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Grade 10 Mathematics Learners’ Book

INNODEMS

Subsection 3.1.3 Measures of Central Tendency

Measures of central tendency are statistical values that describe the center or average of a set of data.
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They help us identify a single number that represents the entire distribution of data, giving us an idea of what is “typical” or “common.”
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The three main measures are Mean, Median and Mode.
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Together, these measures give us different perspectives of the center of the data.
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Subsubsection 3.1.3.1 Ungrouped Data for Measures of Central Tendency

Ungrouped data is the raw form of data, where individual observations are listed as they are collected, without being organized into groups or classes
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It is simply a list of numbers, facts, or values recorded in the order they are obtained
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Activity 3.1.4.

\(\textbf{Work in groups}\)
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  1. Consider the following data set:
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    \(32, 33, 35,36, 38, 40, 41, 42, 44, 45, 47, 48, 50,52, 54, 55, 56, 57,\) \(58, 60, 62, 63, 65,66, 68, 70, 72, 74, 75, 78, 80, 82, 85\)
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  2. Construct a frequency distribution table for the data.
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  3. Find the Mean, Mode and Median for the data.
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  4. Discuss with other groups
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Key Takeaway 3.1.18.

There are three main measures of central tendency:
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  1. \(\textbf{Mean}\)
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    Mean is the sum of all values divided by the total number of values. It is also known as Arithmetic Average.
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    \begin{align*} \textbf{Mean} = \amp \frac{ \textbf{∑X}}{ \textbf{N}} \end{align*}
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    Where \(\textbf{X}\) represents the values in the dataset and \(\textbf{N}\) is the total number of values.
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    Frequency distribution table can be used to find the mean for the ungrouped data. Using the formula below:
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    \begin{align*} \overline{\textbf{x}} = \amp \frac{ \textbf{∑fx}}{ \textbf{∑f}} \end{align*}
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    Where;
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    • \(\textbf{x}\) represents the values in the dataset
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    • \(\overline{\textbf{x}}\) is the mean
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    • \(\textbf{∑fx}\) is the sum of products of \(\textbf{ x}\) and \(\textbf{ f}\)
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    • \(\textbf{∑f}\) is the Sum of frequencies
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  2. \(\textbf{Median}\)
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    Median is the middle value when the data is arranged in ascending or descending order. If the dataset has an even number of values, the median is the average of the two middle values.
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    To find the median for even numbers, the formula is;
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    \(\textbf{Median} \) = \(\frac{\left(\frac{n}{2}\right)^{\text{th}} + \left(\frac{n}{2} + 1\right)^{\text{th}}}{2}\)
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  3. \(\textbf{Mode}\)
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    Mode is the most frequently occurring value in the dataset. A dataset can have:
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    • No mode: if no value repeats.
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    • One mode (Unimodal): if one value appears most frequently.
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    • Two modes (Bimodal): if two values appear equally most frequently.
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    • Multimodal: if more than two values appear frequently.
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Example 3.1.19.

