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Chapter 2: Problem 16

A data set on money spent on lottery tickets during the past year by 200 households has a lowest value of \(\$ 1\) and a highest value of \(\$ 1167\). Suppose we want to group these data into six classes of equal widths. a. Assuming that we take the lower limit of the first class as \(\$ 1\) and the width of each class equal to \(\$ 200\), write the class limits for all six classes. b. What are the class boundaries and class midpoints?

Short Answer

Expert verified
Class limits: \( \$1-\$200, \$201-\$400, \$401-\$600, \$601-\$800, \$801-\$1000, \$1001-\$1200 \), Class Boundaries: \( \$0.5-\$200.5, \$200.5-\$400.5, \$400.5-\$600.5, \$600.5-\$800.5, \$800.5-\$1000.5, \$1000.5-\$1200.5 \), and class midpoints: \( \$100.5, \$300.5, \$500.5, \$700.5, \$900.5, \$1100.5 \).

Step by step solution

01

Class Limits

The class limits given the width of each class as \( \$200 \) and the lower limit of the first class as \( \$1 \) are:1. \( \$1 - \$200 \) 2. \( \$201 - \$400 \) 3. \( \$401 - \$600 \)4. \( \$601 - \$800 \)5. \( \$801 - \$1000 \)6. \( \$1001 - \$1200 \)
02

Class Boundaries

Class boundaries are the halfway points between the upper limit of one class and the lower limit of the next class. This typically adds a .5 to the upper class limit and subtracts a .5 from the lower class limit. For this exercise, using \( \$ \) and rounded up to next integer, we have:1. \( \$0.5 - \$200.5 \)2. \( \$200.5 - \$400.5 \) 3. \( \$400.5 - \$600.5 \)4. \( \$600.5 - \$800.5 \)5. \( \$800.5 - \$1000.5 \)6. \( \$1000.5 - \$1200.5 \)
03

Class Midpoints

The class midpoint is the value halfway between the lower and upper limit of a class. It can be found by averaging the two values. For this exercise, we have:1. \( \$((1+200)/2) = \$100.5 \)2. \( \$((201+400)/2) = \$300.5 \) 3. \( \$((401+600)/2) = \$500.5 \)4. \( \$((601+800)/2) = \$700.5 \)5. \( \$((801+1000)/2) = \$900.5 \)6. \( \$((1001+1200)/2) = \$1100.5 \)

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Key Concepts

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

Class Limits
Class limits are the values that define the range of each class in a data set. They indicate where a class starts and ends.
  • The lower class limit is the smallest value that can belong to a class.
  • The upper class limit is the largest value that can be included in a class.
To find class limits, you first need to determine the number of classes and then the width of each class. In our example, each class has a width of $200, starting from $1 for the first class. Therefore, the class limits are as follows:
  • Class 1: $1 - $200
  • Class 2: $201 - $400
  • Class 3: $401 - $600
  • Class 4: $601 - $800
  • Class 5: $801 - $1000
  • Class 6: $1001 - $1200
These limits help us group data into intervals, making it easier to visualize data distribution through tools like histograms.
Class Boundaries
Class boundaries help to eliminate any gaps between classes. They are especially useful for plotting histograms as they enable a smooth transition from one class to another.
Class boundaries are calculated by finding the midpoint between the upper limit of one class and the lower limit of the next. Typically, this means adding 0.5 to the upper limit and subtracting 0.5 from the lower limit.
  • Class 1 boundary: $0.5 - $200.5
  • Class 2 boundary: $200.5 - $400.5
  • Class 3 boundary: $400.5 - $600.5
  • Class 4 boundary: $600.5 - $800.5
  • Class 5 boundary: $800.5 - $1000.5
  • Class 6 boundary: $1000.5 - $1200.5
By using class boundaries, you can accurately place data points that fall on the exact upper limit of a class into their respective positions in a histogram.
Class Midpoints
Class midpoints, also known as class marks, are the central values of each class interval. They provide a point of reference that represents the data in a class.
Midpoints are found by averaging the lower and upper class limits of each class. Here's how you can calculate them:
  • For Class 1: \((1 + 200) / 2 = 100.5\)
  • For Class 2: \((201 + 400) / 2 = 300.5\)
  • For Class 3: \((401 + 600) / 2 = 500.5\)
  • For Class 4: \((601 + 800) / 2 = 700.5\)
  • For Class 5: \((801 + 1000) / 2 = 900.5\)
  • For Class 6: \((1001 + 1200) / 2 = 1100.5\)
Class midpoints are extremely useful for further statistical analysis, such as calculating the mean or standard deviation of grouped data. They also allow for making comparisons across different data classes.

