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Grade 10 Mathematics Teachers’ Book

INNODEMS

Subsection 3.1.3 Grouped Data

When dealing with small sets of data, it is easy to list and analyze every individual value. However, in the real world, we often encounter datasets with hundreds or thousands of entries, such as the heights of all students in a school or the daily sales of a large store. Listing each number individually becomes impractical and difficult to interpret. To manage this, we organize data into groups or class intervals. In this section, we will learn how to summarize large datasets using frequency distribution tables and how to estimate measures of central tendency the mean, median, and mode for data that has been grouped.
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Subsubsection 3.1.3.1 Frequency Distribution Tables (Grouped)

Before we can analyze large datasets, we must first organize them. A Frequency Distribution Table simplifies raw data by sorting it into specific ranges called class intervals (such as \(0–10, 10–20,\) etc.).
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Instead of looking at a long list of scattered numbers, we count how many data points fall into each interval. This count is known as the frequency. By arranging data this way, we can quickly identify patterns, such as where the majority of the data lies or if there are any outliers, without needing to process every single number individually.
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Learner Experience 3.1.7.

A mathematics teacher recorded the test scores (out of \(50\)) for a class of \(25\) students. The raw data is listed below.
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Raw Scores:
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12, 34, 45, 23, 8, 36, 41, 29, 15, 48, 33, 25, 38, 19, 44, 27, 31, 49, 22, 14, 5, 42, 39, 28, 35.
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It is difficult to see how the class performed just by looking at this jumbled list. Let’s organize this data into a grouped frequency table to better understand the class performance.
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(a)
First, determine the range of the data. Identify the lowest score and the highest score in the list.
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(b)
Construct a frequency distribution table using class intervals of width \(10\text{.}\) Use the following intervals: \(0–9, 10–19, 20–29, 30–39, 40–49\text{.}\)
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Create three columns in your table:
  • Class Interval (The groups listed above)
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  • Tally (Mark each score in the correct group)
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  • Frequency (Count the tallies for the final number)
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(c)
Look at your completed table. Which class interval has the highest frequency? What does this tell you about the overall performance of the class?
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Key Takeaway 3.1.45.

Grouping data helps us see patterns and trends in large datasets that would otherwise be hidden. However, once data is grouped into class intervals, we lose the specific identity of the individual numbers. For this reason, calculations performed on grouped data are always considered estimates rather than exact values.
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Example 3.1.46.

The following data represents the daily earnings (in dollars) of \(20\) freelance workers:
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\(45, 62, 58, 41, 75, 50, 48, 66, 72, 55, 43, 60, 52, 78, 49, 64, 57, 70, 44, 69\)
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Construct a grouped frequency distribution table using class intervals of width \(10\text{.}\) Start with the interval \(40–49\text{.}\)
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Solution.
First, we define the intervals. Then, we count how many earnings fall into each group.
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  • 40–49: 45, 41, 48, 43, 49, 44 (\(6\) values)
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  • 50–59: 58, 50, 55, 52, 57 (\(5\) values)
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  • 60–69: 62, 66, 60, 64, 69 (\(5\) values)
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  • 70–79: 75, 72, 78, 70 (\(4\) values)
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The resulting frequency table is:
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Table 3.1.47. Daily Earnings Distribution
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Class Interval ($) Frequency (\(f\))
\(40–49\) \(6\)
\(50–59\) \(5\)
\(60–69\) \(5\)
\(70–79\) \(4\)
Total 20
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Example 3.1.48.

