/*! 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} Q 7E Suppose that the n random vari... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that thenrandom variablesX1, . . . , Xnform a random sample from a continuous distribution for which the p.d.f. isf. Determine the probability that at leastk of thesenrandom variables will lie in a specified intervala≤x≤b.

Short Answer

Expert verified

The probability that at least \(k\) is \(\frac{2}{{\left( {b - a} \right)}}\) .

Step by step solution

01

Given Information

Here given distribution is continuous distribution for \(n\) random variables.

02

State the random variables 

For \(n\) random variables we take let \({x_1}, \ldots ,{x_n} \sim u\left( {a,b} \right)\) . Here we assume that the random variables follow uniform distribution.

03

Compute the probability at least \(k\) 

For \({x_1}, \ldots ,{x_n}\) the pdf is given by

\(f\left( {{x_1}, \ldots ,{x_n}} \right) = \frac{k}{{{{\left( {b - a} \right)}^n}}}\) where the range is \(a \le {x_1}, \ldots ,{x_n} \le b\) and \(k\) is any constant .

We know that sum of all pdf is\(1\). So here we first calculate the constant value is

\(\begin{align}\int_{ - \infty }^\infty {f\left( {{x_1}, \ldots ,{x_n}} \right) = 1} \\\int_a^b {\frac{k}{{{{\left( {b - a} \right)}^n}}}dx = 1} \\k \times \frac{{\left( {b - a} \right)}}{{{{\left( {b - a} \right)}^n}}} &= 1\\k &= {\left( {b - a} \right)^{n - 1}}\end{align}\)

Then find the probability at least\(k\)is

\(\begin{align}p\left( {x \ge k} \right) &= p\left( {x \ge {{\left( {b - a} \right)}^{n - 1}}} \right)\\ &= p\left( {x = {{\left( {b - a} \right)}^n}} \right) + p\left( {x = \frac{1}{{\left( {b - a} \right)}}} \right)\\ &= \int_a^b {{{\left( {b - a} \right)}^{n - 1}}dx + \int_a^b {dx} } \\ &= \frac{2}{{\left( {b - a} \right)}}\end{align}\)

Hence the value is \(\frac{2}{{\left( {b - a} \right)}}\) .

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!

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

Suppose that the joint distribution of X and Y is uniform over a set A in the xy-plane. For which of the following sets A are X and Y independent?

a. A circle with a radius of 1 and with its center at the origin

b. A circle with a radius of 1 and with its center at the point (3,5)

c. A square with vertices at the four points (1,1), (1,−1), (−1,−1), and (−1,1)

d. A rectangle with vertices at the four points (0,0), (0,3), (1,3), and (1,0)

e. A square with vertices at the four points (0,0), (1,1),(0,2), and (−1,1)

Suppose that a Markov chain has four states 1, 2, 3, 4, and stationary transition probabilities as specified by the following transition matrix

\(p = \left[ {\begin{array}{*{20}{c}}{\frac{1}{4}}&{\frac{1}{4}}&0&{\frac{1}{2}}\\0&1&0&0\\{\frac{1}{2}}&0&{\frac{1}{2}}&0\\{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\end{array}} \right]\):

a.If the chain is in state 3 at a given timen, what is the probability that it will be in state 2 at timen+2?

b.If the chain is in state 1 at a given timen, what is the probability it will be in state 3 at timen+3?

Suppose that the p.d.f. of X is as follows:

\(\begin{aligned}f\left( x \right) &= e{}^{ - x},x > 0\\ &= 0,x \le 0\end{aligned}\)

Determine the p.d.f. of \({\bf{Y = }}{{\bf{X}}^{\frac{{\bf{1}}}{{\bf{2}}}}}\)

Suppose that either of two instruments might be used for making a certain measurement. Instrument 1 yields a measurement whose p.d.f.\({{\bf{h}}_{\bf{1}}}\)is

\({{\bf{h}}_{\bf{1}}}\left( {\bf{x}} \right){\bf{ = }}\left\{ {\begin{align}{}{{\bf{2x}}}&{{\bf{for}}\,{\bf{0 < x < 1}}}\\{\bf{0}}&{{\bf{otherwise}}}\end{align}} \right.\)

Instrument 2 yields a measurement whose p.d.f.\({{\bf{h}}_2}\)is

\({{\bf{h}}_{\bf{2}}}\left( {\bf{x}} \right){\bf{ = }}\left\{ {\begin{align}{}{{\bf{3}}{{\bf{x}}^{\bf{2}}}}&{{\bf{for}}\,{\bf{0 < x < 1}}}\\{\bf{0}}&{{\bf{otherwise}}}\end{align}} \right.\)

Suppose that one of the two instruments is chosen randomly, and a measurement X is made with it.

  1. Determine the marginal p.d.f. of X.
  2. If the measurement value is\({\bf{X = }}{\raise0.7ex\hbox{\({\bf{1}}\)} \!\mathord{\left/ {\vphantom {{\bf{1}} {\bf{4}}}}\right.\ } \!\lower0.7ex\hbox{\({\bf{4}}\)}}\), what is the probability that instrument 1 was used?

Suppose that a coin is tossed repeatedly in such a way that heads and tails are equally likely to appear on any given toss and that all tosses are independent, with the following exception: Whenever either three heads or three tails have been obtained on three successive tosses, then the outcome of the next toss is always of the opposite type. At time\(n\left( {n \ge 3} \right)\)let the state of this process be specified by the outcomes on tosses\(n - 2\),\(n - 1\)and n. Show that this process is a Markov chain with stationary transition probabilities and construct the transition matrix.

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.