/*! 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} Q10E A family of pdf鈥檚 that has bee... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Q10E (page 147)

A family of pdf鈥檚 that has been used to approximate the distribution of income, city population size, and size of firms is the Pareto family. The family has two parameters, \({\rm{k}}\) and \({\rm{\theta }}\), both\({\rm{ > 0}}\), and the pdf is

\({\rm{f(x;\theta ) = \{ }}\begin{array}{*{20}{c}}{\frac{{{\rm{k}} \cdot {{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{x}}^{{\rm{k + 1}}}}}}}&{{\rm{x}} \ge {\rm{\theta }}}\\{\rm{0}}&{{\rm{x < \theta }}}\end{array}\)

a. Sketch the graph of \({\rm{f(x;\theta )}}\).

b. Verify that the total area under the graph equals \({\rm{1}}\).

c. If the rv \({\rm{X}}\) has pdf \({\rm{f(x;\theta )}}\), for any fixed \({\rm{b > \theta }}\), obtain an expression for \({\rm{P(X}} \le {\rm{b)}}\).

d. For \({\rm{\theta < a < b}}\) obtain an expression for the probability \({\rm{P(a}} \le {\rm{X}} \le {\rm{b)}}\).

Short Answer

Expert verified

(a) The graph for\({\rm{f(x;\theta )}}\)is -

(b) It is verified that the total area under the graph equals \({\rm{1}}\).

(c) The expression for\({\rm{P(x}} \le {\rm{b)}}\)is obtained as\(\frac{{{{\rm{b}}^{\rm{k}}}{\rm{ - }}{{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{b}}^{\rm{k}}}}}\).

(d) The expression for \({\rm{P(a}} \le {\rm{x}} \le {\rm{b)}}\) is obtained as \(\frac{{{{\rm{\theta }}^{\rm{k}}} \cdot \left( {{{\rm{b}}^{\rm{k}}}{\rm{ - }}{{\rm{a}}^{\rm{k}}}} \right)}}{{{{\rm{b}}^{\rm{k}}} \cdot {{\rm{a}}^{\rm{k}}}}}\).

Step by step solution

01

Concept Introduction

Probability refers to the likelihood of a random event's outcome. This word refers to determining the likelihood of a given occurrence occurring.

02

The graph for \({\rm{f(x;\theta )}}\)

(a)

The graph of \({\rm{f(x;k,\theta )}}\) is shown below.

Therefore, the graph is obtained.

03

Total Area under the graph

(b)

The pdf of\({\rm{f(x)}}\)is given as 鈥�

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{{\rm{k}} \cdot {{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{x}}^{{\rm{k + 1}}}}}}}&{{\rm{X}} \ge {\rm{\theta }}}\\{\rm{0}}&{{\rm{X < \theta }}}\end{array}} \right.\)

It is known that for any pdf \({\rm{f(x)}}\)鈥�

\({\rm{Area under graph of f(x) = }}\int_\infty ^\infty {\rm{f}} {\rm{(x)}} \cdot {\rm{dx}}\)

From the given pdf, it can be written 鈥�

\(\begin{array}{c}\int_\infty ^\infty {\rm{f}} {\rm{(x)}} \cdot {\rm{dx = }}\int_{\rm{\theta }}^\infty {\frac{{{\rm{k}} \cdot {{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{x}}^{{\rm{k + 1}}}}}}} \cdot {\rm{dx}}\\{\rm{ = k}} \cdot {{\rm{\theta }}^{\rm{k}}}\int_{\rm{\theta }}^\infty {{{\rm{x}}^{{\rm{ - (k + 1)}}}}} \cdot {\rm{dx}}\\{\rm{ = k}} \cdot {{\rm{\theta }}^{\rm{k}}}\left( {\frac{{{{\rm{x}}^{{\rm{ - k}}}}}}{{{\rm{ - k}}}}} \right)_{\rm{\theta }}^\infty \\{\rm{ = }}\frac{{{\rm{k}} \cdot {{\rm{\theta }}^{\rm{k}}}}}{{{\rm{ - k}}}}\left( {\frac{{\rm{1}}}{{{{\rm{x}}^{\rm{k}}}}}} \right)_{\rm{\theta }}^\infty \\{\rm{ = - }}{{\rm{\theta }}^{\rm{k}}}\left( {\frac{{\rm{1}}}{{{\infty ^{\rm{k}}}}}{\rm{ - }}\frac{{\rm{1}}}{{{{\rm{\theta }}^{\rm{k}}}}}} \right)\\{\rm{ = - }}{{\rm{\theta }}^{\rm{k}}}\left( {{\rm{0 - }}\frac{{\rm{1}}}{{{{\rm{\theta }}^{\rm{k}}}}}} \right)\\\int_\infty ^\infty {\rm{f}} {\rm{(x)}} \cdot {\rm{dx = 1}}\end{array}\)

Here \(\frac{{\rm{1}}}{{{\infty ^{\rm{k}}}}}{\rm{ = 0}}\) is used since it is given that \({\rm{k > 0}}\).

