/*! 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 8 Let \(X\) be a continuous random... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 8

Let \(X\) be a continuous random variable and define the standardized version \(X^{*}\) of \(X\) by: $$ X^{*}=\frac{X-\mu}{\sigma} $$ (a) Show that \(P\left(\left|X^{*}\right| \geq a\right) \leq 1 / a^{2}\). (b) If \(X\) is the random variable of Exercise \(1,\) find bounds for \(P\left(\left|X^{*}\right| \geq 2\right)\), \(P\left(\left|X^{*}\right| \geq 5\right),\) and \(P\left(\left|X^{*}\right| \geq 9\right)\)

Short Answer

Expert verified
(a) Proven using Chebyshev's inequality. (b) Bounds: 0.25 for \(a=2\), 0.04 for \(a=5\), 0.0123 for \(a=9\).

Step by step solution

01

Understanding Standardization

The standardized version of a random variable, denoted by \(X^*\), is defined as \(X^* = \frac{X - \mu}{\sigma}\), where \(\mu\) is the mean and \(\sigma\) is the standard deviation of \(X\). This operation transforms \(X\) into a standard normal variable if \(X\) follows a normal distribution.
02

Applying Chebyshev's Inequality

Chebyshev's inequality states that for any random variable \(X\) with mean \(\mu\) and standard deviation \(\sigma\), the probability that \(X\) lies more than \(k\) standard deviations away from the mean is at most \(\frac{1}{k^2}\). That is, \(P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}\).
03

Expressing in Terms of Standardized Variable

Rewriting the probability in terms of the standardized variable, we have \(P\left( |X^*| \geq a \right) = P\left( \frac{|X - \mu|}{\sigma} \geq a \right)\), which simplifies to \(P\left( |X - \mu| \geq a\sigma \right)\).
04

Applying to Show Initial Problem

By applying Chebyshev's inequality, we directly get \(P\left( |X^{*}| \geq a \right) = P(|X - \mu| \geq a\sigma) \leq \frac{1}{a^2}\). This completes part (a) of the question.
05

Calculating Specific Values for Part (b)

For part (b), we will use the results of Step 4. First, for \(a=2\), we have \(P\left( |X^{*}| \geq 2 \right) \leq \frac{1}{4} = 0.25\). For \(a=5\), \(P\left( |X^{*}| \geq 5 \right) \leq \frac{1}{25} = 0.04\). Finally, for \(a=9\), \(P\left( |X^{*}| \geq 9 \right) \leq \frac{1}{81} \approx 0.0123\).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standardization
Standardization is a useful technique in probability and statistics where we transform a given dataset or random variable to have a standard mean and deviation. When we take a random variable, denoted as \(X\), its standardized version \(X^*\) is defined by the formula:
\(X^* = \frac{X - \mu}{\sigma}\)
where \(\mu\) is the mean, and \(\sigma\) is the standard deviation of \(X\).

This formula helps convert any random variable into one that is easier to work with.
  • It centers the variable at zero (mean), suggesting that it is balanced around the center.
  • It scales the variable according to its standard deviation, which ensures uniformity regardless of the original range.
By standardizing, we make different variables directly comparable even if they were originally on different scales or units.
Continuous Random Variable
A continuous random variable is a type of random variable that can take on an infinite number of possible values. For example, if you measure the height of people in a town, their heights can have an infinite variation within a certain range (like 150 cm to 200 cm).

Continuous random variables are unique because they can assume any value within a given interval. Here's what you need to know
  • Continuous variables are often described using probability density functions (PDFs).
  • The total area under the PDF curve equals 1, ensuring that the sum of all probabilities is complete.
  • Examples include measurements like time, temperature, and height.
In the context of our exercise, \(X\) represents a continuous random variable and needs to be standardized for easier manipulation and analysis.
Standard Normal Distribution
Once we standardize a random variable, it often follows what we call a Standard Normal Distribution. This is a special type of normal distribution where the mean is 0 and the standard deviation is 1.

The standard normal distribution is frequently represented as \(Z\) and used as a reference point for calculating probabilities. Important points include:
  • It is symmetric about the mean, meaning the left and right sides of the distribution are mirror images.
  • Most of the population lies within a few standard deviations from the mean in a normal distribution.
  • This distribution makes probability calculations straightforward using tables, calculators, or software.
By converting any normal distribution to a standard normal distribution through standardization, we can leverage these properties to determine probabilities or make statistical inferences.
Probability Bounds
Probability bounds are essential for making sense of the likely spread or concentration of a random variable's values. In statistics, one practical way to determine the probability bounds is through Chebyshev's Inequality. This method provides a probability estimate for how far a random variable can deviate from its mean.

Chebyshev's inequality says that regardless of its distribution, the probability that a random variable is more than \(k\) standard deviations away from its mean is at most \(\frac{1}{k^2}\). Here's what you can deduce:
  • It's useful for any distribution, not just the normal distribution.
  • It gives a conservative estimate and is quite general.
  • For example, in our problem \(P\left(|X^*| \geq a\right)\) uses this inequality to establish the bounds.
By understanding these bounds, you can assess the probability of extreme deviations and gauge how "spread out" data points might be.

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

(Chebyshev \(^{8}\) ) Assume that \(X_{1}, X_{2}, \ldots, X_{n}\) are independent random variables with possibly different distributions and let \(S_{n}\) be their sum. Let \(m_{k}=E\left(X_{k}\right)\), \(\sigma_{k}^{2}=V\left(X_{k}\right),\) and \(M_{n}=m_{1}+m_{2}+\cdots+m_{n} .\) Assume that \(\sigma_{k}^{2}0\) $$ P\left(\left|\frac{S_{n}}{n}-\frac{M_{n}}{n}\right|<\epsilon\right) \rightarrow 1 $$ as \(n \rightarrow \infty\)

We have two coins: one is a fair coin and the other is a coin that produces heads with probability \(3 / 4\). One of the two coins is picked at random, and this coin is tossed \(n\) times. Let \(S_{n}\) be the number of heads that turns up in these \(n\) tosses. Does the Law of Large Numbers allow us to predict the proportion of heads that will turn up in the long run? After we have observed a large number of tosses, can we tell which coin was chosen? How many tosses suffice to make us 95 percent sure?

A 1-dollar bet on craps has an expected winning of -.0141 . What does the Law of Large Numbers say about your winnings if you make a large number of 1 -dollar bets at the craps table? Does it assure you that your losses will be small? Does it assure you that if \(n\) is very large you will lose?

Let \(X\) be a random variable with \(E(X)=0\) and \(V(X)=1\). What integer value \(k\) will assure us that \(P(|X| \geq k) \leq .01 ?\)

Let \(S_{n}\) be the number of successes in \(n\) Bernoulli trials with probability \(p\) for success on each trial. Show, using Chebyshev's Inequality, that for any \(\epsilon>0\) $$ P\left(\left|\frac{S_{n}}{n}-p\right| \geq \epsilon\right) \leq \frac{p(1-p)}{n \epsilon^{2}} $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Applied 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.