/*! 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 Prices of cars have a distributi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 8

Prices of cars have a distribution that is skewed to the right with outliers in the right tail. Which of the measures of central tendency is the best to summarize this data set? Explain.

Short Answer

Expert verified
In the case of skewed data with outliers in the right tail, the median would be the best measure of central tendency. Because it's not affected by the extreme values or outliers.

Step by step solution

01

Understanding the mean in case of skewed data

The arithmetic mean or the average is influenced by every score in the data set. It is the sum of all the data divided by the count. In a data set with extreme values or outliers, such as with a positively skewed data set, the mean gets dragged in the direction of the skew. So, it is higher than the typical values of the data set, making it not the best option for this data set.
02

Understanding the median in case of skewed data

The median is the middle value that separates the higher half from the lower half of the data set. Being a positional average, the effect of extreme values, outliers, or skewed data is minimum. Therefore, it is usually a better choice for data with extreme outliers or skewed data.
03

Understanding the mode in case of skewed data

The mode is the value that appears most frequently in a data set. In a skewed dataset, the mode represents the most common value but ignores the influence of the outliers or the skewed nature. So, it might not be the best measure of central tendency in this case.

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 following data are the ages (in years) of six students. \(\begin{array}{lll}19 & 19 & 19\end{array}\) \(\begin{array}{lllll}19 & 19 & 19 & 19 & 19\end{array}\) Calculate the standard deviation. Is its value zero? If yes, why?

The mean monthly mortgage paid by all home owners in a town is \(\$ 2365\) with a standard deviation of \(\$ 340\). a. Using Chebyshev's theorem, find at least what percentage of all home owners in this town pay a monthly mortgage of i. \(\$ 1685\) to \(\$ 3045\) ii. \(\$ 1345\) to \(\$ 3385\) "b. Using Chebyshev's theorem, find the interval that contains the monthly mortgage payments of at least \(84 \%\) of all home owners in this town.

Nixon Corporation manufactures computer monitors. The following data give the numbers of computer monitors produced at the company for a sample of 30 days. \(\begin{array}{llllllllll}24 & 32 & 27 & 23 & 33 & 33 & 29 & 25 & 23 & 36 \\\ 26 & 26 & 31 & 20 & 27 & 33 & 27 & 23 & 28 & 29 \\ 31 & 35 & 34 & 22 & 37 & 28 & 23 & 35 & 31 & 43\end{array}\) a. Calculate the values of the three quartiles and the interquartile range. Where does the value of 31 lie in relation to these quartiles? b. Find the (approximate) value of the 65 th percentile. Give a brief interpretation of this percentile. c. For what percentage of the days was the number of computer monitors produced 32 or higher? Answer by finding the percentile rank of 32 .

The following data give the numbers of text messages sent by a high school student on 40 randomly selected days during 2012: \(\begin{array}{llllllllll}32 & 33 & 33 & 34 & 35 & 36 & 37 & 37 & 37 & 37 \\\ 38 & 39 & 40 & 41 & 41 & 42 & 42 & 42 & 43 & 44 \\ 44 & 45 & 45 & 45 & 47 & 47 & 47 & 47 & 47 & 48 \\ 48 & 49 & 50 & 50 & 51 & 52 & 53 & 54 & 59 & 61\end{array}\) a. Calculate the values of the three quartiles and the interquartile range. Where does the value 49 fall in relation to these quartiles? b. Determine the approximate value of the 91 st percentile. Give a brief interpretation of this percentile. c. For what percentage of the days was the number of text messages sent 40 or higher? Answer by finding the percentile rank of 40 .

In a study of distances traveled to a college by commuting students, data from 100 commuters yielded a mean of \(8.73\) miles. After the mean was calculated, data came in late from three students, with respective distances of \(11.5,7.6\), and \(10.0\) miles. Calculate the mean distance for all 103 students.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Discrete 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.