≤ 1)

We need to find the probability that \( X \leq 1 \). Since the density function is continuous over the support \([0, 2]\), we integrate the density from 0 to 1:\[ P(X \leq 1) = \int_{0}^{1} 0.5x \, dx \]
03

Calculate Integral for Part a

Calculate the integral:\[ \int 0.5x \, dx = 0.5 \int x \, dx = 0.5 \left(\frac{x^2}{2}\right) = 0.25x^2 \]Evaluating from 0 to 1:\[ P(X \leq 1) = 0.25[1^2 - 0^2] = 0.25 \]
04

Set Up Part b: Find Probability P(0.5 ≤ X ≤ 1.5)

To calculate the probability that 0.5 ≤ X ≤ 1.5, integrate the density function from 0.5 to 1.5:\[ P(0.5 \leq X \leq 1.5) = \int_{0.5}^{1.5} 0.5x \, dx \]
05

Calculate Integral for Part b

Calculate the integral:\[ \int 0.5x \, dx = 0.25x^2 \]Evaluating from 0.5 to 1.5:\[ P(0.5 \leq X \leq 1.5) = 0.25[1.5^2 - 0.5^2] = 0.25[2.25 - 0.25] = 0.25[2] = 0.5 \]
06

Set Up Part c: Find Probability P(1.5 < X)

To calculate the probability that \( X > 1.5 \), integrate the density function from 1.5 to 2:\[ P(1.5 < X) = \int_{1.5}^{2} 0.5x \, dx \]
07

Calculate Integral for Part c

Calculate the integral:\[ \int 0.5x \, dx = 0.25x^2 \]Evaluating from 1.5 to 2:\[ P(1.5 < X) = 0.25[2^2 - 1.5^2] = 0.25[4 - 2.25] = 0.25[1.75] = 0.4375 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Continuous Random Variable
By definition, a continuous random variable takes on an infinite number of values within a given range. Unlike discrete random variables, which have specific, countable outcomes, continuous random variables can assume any value in an interval and are associated with probabilities described by a probability density function (PDF). In our example, the variable \(X\) represents the time a book is checked out at a library, ranging between 0 and 2 hours.
\[\]
A continuous random variable's PDF gives us a way to determine probabilities over intervals rather than at specific values. For \(X\), the PDF is defined as \(f(x) = 0.5x\), indicating how likely different checkout times are within the range \(0 \leq x \leq 2\). Understanding the behavior of \(X\) helps in solving problems like calculating specific probabilities for various intervals of time.
Integral Calculus
Integral calculus is a powerful mathematical tool used to calculate areas under curves, which in probability gives us the total probability between two points on a continuous scale. When dealing with a continuous random variable, we use integration to find the probability of the variable falling within a specific interval. In our example, to find \(P(X \leq 1)\), we compute the integral of the function \(f(x) = 0.5x\) from 0 to 1.
\[\]
The integral \(\int 0.5x\, dx\) gives us the accumulated probability, which represents the area under the curve of the density function up to that point. This is evaluated to solve each part of the problem:
  • For \(P(X \leq 1)\), we integrate from 0 to 1.
  • For \(P(0.5 \leq X \leq 1.5)\), we integrate from 0.5 to 1.5.
  • For \(P(1.5 < X)\), we integrate from 1.5 to 2.
Each of these integrations tells us the probability of \(X\) being within the respective range, demonstrating the usefulness of integral calculus in probability.
Probability Calculation
Probability calculation for continuous random variables involves determining the likelihood that the variable falls within specified limits. Unlike discrete probabilities, probability calculations for continuous variables use integration due to the infinite range of values. For the function \(f(x) = 0.5x\), integration over a particular interval gives the probability for that interval.
\[\]
We compute the integral as follows for each probability:
  • \(P(X \leq 1)\): \(\int_{0}^{1} 0.5x \, dx = 0.25\), indicating a 25% chance of checkout time being 1 hour or less.
  • \(P(0.5 \leq X \leq 1.5)\): \(\int_{0.5}^{1.5} 0.5x \, dx = 0.5\), meaning a 50% probability that time falls between 0.5 and 1.5 hours.
  • \(P(1.5 < X)\): \(\int_{1.5}^{2} 0.5x \, dx = 0.4375\), showing about a 44% chance of the time being more than 1.5 hours.
This illustrates how we use probability calculations in practical scenarios, transforming theoretical math into real-world context by employing calculus techniques to continuously bridging uncertainty into measurable outcomes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(X=\) the hourly median power (in decibels) of received radio signals transmitted between two cities. The authors of the article "Families of Distributions for Hourly Median Power and Instantaneous Power of Received Radio Signals" (J. Research National Bureau of Standards, vol. 67D, 1963: 753-762) argue that the lognormal distribution provides a reasonable probability model for \(X\). If the parameter values are \(\mu=3.5\) and \(\sigma=1.2\), calculate the following: a. The mean value and standard deviation of received power. b. The probability that received power is between 50 and \(250 \mathrm{~dB} .\) c. The probability that \(X\) is less than its mean value. Why is this probability not \(.5\) ?

The distribution of resistance for resistors of a certain type is known to be normal, with \(10 \%\) of all resistors having a resistance exceeding \(10.256\) ohms and \(5 \%\) having a resistance smaller than \(9.671\) ohms. What are the mean value and standard deviation of the resistance distribution?

The article "The Load-Life Relationship for M50 Bearings with Silicon Nitride Ceramic Balls" (Lubrication Engr., 1984: 153-159) reports the accompanying data on bearing load life (million revs.) for bearings tested at a \(6.45 \mathrm{kN}\) load. \(\begin{array}{rrrrrrr}47.1 & 68.1 & 68.1 & 90.8 & 103.6 & 106.0 & 115.0 \\\ 126.0 & 146.6 & 229.0 & 240.0 & 240.0 & 278.0 & 278.0 \\ 289.0 & 289.0 & 367.0 & 385.9 & 392.0 & 505.0 & \end{array}\) a. Construct a normal probability plot. Is normality plausible? b. Construct a Weibull probability plot. Is the Weibull distribution family plausible?

The error involved in making a certain measurement is a continuous rv \(X\) with pdf $$ f(x)=\left\\{\begin{array}{l} 09375\left(4-x^{2}\right) \quad-2 \leq x \leq 2 \\ \text { otherwise } \end{array}\right. $$ a. Sketch the graph of \(f(x)\). b. Compute \(P(X>0)\). c. Compute \(P(-1.5)\).

Consider the following ten observations on bearing lifetime (in hours): \(\begin{array}{lllll}152.7 & 172.0 & 172.5 & 173.3 & 193.0 \\ 204.7 & 216.5 & 234.9 & 262.6 & 422.6\end{array}\) Construct a normal probability plot and comment on the plausibility of the normal distribution as a model for bearing lifetime (data from "Modified Moment Estimation for the Three-Parameter Lognormal Distribution," J. Quality Technology, 1985: 92-99).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.