Therefore, the area is verified.

04

Expression for \({\rm{P(X}} \le {\rm{b)}}\) 

(c)

From the given pdf, it can be written 鈥�

\(\begin{array}{c}{\rm{P(x}} \le {\rm{b) = }}\int_{\rm{\theta }}^{\rm{b}} {\frac{{{\rm{k}} \cdot {{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{x}}^{{\rm{k + 1}}}}}}} \cdot {\rm{dx}}\\{\rm{ = k}} \cdot {{\rm{\theta }}^{\rm{k}}}\int_{\rm{\theta }}^{\rm{b}} {{{\rm{x}}^{{\rm{ - (k + 1)}}}}} \cdot {\rm{dx}}\\{\rm{ = k}} \cdot {{\rm{\theta }}^{\rm{k}}}\left( {\frac{{{{\rm{x}}^{{\rm{ - k}}}}}}{{{\rm{ - k}}}}} \right)_{\rm{\theta }}^{\rm{b}}\\{\rm{ = }}\frac{{{\rm{k}} \cdot {{\rm{\theta }}^{\rm{k}}}}}{{{\rm{ - k}}}}\left( {\frac{{\rm{1}}}{{{{\rm{x}}^{\rm{k}}}}}} \right)_{\rm{\theta }}^{\rm{b}}\\{\rm{ = - }}{{\rm{\theta }}^{\rm{k}}}\left( {\frac{{\rm{1}}}{{{{\rm{b}}^{\rm{k}}}}}{\rm{ - }}\frac{{\rm{1}}}{{{{\rm{\theta }}^{\rm{k}}}}}} \right)\\{\rm{ = - }}{{\rm{\theta }}^{\rm{k}}}\left( {\frac{{{{\rm{\theta }}^{\rm{k}}}{\rm{ - }}{{\rm{b}}^{\rm{k}}}}}{{{{\rm{\theta }}^{\rm{k}}} \cdot {{\rm{b}}^{\rm{k}}}}}} \right)\\{\rm{ = }}\frac{{{{\rm{b}}^{\rm{k}}}{\rm{ - }}{{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{b}}^{\rm{k}}}}}\end{array}\)

Therefore, the expression is obtained as \(\frac{{{{\rm{b}}^{\rm{k}}}{\rm{ - }}{{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{b}}^{\rm{k}}}}}\).

05

Expression for \({\rm{P(a}} \le {\rm{X}} \le {\rm{b)}}\) 

(d)

From the given pdf, it can be written 鈥�

\(\begin{array}{c}{\rm{P(a}} \le {\rm{x}} \le {\rm{b) = }}\int_{\rm{a}}^{\rm{b}} {\frac{{{\rm{k}} \cdot {{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{x}}^{{\rm{k + 1}}}}}}} \cdot {\rm{dx}}\\{\rm{ = k}} \cdot {{\rm{\theta }}^{\rm{k}}}\int_{\rm{a}}^{\rm{b}} {{{\rm{x}}^{{\rm{ - (k + 1)}}}}} \cdot {\rm{dx}}\\{\rm{ = k}} \cdot {{\rm{\theta }}^{\rm{k}}}\left( {\frac{{{{\rm{x}}^{{\rm{ - k}}}}}}{{{\rm{ - k}}}}} \right)_{\rm{a}}^{\rm{b}}\\{\rm{ = }}\frac{{{\rm{k}} \cdot {{\rm{\theta }}^{\rm{k}}}}}{{{\rm{ - k}}}}\left( {\frac{{\rm{1}}}{{{{\rm{x}}^{\rm{k}}}}}} \right)_{\rm{a}}^{\rm{b}}\\{\rm{ = - }}{{\rm{\theta }}^{\rm{k}}}\left( {\frac{{\rm{1}}}{{{{\rm{b}}^{\rm{k}}}}}{\rm{ - }}\frac{{\rm{1}}}{{{{\rm{a}}^{\rm{k}}}}}} \right)\\{\rm{ = - }}{{\rm{\theta }}^{\rm{k}}}\left( {\frac{{{{\rm{a}}^{\rm{k}}}{\rm{ - }}{{\rm{b}}^{\rm{k}}}}}{{{{\rm{a}}^{\rm{k}}} \cdot {{\rm{b}}^{\rm{k}}}}}} \right)\\{\rm{P(a}} \le {\rm{x}} \le {\rm{b) = }}\frac{{{{\rm{\theta }}^{\rm{k}}} \cdot \left( {{{\rm{b}}^{\rm{k}}}{\rm{ - }}{{\rm{a}}^{\rm{k}}}} \right)}}{{{{\rm{b}}^{\rm{k}}} \cdot {{\rm{a}}^{\rm{k}}}}}\end{array}\)

Therefore, the expression is obtained as \(\frac{{{{\rm{\theta }}^{\rm{k}}} \cdot \left( {{{\rm{b}}^{\rm{k}}}{\rm{ - }}{{\rm{a}}^{\rm{k}}}} \right)}}{{{{\rm{b}}^{\rm{k}}} \cdot {{\rm{a}}^{\rm{k}}}}}\).

