/*! 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 1 The May 1, 2009, issue of The Mo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 1

The May 1, 2009, issue of The Montclarian reported the following home sale amounts for a sample of homes in Alameda, CA that were sold the previous month ( \(1000 \mathrm{~s}\) of $$)\( : \)590$ 815 575 608 350 1285 408 540 555 679 a. Calculate and interpret the sample mean and median.b. Suppose the 6th observation had been 985 rather than 1285\. How would the mean and median change? c. Calculate a 20% trimmed mean by first trimming the two smallest and two largest observations. d. Calculate a 15% trimmed mean.

Short Answer

Expert verified
Mean: $640.5K; Median: $582.5K. Changing to $610.5K for mean; 20% trimmed mean: $591.17K; 15% trimmed mean: $596.25K.

Step by step solution

01

Calculate the Sample Mean

To calculate the sample mean, first find the sum of all the listed amounts and divide by the number of observations (10).\[\text{Mean} = \frac{590 + 815 + 575 + 608 + 350 + 1285 + 408 + 540 + 555 + 679}{10}\]Calculate the sum:\[6405\]Now divide by the number of observations:\[\text{Mean} = \frac{6405}{10} = 640.5\]The sample mean is $640.5\text{K}.
02

Calculate the Sample Median

To calculate the median, list the sale amounts in ascending order:\[350, 408, 540, 555, 575, 590, 608, 679, 815, 1285\]Since there are 10 numbers, the median will be the average of the 5th and 6th numbers.\[\text{Median} = \frac{575 + 590}{2} = 582.5\]The sample median is $582.5\text{K}.
03

Adjusted Mean and Median (changing 1285 to 985)

Now modify the 6th observation from 1285 to 985 and calculate again:**Mean Calculation:**\[\text{Sum with change} = 6405 - 1285 + 985 = 6105\]\[\text{New Mean} = \frac{6105}{10} = 610.5\]**Median Calculation:**Original numbers without 1285 (now 985):\[350, 408, 540, 555, 575, 590, 608, 679, 815, 985\]Again, find the average of 5th and 6th:\[\text{Median remains} = \frac{575 + 590}{2} = 582.5\]The new mean is \(610.5\text{K}, while the median remains \)582.5\text{K}.
04

Calculate a 20% Trimmed Mean

For a 20% trimmed mean, remove the smallest 2 and largest 2 values:List after trimming: 540, 555, 575, 590, 608, 679.Calculate the sum of the trimmed list:\[\text{Sum} = 540 + 555 + 575 + 590 + 608 + 679 = 3547\]The number of terms left are 6, divide the sum by this number:\[\text{Trimmed Mean} = \frac{3547}{6} \approx 591.17\]The 20% trimmed mean is approximately $591.17\text{K}.
05

Calculate a 15% Trimmed Mean

For a 15% trimmed mean, remove 1 smallest and 1 largest value (since trimming 15% involves removing closest integer value from each end in such a small dataset).List after trimming: 408, 540, 555, 575, 590, 608, 679, 815.Calculate the sum of the trimmed list:\[\text{Sum} = 408 + 540 + 555 + 575 + 590 + 608 + 679 + 815 = 4770\]The number of terms left are 8, divide the sum by this number:\[\text{Trimmed Mean} = \frac{4770}{8} = 596.25\]The 15% trimmed mean is $596.25\text{K}.

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.

Sample Mean
Understanding the sample mean is crucial when analyzing data sets. The sample mean, often just called the mean, is the average of all the data points in a sample. To find it, you need to sum all the individual values within your data set and then divide this sum by the total number of values. For example, in a list of home sale prices, compute the mean by adding up all the sale amounts and dividing by the number of sales.
Here's a simple step-by-step process:
  • Add all the individual sale prices together.
  • Count the number of sales.
  • Divide the total sale amount by the number of sales.
The sample mean provides a central value that represents the entire dataset. It's a useful measure of the overall trend, but can be affected by extremely high or low values, known as outliers.
Sample Median
The sample median is another key measure in descriptive statistics. Unlike the mean, the median is not affected by outliers and gives a better sense of the central location for skewed data. To find the median, you first need to put all your data points in numerical order.
In a list of home sale prices:
  • Order the sales from smallest to largest.
  • If there is an odd number of data points, the median is the middle number.
  • If there is an even number of data points, the median is the average of the two middle numbers.
The median shows the middle point of a data set, dividing it into two equal halves. It portrays a more accurate central tendency in datasets with skewed distributions or outliers.
Trimmed Mean
A trimmed mean provides a way to calculate an average that is less influenced by extreme values, offering a more robust measure of central tendency. Trimming involves removing a certain percentage of the smallest and largest values from the dataset. This process is useful in datasets with significant outliers that can skew results.
Steps for calculating a trimmed mean:
  • Decide on the percentage of data to trim, often 5%, 10%, or 20%.
  • Rank the data in order.
  • Remove the specified percentage of the smallest and largest data points.
  • Compute the mean of the remaining values.
A trimmed mean can provide a more reliable measure of central tendency, especially in datasets where extreme values could distort the analysis.
Outlier Adjustment
Outliers are data points that significantly differ from other observations and can affect statistical analyses such as means. An outlier adjustment seeks to minimize their impact, allowing for a more accurate representation of central tendency.
You can handle outliers in several ways:
  • Remove the outlier if it is a result of data entry error or doesn't belong to the dataset.
  • Use a trimmed mean, which naturally discards outlying data points.
  • Transform the data, possibly by applying mathematical functions to reduce the effect of outliers.