The marks obtained by 40 students in a mathematics test are:
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\(42, 55, 47, 61, 50, 38, 44, 49, 54, 60,\)
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\(41, 48, 52, 46, 57, 51, 39, 62, 45, 58,\)
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\(43, 56, 54, 59, 63, 40, 37, 64, 36, 65,\)
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\(47, 49, 50, 52, 55, 51, 48, 53, 60, 61\)
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Using \(6\) class intervals, construct a frequency distribution table.
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Solution.
To construct a frequency distribution table, we do the following
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  1. Determine the range
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    Maximum value \(= 65\)
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    Minimum value \(= 36\)
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    Range \(= 65 - 36 = 29\)
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  2. Decide class width
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    Number of classes \(= 6\)
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    Class width \(=\) Range \(\div\) Number of classes \(= 29 \div 6 \approx 4.83 \)
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    We can therefore use a class width of \(5\)
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  3. Complete the frequency table
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    Class Interval Frequency (f)
    \(36-40\) \(5\)
    \(14-45\) \(5\)
    \(46-50\) \(9\)
    \(51-55\) \(9\)
    \(56-60\) \(6\)
    \(61-65\) \(6\)
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Exercises Exercises

1.
Consider the following frequency distribution table representing the ages of members in a chess club:
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Table 3.1.49. Chess Club Ages
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Age Group Frequency
\(10-19\) \(8\)
\(20–29\) \(12\)
\(30–39\) \(5\)
How many members are in the chess club in total, and how many members are aged 20 or older?
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Answer.
Total members: \(8 + 12 + 5 = 25\text{.}\)
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Members aged \(20\) or older: \(12 + 5 = 17\text{.}\)
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2.
A dataset is grouped into the following class intervals: \(0–4\text{,}\) \(5–9\text{,}\) \(10–14\text{,}\) and \(15–19\text{.}\)
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Determine the class width of these intervals. (Hint: Be careful not to just subtract the limits; count the number of values included in the interval).
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Answer.
The class width is \(5\text{.}\)
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(For example, the interval \(0–4\) includes the values \(0, 1, 2, 3,\) and \(4\)).
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3.
The following is a list of distances (in km) cycled by a group of students over a weekend:
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5, 12, 8, 24, 15, 9, 28, 11, 6, 21
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Construct a frequency table with the class intervals: \(0–9\text{,}\) \(10–19\text{,}\) and \(20–29\text{.}\)
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Answer.
Table 3.1.50. Cycling Distances
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Distance (km) Frequency
\(0–9\) \(4\)
\(10–19\) \(3\)
\(20–29\) \(3\)
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4.
A researcher surveyed \(50\) households about their monthly electricity usage. The data was grouped into a frequency table, but one value was erased by accident.
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Table 3.1.51. Electricity Usage
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Usage (kWh) Frequency
\(0–100\) \(5\)
\(101–200\) \(18\)
\(201–300\) \(x\)
\(301–400\) \(7\)
Calculate the missing frequency \(x\text{.}\)
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Answer.
We know the total frequency is \(50\text{.}\)
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\(5 + 18 + x + 7 = 50\)
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\(30 + x = 50\)
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\(x = 20\)
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Subsubsection 3.1.3.2 Mean (Grouped)

Grouped data is data that has been organized into classes or intervals together with their frequencies. Instead of listing every single value, the observations are arranged into groups to make large data sets easier to analyze.
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Learner Experience 3.1.8.

Work in groups
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The table below shows the amount of pocket money, in shillings, that parents give to students per week.
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Table 3.1.53. Pocket Money Distribution
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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. Determine the midpoint for each class interval.
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  2. Calculate the mean of pocket money given to students per week.
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Key Takeaway 3.1.54.

Calculating Mean
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A frequency distribution table can be used to find the mean for grouped data using the formula:
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\begin{align*} \overline{x} = \amp \frac{ \sum fx}{ \sum f} \end{align*}
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Where:
  • \(x\) is the Midpoint of the class interval.
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  • \(\overline{x}\) is the mean.
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  • \(\sum fx\) is the sum of the products of the midpoint and frequency.
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  • \(\sum f\) is the sum of frequencies (total number of data points).
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Midpoint is the average of the lower and upper boundaries of a class interval.
\begin{align*} \text{Midpoint} = \amp \frac{\text{Lower boundary + Upper boundary}}{2} \end{align*}
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Example 3.1.55.