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 91影视!

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

The authors of the article "A Probabilistic Insulation Life Model for Combined Thermal-Electrical Stresses" (IEEE Trans. on Elect. Insulation, \({\rm{1985: 519 - 522}}\)) state that "the Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to aging and stress." They propose the use of the distribution as a model for time (in hours) to failure of solid insulating specimens subjected to \({\rm{AC}}\) voltage. The values of the parameters depend on the voltage and temperature; suppose \({\rm{\alpha = 2}}{\rm{.5}}\) and \({\rm{\beta = 200}}\) (values suggested by data in the article).

a. What is the probability that a specimen's lifetime is at most \({\rm{250}}\)? Less than\({\rm{250}}\)? More than\({\rm{300}}\)?

b. What is the probability that a specimen's lifetime is between \({\rm{100}}\) and \({\rm{250}}\)?

c. What value is such that exactly \({\rm{50\% }}\) of all specimens have lifetimes exceeding that value?

Let\({\bf{X}}\)denote the data transfer time (ms) in a grid computing system (the time required for data transfer between a 鈥渨orker鈥� computer and a 鈥渕aster鈥� computer. Suppose that X has a gamma distribution with mean value\({\bf{37}}.{\bf{5}}{\rm{ }}{\bf{ms}}\)and standard deviation\({\bf{21}}.{\bf{6}}\)(suggested by the article 鈥淐omputation Time of Grid Computing with Data Transfer Times that Follow a Gamma Distribution,鈥� Proceedings of the First International Conference on Semantics, Knowledge, and Grid, 2005).

a. What are the values of\({\rm{\alpha }}\)and\({\rm{\beta }}\)?

b. What is the probability that data transfer time exceeds\({\bf{50}}{\rm{ }}{\bf{ms}}\)?

c. What is the probability that data transfer time is between\({\bf{50}}\)and\({\bf{75}}{\rm{ }}{\bf{ms}}\)?

Sales delay is the elapsed time between the manufacture of a product and its sale. According to the article 鈥淲arranty Claims Data Analysis Considering Sales Delay鈥� (Quality and Reliability Engr. Intl., \({\rm{2013:113 - 123}}\)), it is quite common for investigators to model sales delay using a lognormal distribution. For a particular product, the cited article proposes this distribution with parameter values \({\rm{\mu = 2}}{\rm{.05 and }}{{\rm{\sigma }}^{\rm{2}}}{\rm{ = 0}}{\rm{.06}}\) (here the unit for delay is months).

a. What are the variance and standard deviation of delay time?

b. What is the probability that delay time exceeds \({\rm{12}}\) months?

c. What is the probability that delay time is within one standard deviation of its mean value?

d. What is the median of the delay time distribution?

e. What is the \({\rm{99th}}\) percentile of the delay time distribution?

f. Among \({\rm{10}}\) randomly selected such items, how many would you expect to have a delay time exceeding \({\rm{8}}\) months?

A \(12\)-in. bar that is clamped at both ends is to be subjected to an increasing amount of stress until it snaps. Let Y = the distance from the left end at which the break occurs. Suppose Y has pdf

\(f\left( y \right) = \left\{ {\begin{array}{*{20}{c}}{\left( {\frac{1}{{24}}} \right)y\left( {1 - \frac{y}{{12}}} \right)\,\,\,\,\,0 \le y \le 12}\\{0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,otherwise}\end{array}} \right.\)

Compute the following: a. The cdf of Y, and graph it. b.\(P\left( {Y \le 4} \right), P\left( {Y > 6} \right)\), and\(P\left( {4 \le Y \le 6} \right)\)c. E(Y), E(Y2 ), and V(Y) d. The probability that the breakpoint occurs more than \(2\;\)in. from the expected breakpoint. e. The expected length of the shorter segment when the break occurs.

The completion time X for a certain task has cdf F(x) given by

\(\left\{ {\begin{array}{*{20}{c}}{0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x < 0}\\{\frac{{{x^3}}}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \le x \le \frac{7}{3}}\\{1 - \frac{1}{2}\left( {\frac{7}{3} - x} \right)\left( {\frac{7}{4} - \frac{3}{4}x} \right)\,\,\,\,\,\,1 \le x \le \frac{7}{3}}\\{1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x > \frac{7}{3}}\end{array}} \right.\)

a. Obtain the pdf f (x) and sketch its graph.

b. Compute\({\bf{P}}\left( {.{\bf{5}} \le {\bf{X}} \le {\bf{2}}} \right)\). c. Compute E(X).
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation
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.