/*! 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 4 Verify Property 2 of the definit... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Verify Property 2 of the definition of a probability density function over the given interval. $$ f(x)=\frac{1}{5}, \quad[3,8] $$

Short Answer

Expert verified
The integral of \( f(x) = \frac{1}{5} \) over [3, 8] is 1, verifying Property 2.

Step by step solution

01

Understand the Task

Property 2 of a probability density function states that the integral of the function over its given interval should equal 1. This integral represents the total probability over the interval.
02

Set Up the Integral

Set up the integral of the function over the interval [3, 8]. The function given is \( f(x) = \frac{1}{5} \). The integral to compute is \( \int_{3}^{8} \frac{1}{5} \, dx \).
03

Evaluate the Integral

Evaluate the integral \( \int \frac{1}{5} \, dx \). The antiderivative of the constant \( \frac{1}{5} \) is \( \frac{1}{5}x \).
04

Apply the Limits of Integration

Apply the limits of integration by substituting \(x = 8\) and \(x = 3\) into the antiderivative. This gives \( \frac{1}{5}(8) - \frac{1}{5}(3) \).
05

Calculate the Result

Compute the expression from Step 4: \( \frac{1}{5}(8) - \frac{1}{5}(3) = \frac{8}{5} - \frac{3}{5} = \frac{5}{5} = 1 \).
06

Verify the Property

Confirm that the result of the integral is 1, thereby verifying that Property 2 of the probability density function is satisfied.

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.

Integral Calculus
Integral calculus is a branch of mathematics that deals with the accumulation of quantities. It involves calculating the total size, length, area, or other cumulative aspects over a continuous interval. In the context of probability density functions, integral calculus helps in finding total probabilities over an interval. This is essential for confirming that a function can serve as a valid probability distribution.

When verifying the properties of a probability density function (PDF), we often compute the integral of the function over its given interval. This process involves:
  • Identifying the function to integrate, which in this case, is the constant function \(f(x) = \frac{1}{5}\).
  • Specifying the interval over which to integrate, [3, 8].
  • Evaluating the integral in order to verify it sums to 1, thus validating the PDF.
This use of integral calculus ensures the entire area under the function's curve on a specified interval equates to 1, as it denotes the whole probability sum.
Antiderivative
Antiderivatives, also known as indefinite integrals, are the reverse operation of derivatives. When we have a function whose derivative we know, the antiderivative helps us go back to the original function.

For example, in the provided exercise, the function \( f(x) = \frac{1}{5} \) is a constant function. The antiderivative of a constant \( c \) is given by \( cx + C \), where \( C \) is a constant. Therefore, the antiderivative of \( \frac{1}{5} \) is \( \frac{1}{5}x + C \).

When verifying a probability density function, the antiderivative is critical as it is required to calculate definite integrals over a given interval. Here, it helps transform our integral into a form where limits can be easily applied to extract a numerical probability value.
Limits of Integration
The limits of integration define the interval over which we evaluate a definite integral. In probability density functions, these limits correspond to the endpoints of the interval for which the function is defined to calculate total probability.

For the exercise, the limits of integration are [3, 8]. After finding the antiderivative \( \frac{1}{5}x \), the next step is to substitute these limits into the antiderivative:
  • First, substitute the upper limit into the antiderivative: \( \frac{1}{5}(8) \).
  • Next, substitute the lower limit: \( \frac{1}{5}(3) \).
  • Subtract the value at the lower limit from the value at the upper limit to get the net area under the curve.
This process delivers the definite integral result, ensuring the area under the function's graph over the interval equates to 1, confirming valid calculus and probability assertions. Thus, using limits of integration verifies both mathematical accuracy and conceptual validity within probability theory.

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

Before \(1859,\) rabbits did not exist in Australia. That year, a settler released 24 rabbits into the wild. Without natural predators, the growth of the Australian rabbit population can be modeled by the uninhibited growth model \(d P / d t=k P,\) where \(P(t)\) is the population of rabbits \(t\) years after \(1859 .\) (Source: www dpi.vic.gov.au/agriculture.) a) When the rabbit population was estimated to be \(8900,\) its rate of growth was about 2630 rabbits per year. Use this information to find \(k,\) and then find the particular solution of the differential equation. b) Find the rabbit population in \(1900(t=41)\) and the rate at which it was increasing in that year. c) Without using a calculator, find \(P^{\prime}(41) / P(41)\).

For each probability density function, over the given interval, find \(\mathrm{E}(x), \mathrm{E}\left(x^{2}\right),\) the mean, the variance, and the standard deviation. $$ f(x)=\frac{1}{4}, \quad[3,7] $$

(a) write a differential equation that models the situation, and (b) find the general solution. If an initial condition is given, find the particular solution. Recall that when \(y\) is directly proportional to \(x,\) we have \(y=k x\), and when \(y\) is inversely proportional to \(x,\) we have \(y=k / x,\) where \(k\) is the constant of proportionality. In these exercises, let \(k=1\). The rate of change of \(y\) with respect to \(x\) is inversely proportional to the square root of \(y\).

In \(2013(t=0),\) world consumption of natural gas was approximately 117.2 trillion cubic feet and was growing exponentially at about 1.24\% per year. (Source: U.S. Energy Information Administration.) If the demand continues to grow at this rate, how many cubic feet of natural gas will the world use from 2015 to \(2025 ?\)

Hanna's Hat Company's marginal profit, \(P\), as a function of its total cost, \(C,\) is given by \(\frac{d P}{d C}=\frac{-200}{(C+3)^{3 / 2}}\) a) Find the profit function, \(P(C),\) if \(P=\$ 10\) when \(C=\$ 61\) b) At what cost will the firm break even \((P=0)\) ?

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.