In a class of \(30\) students, the test scores of students are:
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\(45, 67, 89, 56, 45, 78, 90, 67, 81, 73, 55, 62, 77, 84, 91,\) \(69, 58, 72, 88, 95, 60, 75, 45, 67, 80, 92, 87, 79, 68, 55\)
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Find the:
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  1. Mean
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  2. Median
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  3. Mode
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Solution.
  1. The formula for the mean is:
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    \begin{align*} \textbf{Mean} = \amp \frac{ \textbf{∑X}}{ \textbf{N}} \end{align*}
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    Where;
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    • \(∑X\) is the sum of all values
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    • \(N\) is the total number of values
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    \(∑X\) = \(45+67+89+56+45+78+90+67+81+73+55+62\)
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    \(+77+84+91+69+58+72+88+95+60+75+45+67\)
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    \(+80+92+87+79+68+55\)
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    = \(2096\)
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    \begin{align*} = \amp \frac{2096}{30}\\ = \amp 69.87 \end{align*}
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    Therefore, the mean is \(69.87\)
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  2. Median
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    The median is the middle value when data is arranged in ascending or descending order.
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    \(45, 45, 45, 55, 55, 56, 58, 60, 62, 67, 67, 67, 68, 69, 72,\)
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    \(73, 75, 77, 78, 79, 80, 81, 84, 87, 88, 89, 90, 91, 92, 95\)
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    Since there are \(30\) values (even number), the median is;
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    \(\textbf{Median} \) = \(\frac{\left(\frac{n}{2}\right)^{\text{th}} + \left(\frac{n}{2} + 1\right)^{\text{th}}}{2}\)
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    = \(\frac{\left(\frac{30}{2}\right)^{\text{th}} + \left(\frac{30}{2} + 1\right)^{\text{th}}}{2}\)
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    \(15\textbf{th}\) = \(72\)
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    \(16\textbf{th}\) = \(73\)
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    \begin{align*} \textbf{Median} = \amp \frac{72+73}{2}\\ = \amp \frac{145}{2}\\ = \amp 72.5 \end{align*}
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    Therefore the median is \(72.5\)
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  3. Mode
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    The mode is the most frequently occurring values.
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    From the sorted data:
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    Since \(45\) and \(67\) appear most frequently, this dataset is bimodal with modes: \(45\) and \(67\text{.}\)
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Example 3.1.20.

The frequency distribution table below shows marks of \(20\) students in a Grade 10 class.
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Table 3.1.21.
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Marks (x) Frequency (f) fx
\(2\) \(3\) \(6\)
\(4\) \(2\) \(8\)
\(6\) \(4\) \(24\)
\(7\) \(3\) \(21\)
\(9\) \(5\) \(45\)
\(11\) \(2\) \(22\)
\(12\) \(1\) \(12\)
\(Σx = 51\) \(Σf = 20\) \(Σfx = 138\)
  1. Find the mean
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  2. find the mode
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Solution.
  1. To find the mean
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    \begin{align*} \overline{\textbf{x}} = \amp \frac{ \textbf{∑fx}}{ \textbf{∑f}} \\ = \amp \frac{138}{20}\\ = \amp 6.9 \end{align*}
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    Therefore, the mean is \(6.9\)
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  2. The mode is the mark with the highest frequency. In this case, the highest frequency is \(5\text{,}\) which corresponds to \(9\text{.}\)
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    Therefore, Mode is \(9\)
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Exercises Exercises

1.
The number of books borrowed by students from a school library in a week is as follows:
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\(3, 5, 2, 4, 6, 3, 5, 7, 4, 3, 6, 2, 4, 5, 3, 6\)
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  1. Find the Mean (Average) number of books borrowed.
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  2. Find the Median number of books borrowed.
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  3. Find the Mode of the books borrowed.
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2.
The following frequency distribution table represents volume of water (in liters) contained in different bottles:
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Table 3.1.22.
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Volume of water (liters) Number of bottles
\(34.5\) \(3\)
\(35.8\) \(4\)
\(37.2\) \(2\)
\(39.0\) \(3\)
\(40.4\) \(3\)
  1. Find the mean volume of water in the bottles.
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  2. Determine the mode of the data.
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  3. Find the median volume.
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3.
A company records the monthly salaries (in KES) of its employees.
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Table 3.1.23.
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Salary (KES) Frequency
\(25,000\) \(8\)
\(30,000\) \(15\)
\(35,000\) \(12\)
\(40,000\) \(10\)
\(50,000\) \(5\)
Calculate the mean, median and mode.
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4.
A factory records the number of products manufactured in a week:
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\(150, 130, 220, 135, 180, 140, 125, 250, 145, 230, 200, 205, 145, 190, 155,\)
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\(210, 225, 240, 135, 165, 245, 170, 175, 185, 130, 190, 195, 160, 170, 150,\)
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\(160, 120, 200, 210, 220, 235, 180, 230, 240, 215\)
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  1. Make a frequency distribution table for the set of data.
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  2. Calculate the mean, mode and median.
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Checkpoint 3.1.24.

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Checkpoint 3.1.25.