A company records the monthly salaries (in KES) of \(50\) employees in a frequency distribution table below. Calculate the mean salary.
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Table 3.1.56. Employee Salaries
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Salary Range(KES) Number of Employees
\(20,000 - 29,999\) \(3\)
\(30,000 - 39,999\) \(5\)
\(40,000 - 49,999\) \(7\)
\(50,000 - 59,999\) \(10\)
\(60,000 - 69,999\) \(9\)
\(70,000 - 79,999\) \(6\)
\(80,000 - 89,999\) \(5\)
\(90,000 - 99,999\) \(3\)
\(100,000 - 109,999\) \(2\)
Solution.
To calculate the mean, we first find the midpoint (\(x\)) of each class and multiply it by the frequency (\(f\)).
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Table 3.1.57. Mean Calculation Table
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Salary Range(KES) Midpoint (\(x\)) Frequency (\(f\)) \(fx\)
\(20,000 - 29,999\) \(25,000\) \(3\) \(75,000\)
\(30,000 - 39,999\) \(35,000\) \(5\) \(175,000\)
\(40,000 - 49,999\) \(45,000\) \(7\) \(315,000\)
\(50,000 - 59,999\) \(55,000\) \(10\) \(550,000\)
\(60,000 - 69,999\) \(65,000\) \(9\) \(585,000\)
\(70,000 - 79,999\) \(75,000\) \(6\) \(450,000\)
\(80,000 - 89,999\) \(85,000\) \(5\) \(425,000\)
\(90,000 - 99,999\) \(95,000\) \(3\) \(285,000\)
\(100,000 - 109,999\) \(105,000\) \(2\) \(210,000\)
Total \(\sum f = 50\) \(\sum fx = 3,070,000\)
\begin{align*} \overline{x} = \amp \frac{ \sum fx}{ \sum f} \\ = \amp \frac{ 3,070,000}{ 50}\\ = \amp 61,400 \end{align*}
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Therefore, the mean salary is \(61,400\) KES.
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Example 3.1.58.

The following data shows the number of minutes \(50\) students spent studying for a mathematics test.
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\(68, 34, 52, 41, 77, 29, 63, 55, 46, 72,\)
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\(38, 64, 51, 83, 47, 36, 59, 74, 62, 44,\)
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\(57, 69, 31, 48, 66, 53, 39, 71, 42, 60,\)
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\(75, 58, 37, 49, 65, 54, 32, 70, 45, 61,\)
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\(56, 43, 33, 78, 67, 35, 73, 40, 50, 76\)
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Find the mean time taken by the students (use a class width of \(10\))
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Solution.
Since the class width is \(10\text{,}\) the classes are
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Class Frequency (f) Midpoint (x) fx
\(20-29\) \(1\) \(24.5\) \(24.5\)
\(30-39\) \(9\) \(34.5\) \(310.5\)
\(40-49\) \(10\) \(44.5\) \(445\)
\(50-59\) \(10\) \(54.5\) \(545\)
\(60-69\) \(10\) \(64.5\) \(645\)
\(70-79\) \(9\) \(74.5\) \(670.5\)
\(80-89\) \(1\) \(84.5\) \(84.5\)
\(\sum \text{f} = 50\) \(\) \(\sum \text{fx} = 2725\)
Apply the mean formula
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\(\text {Mean} = \frac{\sum fx}{\sum f}\)
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Mean \(=\frac{2725}{50} = 54.5\; \text{minutes}\)
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Therefore every student take an average of \(54.5\) minutes to study for a mathematics test.
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Exercises Exercises

