/*! 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 random sample of n items is to... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Q10E (page 359)

A random sample of n items is to be taken from a distribution with mean 渭 and standard deviation 蟽.

a. Use the Chebyshev inequality to determine the smallest number of items n that must be taken to satisfy the following relation:

\({\bf{Pr}}\left( {\left| {{{{\bf{\bar X}}}_{\bf{n}}}{\bf{ - \mu }}} \right| \le \frac{{\bf{\sigma }}}{{\bf{4}}}} \right) \ge {\bf{0}}{\bf{.99}}\)

b. Use the central limit theorem to determine the smallest number of items n that must be taken to satisfy the relation in part (a) approximately

Short Answer

Expert verified

a. The smallest number of items is\(n \ge 1600\)

b. The smallest possible sample size is 106

Step by step solution

01

Given information

A random sample of n items with a mean of\(\mu \)and a standard deviation of\(\sigma \)

The relation is,

\(\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \le \frac{\sigma }{4}} \right) \ge 0.99\)

02

(a) Finding the smallest number of n

Using the Chebyshev inequality,

\(\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \ge t} \right) \le \frac{{{\sigma ^2}}}{{n{t^2}}}\)

And

\(\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \le t} \right) \ge 1 - \frac{{{\sigma ^2}}}{{n{t^2}}}\)

So,

\(\begin{array}{c}\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \ge \frac{\sigma }{4}} \right) \le \frac{{{\sigma ^2}}}{n} \times \frac{{16}}{{{\sigma ^2}}}\\ = \frac{{16}}{n}\end{array}\)

\(\begin{array}{c}\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \le \frac{\sigma }{4}} \right) \ge 1 - \left( {\frac{{{\sigma ^2}}}{n} \times \frac{{16}}{{{\sigma ^2}}}} \right)\\ = 1 - \frac{{16}}{n}\end{array}\)

We know that,

\(\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \le \frac{\sigma }{4}} \right) \ge 0.99\)

i.e.,

\(\begin{array}{c}1 - \frac{{16}}{n} \ge 0.99\\\frac{{16}}{n} \ge 1 - 0.99\\\frac{n}{{16}} \ge 100\\n \ge 1600\end{array}\)

Therefore, the smallest number is, \(n \ge 1600\)

03

(b) Finding the smallest number of n

Using the central limit theorem,

\(\begin{array}{c}Z = \frac{{{{\bar X}_n} - \mu }}{{{\raise0.7ex\hbox{$\sigma $} \!\mathord{\left/{\vphantom {\sigma {\sqrt n }}}\right.}\!\lower0.7ex\hbox{${\sqrt n }$}}}}\\ = \frac{{\sqrt n }}{\sigma }\left( {{{\bar X}_n} - \mu } \right)\end{array}\)

It will be approximately a standard normal distribution.

\(\begin{array}{c}\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \le \frac{\sigma }{4}} \right) = \Pr \left( {\left| Z \right| \le \frac{{\sqrt n }}{4}} \right)\\ = 2\Phi \left( {\frac{{\sqrt n }}{4}} \right) - 1\end{array}\)

Now,

\(\begin{array}{c}2\Phi \left( {\frac{{\sqrt n }}{4}} \right) - 1 \ge 0.99\\2\Phi \left( {\frac{{\sqrt n }}{4}} \right) \ge 1 + 0.99\\\Phi \left( {\frac{{\sqrt n }}{4}} \right) \ge \frac{{1.99}}{2}\\\Phi \left( {\frac{{\sqrt n }}{4}} \right) \ge 0.995\end{array}\)

\(\begin{array}{c}\frac{{\sqrt n }}{4} \ge {\Phi ^{ - 1}}\left( {0.995} \right)\\\frac{{\sqrt n }}{4} \ge 2.567\\\sqrt n \ge 10.268\\n \ge 105.431824\end{array}\)

Therefore, the smallest possible sample size is 106

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

Let\({Z_1},{Z_2},...\)be a sequence of random variables, and suppose that for \(n = 2,3,...\)the distribution of \({Z_n}\)is as follows: \(\Pr \left( {{Z_n} = \frac{1}{n}} \right) = 1 - \frac{1}{{{n^2}}}\) and\(\Pr \left( {{Z_n} = n} \right) = \frac{1}{{{n^2}}}\).

  1. Does there exist a constant c to which the sequence converges in probability?
  2. Does there exist a constantc to which the sequence converges in quadratic mean?

Suppose that the distribution of the number of defects on any given bolt of cloth is the Poisson distribution with a mean 5, and the number of defects on each bolt is counted for a random sample of 125 bolts. Determine the probability that the average number of defects per bolt in the sample will be less than 5.5.

Suppose that\({{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\)form a random sample from a normal distribution with mean 0 and unknown variance\({{\bf{\sigma }}^{\bf{2}}}\).

  1. Determine the asymptotic distribution of the statistic

\({\left( {\frac{{\bf{1}}}{{\bf{n}}}\sum\nolimits_{{\bf{i = 1}}}^{\bf{n}} {{{\bf{X}}^{\bf{2}}}_{\bf{i}}} } \right)^{{\bf{ - 1}}}}\).

  1. Find a variance stabilizing transformation for the statistic

\(\left( {\frac{{\bf{1}}}{{\bf{n}}}\sum\nolimits_{{\bf{i = 1}}}^{\bf{n}} {{{\bf{X}}^{\bf{2}}}_{\bf{i}}} } \right)\).

Let f be a p.f. for a discrete distribution. Suppose that\(f\left( x \right) = 0\)for \(x \notin \left[ {0,1} \right]\). Prove that the variance of this distribution is at most\(\frac{1}{4}\). Hint: Prove that there is a distribution supported on just the two points\(\left\{ {0,1} \right\}\)with variance at least as large as f, and then prove that the variance of distribution supported on\(\left\{ {0,1} \right\}\)is at most\(\frac{1}{4}\).

In this exercise, we construct an example of a sequence of random variables\({{\bf{Z}}_{\bf{n}}}\)that\({{\bf{Z}}_{\bf{n}}}\)converges to 0 with probability 1 but\({{\bf{Z}}_{\bf{n}}}\)fails to converge to 0 in a quadratic mean. Let X be a random variable with a uniform interval distribution [0, 1]. Define the sequence\({{\bf{Z}}_{\bf{n}}}\)by\({{\bf{Z}}_{\bf{n}}}{\bf{ = }}{{\bf{n}}^{\bf{2}}}\)if\({\bf{0 < X < }}{\raise0.7ex\hbox{\({\bf{1}}\)} \!\mathord{\left/{\vphantom {{\bf{1}} {\bf{n}}}}\right.}\!\lower0.7ex\hbox{\({\bf{n}}\)}}\) and\({{\bf{Z}}_{\bf{n}}}{\bf{ = 0}}\)otherwise.

a. Prove that\({{\bf{Z}}_{\bf{n}}}\)converges to 0 with probability 1.

b. Prove that\({{\bf{Z}}_{\bf{n}}}\)it does not converge to 0 in quadratic mean.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

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.