/*! 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} Q88E Each customer making a particula... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Q88E (page 1)

Each customer making a particular Internet purchase must pay with one of three types of credit cards (think Visa, MasterCard, AmEx). Let \({{\rm{A}}_{\rm{i}}}{\rm{(i = 1,2,3)}}\) be the event that a type \({\rm{i}}\)credit card is used, with\({\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = }}{\rm{.5}}\), \({\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = }}{\rm{.3}}\), and\({\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{ = }}{\rm{.2}}\). Suppose that the number of customers who make such a purchase on a given day is a Poisson \({\rm{rv}}\) with parameter\({\rm{\lambda }}\). Define \({\rm{rv}}\) 's \({{\rm{X}}_{{\rm{1,}}}}{{\rm{X}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}\)by \({{\rm{X}}_{\rm{i}}}{\rm{ = }}\)the number among the \({\rm{N}}\)customers who use a type \({\rm{i}}\)card\({\rm{(i = 1,2,3)}}\). Show that these three \({\rm{rv}}\) 's are independent with Poisson distributions having parameters\({\rm{.5\lambda ,}}{\rm{.3\lambda }}\), and\({\rm{.2\lambda }}\), respectively. (Hint: For non-negative integers\({{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{x}}_{\rm{3}}}\), let\({\rm{n = }}{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{\rm{ + }}{{\rm{x}}_{\rm{3}}}\). Then\({\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}} \right.\), \(\left. {{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}{\rm{ = }}{{\rm{x}}_{\rm{3}}}} \right){\rm{ = P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}{\rm{ = }}{{\rm{x}}_{\rm{3}}}{\rm{,N = n}}} \right)\) (why is this?). Now condition on\({\rm{N = n}}\), in which case the three \({\rm{Xi}}\) 's have a trinomial distribution (multinomial with three categories) with category probabilities\({\rm{.5,}}{\rm{.3}}\), and . \({\rm{2}}\).)

Short Answer

Expert verified

\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}\), and \({{\rm{X}}_{\rm{3}}}\)are independent with\({{\rm{X}}_{\rm{1}}}{\rm{\~Poisson(0}}{\rm{.5\lambda ),}}{{\rm{X}}_{\rm{2}}}{\rm{\~Poisson(0}}{\rm{.3\lambda )}}\), and\({{\rm{X}}_{\rm{3}}}{\rm{\~Poisson(0}}{\rm{.2\lambda )}}\).

Step by step solution

01

Definition

The standard deviation is a measurement of a collection of values' variance or dispersion. A low standard deviation implies that the values are close to the set's mean, whereas a high standard deviation suggests that the values are dispersed over a larger range.

02

Determining rv’s independent  

Given:

\({{\rm{A}}_{\rm{i}}}{\rm{ = }}\)Type \({\rm{i}}\) credit card is used

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.5}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.3}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{ = 0}}{\rm{.2}}\end{array}\)

\({\rm{X = }}\)number of customers who make a purchase

\({\rm{X\sim Poisson(\lambda )}}\)

\({{\rm{X}}_{\rm{i}}}{\rm{ = }}\)number of customers who make a purchase with a type\({\rm{i}}\)card

Formula Poisson probability:

\({\rm{P(X = N) = }}\frac{{{{\rm{\lambda }}^{\rm{N}}}{{\rm{e}}^{{\rm{ - \lambda }}}}}}{{{\rm{N!}}}}\)

A trinomial distribution describes the number of successes for three separate categories among a certain number of independent trials with a constant chance of success.

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}{\rm{ = }}{{\rm{x}}_{\rm{3}}}\mid {\rm{X = }}{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{\rm{ + }}{{\rm{x}}_{\rm{3}}}} \right)\\{\rm{ = }}\frac{{\left( {{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{\rm{ + }}{{\rm{x}}_{\rm{3}}}} \right){\rm{!}}}}{{{{\rm{x}}_{\rm{1}}}{\rm{!}}{{\rm{x}}_{\rm{2}}}{\rm{!}}{{\rm{x}}_{\rm{3}}}{\rm{!}}}}{\rm{0}}{\rm{.}}{{\rm{5}}^{{{\rm{x}}_{\rm{1}}}}}{\rm{0}}{\rm{.}}{{\rm{3}}^{{{\rm{x}}_{\rm{2}}}}}{\rm{0}}{\rm{.}}{{\rm{2}}^{{{\rm{x}}_{\rm{3}}}}}\end{array}\)

Use the General multiplication rule: \({\rm{P(A}}\)and \({\rm{B) = P(A) \times P(B}}\mid {\rm{A) = P(B) \times P(A}}\mid {\rm{B)}}\)