1.
A tea factory recorded the weight of tea leaves (in kg) delivered by \(40\) small-scale farmers in a single morning.
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Table 3.1.59. Tea Weight Deliveries
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Weight (kg) Frequency (Farmers)
\(10 - 19\) \(5\)
\(20 - 29\) \(8\)
\(30 - 39\) \(15\)
\(40 - 49\) \(10\)
\(50 - 59\) \(2\)
Calculate the mean weight of tea leaves delivered per farmer.
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Answer.
Midpoints (\(x\)): \(14.5, 24.5, 34.5, 44.5, 54.5\)
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\(\sum f = 40\)
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\(\sum fx = (14.5 \times 5) + (24.5 \times 8) + (34.5 \times 15) + (44.5 \times 10) + (54.5 \times 2) = 1330\)
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Mean \(= \frac{1330}{40} = 33.25\) kg.
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2.
A Sacco operating on the Nairobi-Thika route recorded the fares paid by passengers during peak hours on a rainy day.
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Table 3.1.60. Fare Distribution
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Fare (KES) No. of Passengers
\(50 - 99\) \(20\)
\(100 - 149\) \(45\)
\(150 - 199\) \(30\)
\(200 - 249\) \(5\)
Determine the mean fare paid by the passengers.
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Answer.
Midpoints (\(x\)): \(74.5, 124.5, 174.5, 224.5\)
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\(\sum f = 100\)
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\(\sum fx = (74.5 \times 20) + (124.5 \times 45) + (174.5 \times 30) + (224.5 \times 5) = 13,450\)
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Mean \(= \frac{13,450}{100} = 134.50\) KES.
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3.
The mathematics department at a High School in Nakuru analyzed the percentage scores of \(50\) students in a mock exam.
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Table 3.1.61. Maths Scores
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Marks (%) Frequency
\(0 - 20\) \(4\)
\(20 - 40\) \(12\)
\(40 - 60\) \(18\)
\(60 - 80\) \(10\)
\(80 - 100\) \(6\)
Calculate the mean score for the class.
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Answer.
Midpoints (\(x\)): \(10, 30, 50, 70, 90\)
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\(\sum f = 50\)
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\(\sum fx = (10 \times 4) + (30 \times 12) + (50 \times 18) + (70 \times 10) + (90 \times 6) = 2,540\)
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Mean \(= \frac{2,540}{50} = 50.8\%\text{.}\)
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4.
An M-Pesa agent in Kisumu recorded the value of transactions performed in the first hour of business.
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Table 3.1.62. Transaction Values
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Value (KES) No. of Transactions
\(0 - 1000\) \(12\)
\(1000 - 2000\) \(8\)
\(2000 - 3000\) \(6\)
\(3000 - 4000\) \(4\)
Estimate the mean transaction value.
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Answer.
Midpoints (\(x\)): \(500, 1500, 2500, 3500\)
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\(\sum f = 30\)
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\(\sum fx = (500 \times 12) + (1500 \times 8) + (2500 \times 6) + (3500 \times 4) = 47,000\)
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Mean \(= \frac{47,000}{30} \approx 1,566.67\) KES.
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Subsubsection 3.1.3.3 Median (Grouped)

When data is grouped, we cannot determine the exact middle value by simply counting. Instead, we use cumulative frequencies to locate the median class and then use interpolation to estimate the median value.
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Learner Experience 3.1.9.

Work in groups
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Using the pocket money data provided below:
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Table 3.1.64. Pocket Money Distribution
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Pocket Money (Ksh) \(100 - 199\) \(200 - 299\) \(300 - 399\) \(400 - 499\) \(500 - 599\)
Frequency \(8\) \(15\) \(22\) \(20\) \(10\)
  1. Construct a Cumulative Frequency column for this data.
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  2. Identify the class interval where the middle student lies.
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Key Takeaway 3.1.65.

Calculating 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*} \text{Median} = \amp L + \left(\frac{\frac{n}{2} - CF}{F}\right)\times C \end{align*}
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Where:
  • \(L\) = Lower boundary of the median class.
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  • \(n\) = Sum of all frequencies (\(\sum f\)).
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  • \(CF\) = Cumulative frequency before the median class.
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  • \(F\) = Frequency of the median class itself.
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  • \(C\) = Class width.
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Example 3.1.66.