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 article "Effects of Short-Term Warming on Low and High Latitude Forest Ant Communities" (Ecoshpere, May 2011, Article 62) described an experiment in which observations on various characteristics were made using minichambers of three different types: (1) cooler (PVC frames covered with shade cloth), (2) control (PVC frames only), and (3) warmer (PVC frames covered with plastic). One of the article's authors kindly supplied the accompanying data on the difference between air and soil temperatures \(\left({ }^{\circ} \mathrm{C}\right)\). \(\begin{array}{ccc}\text { Cooler } & \text { Control } & \text { Warmer } \\\ 1.59 & 1.92 & 2.57 \\ 1.43 & 2.00 & 2.60 \\ 1.88 & 2.19 & 1.93 \\ 1.26 & 1.12 & 1.58 \\ 1.91 & 1.78 & 2.30 \\ 1.86 & 1.84 & 0.84 \\ 1.90 & 2.45 & 2.65 \\ 1.57 & 2.03 & 0.12 \\ 1.79 & 1.52 & 2.74 \\ 1.72 & 0.53 & 2.53 \\\ 2.41 & 1.90 & 2.13 \\ 2.34 & & 2.86 \\ 0.83 & & 2.31 \\ 1.34 & & 1.91 \\\ 1.76 & & \end{array}\) a. Compare measures of center for the three different samples. b. Calculate, interpret, and compare the standard deviations for the three different samples. c. Do the fourth spreads for the three samples convey the same message as do the standard deviations about relative variability? d. Construct a comparative boxplot (which was included in the cited article) and comment on any interesting features.

In 1997 a woman sued a computer keyboard manufacturer, charging that her repetitive stress injuries were caused by the keyboard (Genessy \(v\). Digital Equipment Corp.). The injury awarded about \(\$ 3.5\) million for pain and suffering, but the court then set aside that award as being unreasonable compensation. In making this determination, the court identified a "normative" group of 27 similar cases and specified a reasonable award as one within two standard deviations of the mean of the awards in the 27 cases. The 27 awards were (in \(\$ 1000\) s) 37,60 , \(75,115,135,140,149,150,238,290,340,410,600,750\), \(750,750,1050,1100,1139,1150,1200,1200,1250\), \(1576,1700,1825\), and 2000 , from which \(\Sigma x_{i}=20,179\), \(\Sigma x_{i}^{2}=24,657,511\). What is the maximum possible amount that could be awarded under the two-standarddeviation rule?

The three measures of center introduced in this chapter are the mean, median, and trimmed mean. Two additional measures of center that are occasionally used are the midrange, which is the average of the smallest and largest observations, and the midfourth, which is the average of the two fourths. Which of these five measures of center are resistant to the effects of outliers and which are not? Explain your reasoning.

A certain city divides naturally into ten district neighborhoods. How might a real estate appraiser select a sample of single-family homes that could be used as a basis for developing an equation to predict appraised value from characteristics such as age, size, number of bathrooms, distance to the nearest school, and so on? Is the study enumerative or analytic?

The following categories for type of physical activity involved when an industrial accident occurred appeared in the article "Finding Occupational Accident Patterns in the Extractive Industry Using a Systematic Data Mining Approach" (Reliability Engr. and System Safety, 2012: 108-122): A. Working with handheld tools B. Movement C. Carrying by hand D. Handling of objects E. Operating a machine F. Other Construct a frequency distribution, including relative frequencies, and histogram for the accompanying data from 100 accidents (the percentages agree with those in the cited article): \(\begin{array}{llllllllllllll}\mathrm{A} & \mathrm{B} & \mathrm{D} & \mathrm{A} & \mathrm{A} & \mathrm{F} & \mathrm{C} & \mathrm{A} & \mathrm{C} & \mathrm{B} & \mathrm{E} & \mathrm{B} & \mathrm{A} & \mathrm{C} \\ \mathrm{F} & \mathrm{D} & \mathrm{B} & \mathrm{C} & \mathrm{D} & \mathrm{A} & \mathrm{A} & \mathrm{C} & \mathrm{B} & \mathrm{E} & \mathrm{B} & \mathrm{C} & \mathrm{E} & \mathrm{A} \\ \mathrm{B} & \mathrm{A} & \mathrm{A} & \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{C} & \mathrm{D} & \mathrm{F} & \mathrm{D} & \mathrm{B} & \mathrm{B} & \mathrm{A} & \mathrm{F} \\\ \mathrm{C} & \mathrm{B} & \mathrm{A} & \mathrm{C} & \mathrm{B} & \mathrm{E} & \mathrm{B} & \mathrm{D} & \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{E} & \mathrm{A} & \mathrm{A} \\ \mathrm{F} & \mathrm{C} & \mathrm{B} & \mathrm{D} & \mathrm{D} & \mathrm{D} & \mathrm{B} & \mathrm{D} & \mathrm{C} & \mathrm{A} & \mathrm{F} & \mathrm{A} & \mathrm{A} & \mathrm{B} \\\ \mathrm{D} & \mathrm{B} & \mathrm{A} & \mathrm{E} & \mathrm{D} & \mathrm{B} & \mathrm{C} & \mathrm{A} & \mathrm{F} & \mathrm{A} & \mathrm{C} & \mathrm{D} & \mathrm{D} & \mathrm{A} \\ \mathrm{A} & \mathrm{B} & \mathrm{A} & \mathrm{F} & \mathrm{D} & \mathrm{C} & \mathrm{A} & \mathrm{C} & \mathrm{B} & \mathrm{F} & \mathrm{D} & \mathrm{A} & \mathrm{E} & \mathrm{A} \\ \mathrm{C} & \mathrm{D} & & & & & & & & & & & & \end{array}\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Geometry

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.