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Subsubsection 3.1.3.2 Grouped Data for Measures of Central Tendency

Grouped data is data that has been organized into classes or intervals together with their frequencies.
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Instead of listing every single value (as in ungrouped data), the observations are arranged into groups (class intervals) to make large data sets easier to understand, analyze, and interpret.
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Activity 3.1.5.

\(\textbf{Work in groups}\)
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The amount of pocket money, in shillings, that parents give to students per week.
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Table 3.1.26.
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Pocket Money (Ksh) \(100 - 199\) \(200 - 299\) \(300 - 399\) \(400 - 499\) \(500 - 599\)
Number of students \(8\) \(15\) \(22\) \(20\) \(10\)
  1. What is the modal class of pocket money given to students?
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  2. Calculate the mean of pocket money given to students per week.
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  3. Find the median amount of pocket money from the given data.
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  4. Discuss with other groups.
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Key Takeaway 3.1.27.

  1. \(\textbf{Mean}\)
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    Frequency distribution table can be used to find the mean for the grouped data. Using the formula below:
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    \begin{align*} \overline{\textbf{x}} = \amp \frac{ \textbf{∑fx}}{ \textbf{∑f}} \end{align*}
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    Where;
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    • \(\textbf{x}\) is the midpoint
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    • \(\overline{\textbf{x}}\) is the mean
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    • \(\textbf{∑fx}\) is the sum of products of \(\textbf{ x}\) and \(\textbf{ f}\)
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    • \(\textbf{∑f}\) is the Sum of frequencies
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    \(\textbf{Midpoint}\) is the average of the lower and upper boundaries of a class interval. It represents the central value of each class.
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    \begin{align*} \textbf{Midpoint} = \amp \frac{\textbf{Lower boundary + Upper boundary}}{2} \end{align*}
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  2. \(\textbf{Median}\)
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    For grouped frequency data, we use interpolation to estimate the median using the following formula;
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    \begin{align*} \textbf{Median} = \amp \textbf{L} + (\frac{\frac{n}{2} - \textbf{CF}}{\textbf{F}})\times \textbf{C} \end{align*}
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    Where;
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    • \(\textbf{L}\) = Lower boundary of the median class.
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    • \(\textbf{n}\) = sum of all frequencies.
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    • \(\textbf{CF}\) = Cumulative frequency before the median class.
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    • \(\textbf{F}\) = Frequency of the median class.
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    • \(\textbf{C}\) = class width
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  3. \(\textbf{Mode}\)
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    The Modal class is the class with highest frequency.
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Example 3.1.28.

A company records the monthly salaries (in KES) of \(50\) employees in a frequency distribution table below:
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Table 3.1.29.
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Salary Range(KES) Number of Employees
\(20,000 - 29,999\) \(3\)
\(30,000 - 39,999\) \(5\)
\(40,000 - 49,999\) \(7\)
\(50,000 - 59,999\) \(10\)
\(60,000 - 69,999\) \(9\)
\(70,000 - 79,999\) \(6\)
\(80,000 - 89,999\) \(5\)
\(90,000 - 99,999\) \(3\)
\(100,000 - 109,999\) \(2\)
  1. Find the mean and mode of the data.
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  2. Find the modal class of the data.
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Solution.
  1. To calculate the mean of the grouped data we need to find;
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    • \(\textbf{∑fx}\) which is the sum of products of \(\textbf{x}\) and \(\textbf{f}\)
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    • \(\textbf{∑f}\) is the sum of frequencies
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    Table 3.1.30.
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    Salary Range(KES) Midpoint(x) Number of Employees(f) xf
    \(20,000 - 29,999\) \(25,000\) \(3\) \(25,000 \times 3 = 75,000\)
    \(30,000 - 39,999\) \(35,000\) \(5\) \(35,000 \times 5 = 175,000\)
    \(40,000 - 49,999\) \(45,000\) \(7\) \(45,000 \times 7 = 315,000\)
    \(50,000 - 59,999\) \(55,000\) \(10\) \(55,000 \times 10 = 550,000\)
    \(60,000 - 69,999\) \(65,000\) \(9\) \(65,000 \times 9 = 585,000\)
    \(70,000 - 79,999\) \(75,000\) \(6\) \(75,000 \times 6 = 450,000\)
    \(80,000 - 89,999\) \(85,000\) \(5\) \(85,000 \times 5 = 425,000\)
    \(90,000 - 99,999\) \(95,000\) \(3\) \(95,000 \times 3 = 285,000\)
    \(100,000 - 109,999\) \(105,000\) \(2\) \(105,000 \times 2 = 210,000\)
    \(\textbf{Total (∑)}\) \(585,000\) \(50\) \(3,070,000\)
    \begin{align*} \overline{\textbf{x}} = \amp \frac{ \textbf{∑fx}}{ \textbf{∑f}} \\ = \amp \frac{ 3,070,000}{ 50}\\ = \amp 61,400 \end{align*}
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    Therefore, the mean is \(61,400\)
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  2. The modal class is \(50,000 - 59,999 \textbf{ KES}\text{,}\) since is the one with the highest frequency.
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Example 3.1.31.