\(\begin{array}{c}{\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}{\rm{ = }}{{\rm{x}}_{\rm{3}}}} \right)\\{\rm{ = P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}{\rm{ = }}{{\rm{x}}_{\rm{3}}}\mid {\rm{X = }}{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{\rm{ + }}{{\rm{x}}_{\rm{3}}}} \right){\rm{P}}\left( {{\rm{X = }}{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{\rm{ + }}{{\rm{x}}_{\rm{3}}}} \right)\\{\rm{ = }}\frac{{\left( {{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{\rm{ + }}{{\rm{x}}_{\rm{3}}}} \right){\rm{!}}}}{{{{\rm{x}}_{\rm{1}}}{\rm{!}}{{\rm{x}}_{\rm{2}}}{\rm{!}}{{\rm{x}}_{\rm{3}}}{\rm{!}}}}{\rm{0}}{\rm{.}}{{\rm{5}}^{{{\rm{x}}_{\rm{1}}}}}{\rm{0}}{\rm{.}}{{\rm{3}}^{{{\rm{x}}_{\rm{2}}}}}{\rm{0}}{\rm{.}}{{\rm{2}}^{{{\rm{x}}_{\rm{3}}}}}\frac{{{{\rm{\lambda }}^{{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{\rm{ + }}{{\rm{x}}_{\rm{3}}}}}{{\rm{e}}^{{\rm{ - \lambda }}}}}}{{\left( {{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{\rm{ + }}{{\rm{x}}_{\rm{3}}}} \right){\rm{!}}}}\\{\rm{ = }}\frac{{\rm{1}}}{{{{\rm{x}}_{\rm{1}}}{\rm{!}}{{\rm{x}}_{\rm{2}}}{\rm{!}}{{\rm{x}}_{\rm{3}}}{\rm{!}}}}{\rm{0}}{\rm{.}}{{\rm{5}}^{{{\rm{x}}_{\rm{1}}}}}{\rm{0}}{\rm{.}}{{\rm{3}}^{{{\rm{x}}_{\rm{2}}}}}{\rm{0}}{\rm{.}}{{\rm{2}}^{{{\rm{x}}_{\rm{3}}}}}{{\rm{\lambda }}^{{{\rm{x}}_{\rm{1}}}}}{{\rm{\lambda }}^{{{\rm{x}}_{\rm{2}}}}}{{\rm{\lambda }}^{{{\rm{x}}_{\rm{3}}}}}{{\rm{e}}^{{\rm{ - \lambda (0}}{\rm{.5 + 0}}{\rm{.3 + 0}}{\rm{.2)}}}}\\{\rm{ = }}\frac{{\rm{1}}}{{{{\rm{x}}_{\rm{1}}}{\rm{!}}{{\rm{x}}_{\rm{2}}}{\rm{!}}{{\rm{x}}_{\rm{3}}}{\rm{!}}}}{{\rm{(0}}{\rm{.5\lambda )}}^{{{\rm{x}}_{\rm{1}}}}}{{\rm{(0}}{\rm{.3\lambda )}}^{{{\rm{x}}_{\rm{2}}}}}{{\rm{(0}}{\rm{.2\lambda )}}^{{{\rm{x}}_{\rm{3}}}}}{{\rm{e}}^{{\rm{ - 0}}{\rm{.5\lambda }}}}{{\rm{e}}^{{\rm{ - 0}}{\rm{.3\lambda }}}}{{\rm{e}}^{{\rm{ - 0}}{\rm{.2\lambda }}}}\\{\rm{ = }}\frac{{{{{\rm{(0}}{\rm{.5\lambda )}}}^{{{\rm{x}}_{\rm{1}}}}}{{\rm{e}}^{{\rm{0}}{\rm{.5\lambda }}}}}}{{{{\rm{x}}_{\rm{1}}}{\rm{!}}}}{\rm{ \times }}\frac{{{{{\rm{(0}}{\rm{.3\lambda )}}}^{{{\rm{x}}_{\rm{2}}}}}{{\rm{e}}^{{\rm{0}}{\rm{.3\lambda }}}}}}{{{{\rm{x}}_{\rm{2}}}{\rm{!}}}}{\rm{ \times }}\frac{{{{{\rm{(0}}{\rm{.2\lambda )}}}^{{{\rm{x}}_{\rm{3}}}}}{{\rm{e}}^{{\rm{0}}{\rm{.2\lambda }}}}}}{{{{\rm{x}}_{\rm{3}}}{\rm{!}}}}\end{array}\)

03

Determining rv’s independent  

Individual probability distributions are calculated by adding all potential values of the other variables:

\(\begin{aligned}P\left( {{X_1} = {x_1}} \right) &= \sum\limits_{{x_2}}^{ + \yen} {\sum\limits_{{x_3}}^{ + \yen} {\left( {\frac{{{{(0.5\lambda )}^{{x_1}}}{e^{0.5\lambda }}}}{{{x_1}!}} \times \frac{{{{(0.3\lambda )}^{{x_2}}}{e^{0.3\lambda }}}}{{{x_2}!}} \times \frac{{{{(0.2\lambda )}^{{x_3}}}{e^{0.2\lambda }}}}{{{x_3}!}}} \right)} } \\ &= \frac{{{{(0.5\lambda )}^{{x_1}}}{e^{0.5\lambda }}}}{{{x_1}!}} \times \sum\limits_{{x_2}}^{ + \yen} {\frac{{{{(0.3\lambda )}^{{x_2}}}{e^{0.3\lambda }}}}{{{x_2}!}}} \times \sum\limits_{{x_3}}^{ + \yen} {\frac{{{{(0.2\lambda )}^{{x_3}}}{e^{0.2\lambda }}}}{{{x_3}!}}} \\ &= \frac{{{{(0.5\lambda )}^{{x_1}}}{e^{0.5\lambda }}}}{{{x_1}!}} \times 1 \times 1 \\ & = \frac{{{{(0.5\lambda )}^{{x_1}}}{e^{0.5\lambda }}}}{{{x_1}!}}\sim{\text{ }}Poisson{\text{ }}(0.5\lambda ){\text{ }} \\

Similarly:{\text{ }}P\left( {{X_2} = {x_2}} \right) &= \frac{{{{(0.3\lambda )}^{{x_2}}}{e^{0.3\lambda }}}}{{{x_2}!}}\sim{\text{ }}Poisson{\text{ }}(0.3\lambda )P\left( {{X_3} = {x_3}} \right) \\ &= \frac{{{{(0.2\lambda )}^{{x_3}}}{e^{0.2\lambda }}}}{{{x_3}!}}\sim{\text{ }}Poisson{\text{ }}(0.2\lambda ) \\ \end{aligned} \)

Finally, we note that

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}{\rm{ = }}{{\rm{x}}_{\rm{3}}}} \right)\\{\rm{ = P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}} \right){\rm{P}}\left( {{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}} \right){\rm{P}}\left( {{{\rm{X}}_{\rm{3}}}{\rm{ = }}{{\rm{x}}_{\rm{3}}}} \right)\end{array}\)

, which implies that \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, and }}{{\rm{X}}_{\rm{3}}}\)are all independent.

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 propagation of fatigue cracks in various aircraft parts has been the subject of extensive study in recent years. The accompanying data consists of propagation lives(flight hours/\({\bf{1}}{{\bf{0}}^{\bf{4}}}\)) to reach a given crack size in fastener holes intended for use in military aircraft (鈥淪tatistical Crack Propagation in Fastener Holes Under Spectrum Loading,鈥� J. Aircraft,1983: 1028鈥�1032):

.736 .863 .865 .913 .915 .937 .983 1.007

1.011 1.064 1.109 1.132 1.140 1.153 1.253 1.394

  1. Compute and compare the values of the sample mean and median.
  2. By how much could the largest sample observation be decreased without affecting the value of the median?

Here is a description from Minitab of the strength datagiven in Exercise 13.


Variable N Mean Median TrMean StDev SE Mean
strength 153 135.39 135.40 135.41 4.59 0.37
Variable Minimum Maximum Q1Q3
strength 122.20 147.70 132.95 138.25

  1. Comment on any interesting features (the quartiles and fourths are virtually identical here).
  2. Construct a boxplot of the data based on the quartiles, and comment on what you see.

For each of the following hypothetical populations, give

a plausible sample of size 4:

a. All distances that might result when you throw a football

b. Page lengths of books published 5 years from now

c. All possible earthquake-strength measurements (Richter scale) that might be recorded in California during the next year

d. All possible yields (in grams) from a certain chemical reaction carried out in a laboratory.

In a famous experiment carried out in 1882, Michelson,and Newcomb obtained 66 observations on the time it took for light to travel between two locations in Washington, D.C. A few of the measurements(coded in a certain manner) were 31, 23, 32, 36, 22, 26, 27, and 31.

a. Why are these measurements not identical?

b. Is this an enumerative study? Why or why not?

How does the speed of a runner vary over the course of a marathon (a distance of 42.195 km)? Consider determining both the time to run the first 5 km and the time to run between the 35-km and 40-km points, and then subtracting the former time from the latter time. A positive value of this difference corresponds to a runner slowing down toward the end of the race. The accompanying histogram is based on times of runners who participated in several different Japanese marathons (鈥淔actors Affecting Runners鈥� Marathon Performance,鈥� Chance, Fall, 1993: 24鈥�30).What are some interesting features of this histogram? What is a typical difference value? Roughly what proportion of the runners ran the late distance more quickly than the early distance?
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

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.