/*! 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 63 Suppose that you roll a fair 10 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 63

Suppose that you roll a fair 10 -sided die \((0,1,2, \ldots, 9) 500\) times. Using Chebyshev's theorem, compute the probability that the sample mean, \(X,\) is between 4 and 5 .

Short Answer

Expert verified
Chebyshev's theorem doesn't provide a useful result in this scenario, as it generates a negative probability when its result is intended as a lower bound not an exact value.

Step by step solution

01

Calculate the Mean

The mean \(\mu\) of a fair 10-sided die (values from 0 to 9) can be calculated using the formula \(\mu = \frac{(0+1+2+3+...+9)}{10}\). This equals to 4.5 .
02

Calculate the Variance and Standard Deviation

The variance \(\sigma^2\) of a 10-sided die can be calculated by using the formula \(\sigma^2 = \frac{ \sum_{i=0}^{9} i^2}{10} - \mu^2\). Calculating this yields a variance of \( \frac{285}{10} - 20.25 = 8.25\). The standard deviation \(\sigma\) is the square root of the variance, so \(\sigma = \sqrt{8.25} \approx 2.87\).
03

Apply Chebyshev's theorem

Chebyshev's theorem states that the proportion of any distribution that lies within k standard deviations of the mean is at least \(1 - \frac{1}{k^2}\). Between 4 and 5, lie 0.5 values on either side of the mean (\(4.5\)). This is approximately \(\frac{0.5}{\sigma}=0.174\) standard deviations away from the mean. So, \(k = 0.174\). Plug this into Chebyshev's inequality to find the probability.
04

Compute the probability

Plugging into Chebyshev's theorem yields a minimum probability of \(1 - \frac{1}{(0.174)^2} \approx -33.24\). However, probabilities cannot be negative. This seeming paradox is because Chebyshev's theorem provides a lower bound on the probability, not an exact probability. In cases where k is small (as in this case), the theorem doesn't provide useful results.

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.

Probability Distribution
Understanding probability distribution is crucial when dealing with random phenomena. It represents how the probabilities of a random variable's possible values are dispersed. Specifically, it tells us the likelihood of each outcome.

A ten-sided die, as in our exercise, has equally likely outcomes when rolled. We're considering a uniform distribution because every outcome 鈥� from 0 to 9 鈥� has an equal chance of occurring, which is 1/10. Now, if you roll the die multiple times, the collection of results forms a sample from this distribution.

The sample mean, then, will give us an average of all the outcomes. This is especially useful in predicting the behavior of the die over the long term. The more rolls you make, the closer your sample mean is likely to get to the expected value, provided the die is fair.
Sample Mean Calculation
Calculating the sample mean is a fundamental concept in statistics, representing the average of a set of numbers. For the ten-sided die, our possible values range from 0 to 9, and the mean (expected value) is the middle ground of these outcomes.

To compute the mean \(\mu\), add all the possible outcomes and divide by the number of outcomes, which is ten in this case. The result is 4.5, an important figure when assessing the outcomes of repeated rolls. Understanding how to calculate the sample mean is essential in various fields, including quality control, research analysis, and even games involving probability.

When dealing with a large number of rolls or a significant data set, knowing the sample mean allows to estimate the central tendency of the outcomes quickly and helps in predicting future events based on past occurrences.
Standard Deviation and Variance
Standard deviation and variance are measures of dispersion that tell us how spread out our values are in a data set. Variance is the average of the squared differences from the mean, and standard deviation is the square root of this figure.

In the context of our die, calculating the variance \(\sigma^2\) involves a mathematical formula that considers each possible outcome's squared difference from the mean. With the ten-sided die, this calculation gives us a variance of 8.25. Taking the square root gives us the standard deviation, approximately 2.87, which reflects the amount of variability or spread in the outcomes of the die rolls.

Recognizing these two concepts is essential when doing any data analysis because they offer insights into the consistency of the data. Low variance and standard deviation indicate that the data points tend to be close to the mean, while higher numbers suggest a more widespread set of data. Also, these values are instrumental when applying statistical theorems, such as Chebyshev's theorem, to estimate the probabilities of outcomes within certain ranges.

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

Consider a ferry that can carry both buses and cars on a trip across a waterway. Each trip costs the owner approximately \(\$ 10 .\) The fee for cars is \(\$ 3\) and the fee for buses is \(\$ 8 .\) Let \(X\) and \(Y\) denote the number of buses and cars, respectively, carried on a given trip. The joint distribution of \(X\) and \(Y\) is given by $$\begin{array}{cccc} & & x & \\\y & 0 & 1 & 2 \\\\\hline 0 & 0.01 & 0.01 & 0.03 \\\1 & 0.03 & 0.08 & 0.07 \\ 2 & 0.03 & 0.06 & 0.06 \\\3 & 0.07 & 0.07 & 0.13 \\\4 & 0.12 & 0.04 & 0.03 \\\5 & 0.08 & 0.06 & 0.02\end{array}$$ Compute the expected profit for the ferry trip.

Consider Review Exercise 3.79 on page \(105 .\) The random variables \(X\) and \(Y\) represent the number of vehicles that arrive at two separate street corners during a certain 2-minute period in the day. The joint distribution is $$f(x, y)-\frac{1}{4^{(x+y)}} \cdot \frac{9}{16}$$ for \(x=0,1,2, \ldots,\) and \(y=0,1,2, \ldots\) (a) Give \(E(X), E(Y): \operatorname{Var}(X),\) and \(\operatorname{Var}(Y)\). (b) Consider \(Z=X+Y,\) the sum of the two. Find \(E(Z)\) and \(\operatorname{Var}(Z)\).

A coin is biased so that a head is three time:s as likely to occur as a tail. Find the expected number of tails when this coin is tossed twice.

The probability distribution of \(X,\) the number of imperfections per 10 meters of a synthetic fabric in continuous rolls of uniform width, is given in Exercise 3.13 on page 89 as $$\begin{array}{cccccc}x & \mathrm{I} & 0 & 1 & 2 & 3 & 4 \\\\\hline f(x) & 0.41 & 0.37 & 0.16 & 0.05 & 0.01\end{array}$$ Find the average number of imperfections per 10 meters of this fabric.

An attendant at a car wash is paid according to the number of cars that pass through. Suppose the probabilities are \(1 / 12,1 / 12,1 / 4,1 / 4,1 / 6,\) and \(1 / 6,\) respectively, that the attendant receives \(\$ 7, \$ 9 . \$ 11\) \$13, \$15, or \$17 between 4:00 P.M. and 5:00 P.M. on any sunny Friday. Find the attendant's expected earnings for this particular period.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Statistics

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.