/*! 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 3 Suppose the error involved in ma... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Short Answer

Expert verified
a) Symmetric, downward parabola, b) 0.5, c) Integral of f(x) from -1 to 1, d) 1 minus the integral from -0.5 to 0.5.

Step by step solution

01

Understanding the PDF

The probability density function (pdf) given is a piecewise function where it defines the density as 0.09375 multiplied by the expression \(4 - x^2\) for values of \(x\) within the interval \([-2, 2]\). Outside this interval, it is 0. This indicates that the distribution is limited to \([-2, 2]\) and is symmetric around \(x = 0\).
02

Sketching the Graph of f(x)

To sketch the graph of \(f(x)\), note that the function is a quadratic function with a downward opening, defined by \(f(x) = 0.09375(4 - x^2)\). It has its maximum value at \(x = 0\), where \(f(0) = 0.09375 \times 4 = 0.375\), and intercepts on the x-axis at \(x = -2\) and \(x = 2\). Outside \([-2, 2]\), the value is 0.
03

Computing P(X > 0)

Since the distribution is symmetric and centered at 0, \(P(X > 0)\) is simply half of the total probability, assuming this pdf is correctly normalized. Integrate \(f(x)\) from 0 to 2, i.e., \( \int_0^2 0.09375(4 - x^2) \, dx\). Calculate this integral to find \(P(X > 0)\).
04

Computing P(-1 < X < 1)

To find \(P(-1 < X < 1)\), integrate the pdf \(f(x)\) from \(-1\) to \(1\). This gives \( \int_{-1}^1 0.09375(4 - x^2) \, dx\). Calculate this integral to determine the probability.
05

Computing P(X < -0.5 or X > 0.5)

This is equivalent to \(1 - P(-0.5 < X < 0.5)\). First, compute \(P(-0.5 < X < 0.5)\) using the integral \( \int_{-0.5}^{0.5} 0.09375(4 - x^2) \, dx\). Subtract this result from 1 to find \(P(X < -0.5 \) or \(X > 0.5)\).

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
A continuous random variable, such as the error measurement in the given problem, is a variable that can assume an infinite number of values within a given range. Unlike discrete random variables, which have specific values, continuous ones may take any value within an interval. This smooth range allows the variable to be modeled with a probability density function (PDF). In this exercise, the continuous random variable is denoted as \(X\), representing measurement errors. Understanding this concept is crucial as it helps in assessing the likelihood of different outcomes based on the continuous data provided. This continuous nature is captured mathematically in the PDF, which assigns probabilities over an interval rather than points.
Integral Calculus
Integral calculus is key in probability when working with continuous random variables, as it helps to find probabilities over continuous ranges. In our example, we use integration extensively to compute probabilities. The probability density function is sometimes expressed as \(f(x)\), and the area under this curve within a specific interval provides the probability for that range.
  • For \(P(X > 0)\), we calculate the integral of the PDF from 0 to 2.
  • The probability \(P(-1 < X < 1)\) is obtained by integrating from -1 to 1.
  • To determine \(P(X < -0.5\) or \(X > 0.5)\), we compute one integral and subtract another.
These integrals are essential to understanding and calculating how probabilities distribute across the range where the function is defined. It translates how a continuous random variable is applied in real-world data analysis.
Symmetric Distribution
A symmetric distribution is one where the data is mirrored equally around a central point. For the problem at hand, the PDF is mirrored about \(x = 0\), which indicates that the probability of \(X\) assuming any positive value is mirrored within the corresponding negative range. In simpler terms, the probabilities to the left and right of the center are identical, which often simplifies computations. For instance, the question of finding \(P(X > 0)\) becomes straightforward because you only need to compute half of the total distribution, knowing the other half mirrors it perfectly. This symmetry is integral in statistics due to its often simplifying nature when addressing problems related to balance, expectation, and variance.
Probability Computation
Probability computation using continuous random variables and their PDFs involves leveraging integrals to find areas under curves. For example:
  • To find \(P(X > 0)\), integrate the PDF from 0 to the upper bound of 2.
  • For \(P(-1 < X < 1)\), calculate the integral over this range to determine the probability enclosed within these values.
  • When calculating \(P(X < -0.5\) or \(X > 0.5)\), first find the probability for \(-0.5 < X < 0.5\) and subtract from 1.
Each computation involves determining the integral of the function over the specified range, which in essence measures the 'weight' of the distribution over the interval, translating to the probability of the random variable falling within that interval. These computations are foundational in statistics for predicting events and making informed decisions based on the likelihood of various 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

In each case, determine the value of the constant \(c\) that makes the probability statement correct. a. \(\phi(c)=.9838\) b. \(P(0 \leq Z \leq c)=.291\) c. \(P(c \leq Z)=.121\) d. \(P(-c \leq Z \leq c)=.668\) e. \(P(c \leq|Z|)=.016\)

Let \(X=\) the time (in \(10^{-1}\) weeks) from shipment of a defective product until the customer retums the product. Suppose that the minimum return time is \(\gamma=3.5\) and that the excess \(X-3.5\) over the minimum has a Weibull distribution with parameters \(\alpha=2\) and \(\beta=1.5\) (see the article "Practical Applications of the Weibull Distribution," Indust. Qual. Control, 1964: 71-78). a. What is the cdf of \(X\) ? b. What are the expected return time and variance of return time? [Hint: First obtain \(E(X-3.5)\) and \(V(X-3.5)\).] c. Compute \(P(X>5)\). d. Compute \(P(5 \leq X \leq 8)\).

If \(X\) has an exponential distribution with parameter \(\lambda\). derive a general expression for the \((100 \mathrm{p})\) th percentile of the distribution. Then specialize to obtain the median.

If \(X\) is a normal ry with mean 80 and standard deviation 10 , compute the following probabilities by standardizing: a. \(P(X \leq 100)\) b. \(P(X \leq 80)\) c. \(P(65 \leq X \leq 100)\) d. \(P(70 \leq X)\) e. \(P(85 \leq X \leq 95)\) f. \(P(|X-80| \leq 10)\)

The breakdown voltage of a randomly chosen diode of a certain type is known to be normally distributed with mean value \(40 \mathrm{~V}\) and standard deviation \(1.5 \mathrm{~V}\). a. What is the probability that the voltage of a single diode is between 39 and 42 ? b. What value is such that only \(15 \%\) of all diodes have voltages exceeding that value? c. If four diodes are independently selected, what is the probability that at least one has a voltage exceeding 42 ?

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.