The data below represents the times (in seconds) recorded in the heats of a \(100 \textbf{ m}\) race during an athletics event:
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\(14.5, 13.7, 14.8, 15.3, 15.1, 14.2, 14.9, 12.6, 11.9, 13.1, 12.3, 14.7, 14.1, 15.0,\)
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\(14.3, 15.2, 11.7, 12.9, 13.5, 15.4, 12.8, 12.1, 14.4, 13.2, 14.6, 11.6, 12.7, 15.5\)
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\(14.0, 14.9, 13.9, 12.0, 13.8, 15.2, 13.3\)
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  1. Create a frequency distribution table using class intervals:
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    • \(\displaystyle 11.5 - 11.9\)
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    • \(\displaystyle 12.0 - 12.4\)
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    • \(\displaystyle 12.5 - 12.9\)
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    • \(\displaystyle 13.0 - 13.4\)
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    • \(\displaystyle 13.5 - 13.9\)
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    • \(\displaystyle 14.0 - 14.4\)
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    • \(\displaystyle 14.5 - 14.9\)
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    • \(\displaystyle 15.0 - 15.4\)
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    • \(\displaystyle 15.5 - 15.9\)
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  2. Determine the modal class
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  3. Estimate median based on the frequency table.
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  4. Find the mean based on the frequency table.
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Solution.
  1. To create a frequency distribution table We need to count how many values fall into each class interval.
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    Table 3.1.32.
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    Class Interval (seconds) Midpoint(x) Frequency(f) Cumulative frequency(CF) fx
    \(11.5 - 11.9\) \(11.7\) \(2\) \(2\) \(23.4\)
    \(12.0 - 12.4\) \(12.2\) \(4\) \(6\) \(36.6\)
    \(12.5 - 12.9\) \(12.7\) \(5\) \(11\) \(38.1\)
    \(13.0 - 13.4\) \(13.2\) \(4\) \(15\) \(52.8\)
    \(13.5 - 13.9\) \(13.7\) \(5\) \(20\) \(54.8\)
    \(14.0 - 14.4\) \(14.2\) \(6\) \(26\) \(71.0\)
    \(14.5 - 14.9\) \(14.7\) \(5\) \(31\) \(73.5\)
    \(15.0 - 15.4\) \(15.2\) \(4\) \(35\) \(60.8\)
    \(15.5 - 15.9\) \(15.7\) \(1\) \(36\) \(15.7\)
    \(\textbf{Total (∑)}\) \(123.3\) \(36\) \(36\) \(491.2\)
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  2. The modal class is the class interval with the highest frequency.
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    From the table, The highest frequency is \(6\text{,}\) which appears in class interval: \(14.0 - 14.4\text{.}\)
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    So, the modal class is \(14.0 - 14.4\)
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  3. Total frequency (∑f) = \(36\)
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    Median position is \(\frac{36}{2}\) = \(18\)
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    The cumulative frequency just before \(18\) is \(15\text{,}\) and the next class reaches \(20\text{,}\) so the median class is \(13.5 - 13.9\text{.}\)
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    Using the median formula:
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    \begin{align*} \textbf{Median} = \amp \textbf{L} + (\frac{\frac{n}{2} - \textbf{CF}}{\textbf{F}})\times \textbf{C} \end{align*}
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    Where;
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    • L = \(13.5\) (lower boundary of median class)
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    • \(\frac{n}{2}\) = \(18\)
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    • CF = \(15\) (cumulative frequency before the median class)
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    • F = \(5\) (frequency of the median class)
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    • C = \(0.5\) class width
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    \begin{align*} \textbf{Median} = \amp 13.5 + (\frac{18 - 15}{5})\times 0.5\\ = \amp 13.5 + (\frac{3}{5}\times 0.5)\\ = \amp 13.5 + 0.3\\ = \amp 13.8 \end{align*}
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    Thus, the median time is \(13.8 \textbf{ seconds}\text{.}\)
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  4. The mean is given by;
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    \begin{align*} \overline{\textbf{x}} = \amp \frac{ \textbf{∑fx}}{ \textbf{∑f}} \end{align*}
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    From the table;
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    \(\textbf{∑fx}\) = \(491.2\)
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    \(\textbf{∑f}\) = \(36\)
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    \begin{align*} = \amp \frac{ 491.2}{ 36}\\ = \amp 13.64 \end{align*}
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    Thus, the mean time is \(13.64 \textbf{ seconds}\)
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Exercises Exercises