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Most popular questions from this chapter

The following data give the numbers of orders received for a sample of 30 hours at the Timesaver Mail Order Company. \(\begin{array}{llllllllll}34 & 44 & 31 & 52 & 41 & 47 & 38 & 35 & 32 & 39 \\\ 28 & 24 & 46 & 41 & 49 & 53 & 57 & 33 & 27 & 37 \\ 30 & 27 & 45 & 38 & 34 & 46 & 36 & 30 & 47 & 50\end{array}\) a. Construct a frequency distribution table. Take 23 as the lower limit of the first class and 7 as the width of each class. b. Calculate the relative frequencies and percentages for all classes. c. For what percentage of the hours in this sample was the number of orders more than 36 ?

The following data give the repair costs (in dollars) for 30 cars randomly selected from a list of cars that were involved in collisions. \(\begin{array}{lrrrrr}2300 & 750 & 2500 & 410 & 555 & 1576 \\ 2460 & 1795 & 2108 & 897 & 989 & 1866 \\ 2105 & 335 & 1344 & 1159 & 1236 & 1395 \\ 6108 & 4995 & 5891 & 2309 & 3950 & 3950 \\ 6655 & 4900 & 1320 & 2901 & 1925 & 6896\end{array}\) a. Construct a frequency distribution table. Take \(\$ 1\) as the lower limit of the first class and \(\$ 1400\) as the width of each class. b. Compute the relative frequencies and percentages for all classes. c. Draw a histogram and a polygon for the relative frequency distribution. d. What are the class boundaries and the width of the fourth class?

The National Highway Traffic Safety Administration collects information on fatal accidents that occur on roads in the United States. Following are the number of fatal motorcycle accidents that occurred in each of South Carolina's 46 counties during the year 2009 (http://www-fars.nhtsa.dot.gov). \(\begin{array}{rrrrrrrrrrr}3 & 28 & 3 & 35 & 3 & 7 & 13 & 38 & 6 & 44 & 11 & 14 \\ 12 & 18 & 17 & 17 & 6 & 20 & 3 & 7 & 29 & 17 & 51 & 12 \\ 5 & 60 & 12 & 18 & 17 & 21 & 14 & 34 & 3 & 12 & 8 & 5 \\ 11 & 29 & 20 & 40 & 3 & 30 & 23 & 5 & 10 & 23 & & \end{array}\) a. Construct a frequency distribution table using the classes \(1-10,11-20,21-30,31-40,41-50\), and \(51-60\) b. Calculate the relative frequency and percentage for each class. c. Construct a histogram and a polygon for the relative frequency distribution of part b. d. What percentage of the counties had between 21 and 40 fatal motorcycle accidents during \(2009 ?\)

Why do we need to group data in the form of a frequency table? Explain briefly.

The following data give the political party of each of the first 30 U.S. presidents. In the data, D stands for Democrat, DR for Democratic Republican, \(\mathrm{F}\) for Federalist, \(\mathrm{R}\) for Republican, and \(\mathrm{W}\) for Whig. \(\begin{array}{lllllllll}\text { F } & \text { F } & \text { DR } & \text { DR } & \text { DR } & \text { DR } & \text { D } & \text { D } & \text { W } & \text { W } \\ \text { D } & \text { W } & \text { W } & \text { D } & \text { D } & \text { R } & \text { D } & \text { R } & \text { R } & \text { R } \\ \text { R } & \text { D } & \text { R } & \text { D } & \text { R } & \text { R } & \text { R } & \text { D } & \text { R } & \text { R }\end{array}\) a. Prepare a frequency distribution table for these data. b. Calculate the relative frequency and percentage distributions. c. Draw a bar graph for the relative frequency distribution and a pie chart for the percentage distribution. d. What percentage of these presidents were Whigs?
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