1.
Mathematics test scores for \(60\) students in a class from Ichina primary school are:
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\(32, 45, 12, 56, 38, 74, 60, 29, 41, 50, 55, 67, 72, 31, 47,\)
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\(39, 18, 26, 64, 42, 48, 52, 69, 77, 35, 58, 23, 19, 61, 54,\)
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\(70, 33, 28, 37, 44, 46, 30, 49, 79, 62, 21, 16, 53, 57, 40,\)
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\(34, 25, 68, 66, 51, 59, 71, 27, 20, 36, 43, 63, 65, 75, 80\)
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  1. Create a frequency distribution table using class intervals of \(10\text{,}\) starting from \(10 - 19\text{,}\) \(20 - 29\text{,}\) \(30 - 39\text{,}\) ..., \(70 - 79\text{.}\)
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  2. Determine the modal class
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  3. Estimate the mean and median from the frequency table.
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2.
The population of \(50\) towns in Kakamega was recorded as follows:
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\(152, 168, 140, 155, 172, 184, 176, 193, 150, 160, 175, 143, 182, 164, 149,\)
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\(170, 185, 157, 169, 188, 154, 178, 166, 147, 190,\)
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\(145, 180, 158, 137, 174, 192, 141, 165, 187, 144,\)
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\(162, 153, 171, 139, 148, 156, 183, 177, 186, 159\)
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  1. Create a grouped frequency table with class intervals of \(10\text{,}\) starting from \(135 - 144\text{.}\)
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  2. Determine the modal class.
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  3. Estimate the mean and median from the distribution.
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3.
The monthly electricity bills (in KES) of households in a town are recorded in the table below:
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Table 3.1.33.
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Electricity Bill (KES) Frequency (f)
\(1,000 - 1,999\) \(6\)
\(2,000 - 2,999\) \(10\)
\(3,000 - 3,999\) \(14\)
\(4,000 - 4,999\) \(12\)
\(5,000 - 5,999\) \(8\)
  1. Find the median electricity bill.
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  2. Identify the modal class.
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4.
A researcher collects data on daily rainfall (in mm) over a month and organizes it into \(10\) equal class intervals
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Table 3.1.34.
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Rainfall (mm) Frequency (f)
\(0 - 9\) \(2\)
\(10 - 19\) \(4\)
\(20 - 29\) \(6\)
\(30 - 39\) \(8\)
\(40 - 49\) \(10\)
\(50 - 59\) \(12\)
\(60 - 69\) \(9\)
\(70 - 79\) \(6\)
\(80 - 89\) \(4\)
\(90 - 100\) \(2\)
  1. Identify the modal class.
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  2. Calculate the Mean rainfall.
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  3. Determine the Median rainfall.
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Checkpoint 3.1.35.

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Checkpoint 3.1.36.

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