The data below represents the times (in seconds) recorded in the heats of a \(100\) m race. The data has been grouped into intervals of width 0.5 seconds.
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Estimate the median time based on the frequency table below.
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Table 3.1.67. Race Times
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Class Interval Frequency
\(11.5 - 11.9\) \(2\)
\(12.0 - 12.4\) \(4\)
\(12.5 - 12.9\) \(5\)
\(13.0 - 13.4\) \(4\)
\(13.5 - 13.9\) \(5\)
\(14.0 - 14.4\) \(6\)
\(14.5 - 14.9\) \(5\)
\(15.0 - 15.4\) \(4\)
\(15.5 - 15.9\) \(1\)
Solution.
First, we add a Cumulative Frequency (CF) column to find the position of the median.
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Table 3.1.68. Cumulative Frequency Table
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Class Interval Frequency (\(f\)) Cumulative Frequency (\(CF\))
\(11.5 - 11.9\) \(2\) \(2\)
\(12.0 - 12.4\) \(4\) \(6\)
\(12.5 - 12.9\) \(5\) \(11\)
\(13.0 - 13.4\) \(4\) \(15\)
\(13.5 - 13.9\) \(5\) \(20\)
\(14.0 - 14.4\) \(6\) \(26\)
\(14.5 - 14.9\) \(5\) \(31\)
\(15.0 - 15.4\) \(4\) \(35\)
\(15.5 - 15.9\) \(1\) \(36\)
Step 1: Locate the Median Class Total frequency \(n = 36\text{.}\) The median position is \(\frac{36}{2} = 18\text{.}\) Looking at the CF column, the value 18 falls within the interval where the cumulative frequency rises from 15 to 20. Therefore, the median class is 13.5 - 13.9.
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Step 2: Apply the Formula
  • \(L = 13.5\) (Lower boundary)
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  • \(n = 36\text{,}\) so \(\frac{n}{2} = 18\)
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  • \(CF = 15\) (Value 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*} \text{Median} = \amp 13.5 + \left(\frac{18 - 15}{5}\right)\times 0.5\\ = \amp 13.5 + \left(\frac{3}{5}\times 0.5\right)\\ = \amp 13.5 + 0.3\\ = \amp 13.8 \end{align*}
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Thus, the median time is \(13.8\) seconds.
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Example 3.1.69.

The marks obtained by 40 students in a mathematics test are:
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\(42, 55, 47, 61, 50, 38, 44, 49, 53, 60,\)
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\(41, 48, 52, 46, 57, 51, 39, 62, 45, 58,\)
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\(43, 56, 54, 59, 63, 40, 37, 64, 36, 65,\)
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\(47, 49, 50, 52, 55, 51, 48, 53, 60, 62\)
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Using \(6\) class intervals, find the median mark.
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Solution.
  1. Determine the range
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    Maximum value \(= 65\)
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    Minimum value \(= 36\)
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    Range \(= 65 - 36 = 29\)
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  2. Decide class width
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    Number of classes \(= 6\)
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    Class width \(=\) Range \(\div\) Number of classes \(= 29 \div 6 \approx 4.83 \)
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    We can therefore use a class width of \(5\)
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  3. Complete the frequency table
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    Class Interval Frequency (f) Cumulative Frequency (CF)
    \(36-40\) \(5\) \(5\)
    \(14-45\) \(5\) \(10\)
    \(46-50\) \(9\) \(19\)
    \(51-55\) \(9\) \(28\)
    \(56-60\) \(6\) \(34\)
    \(61-65\) \(6\) \(40\)
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  4. Find the median Class
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    Number of observations, \(n = 40\)
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    Median class \(=\) the class containing \(\frac{n}{2} = 20^{th}\) observation.
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    From CF: \(19^{th}\) observation is in \(46-50\text{,}\) while \(28^{th}\) observation is in \(51-55\)
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    So the median class \(= 51-55\) since that is where the \(20^{th}\) observation fall.
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  5. Apply the median formula
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    \begin{align*} \text{Median} = \amp L + \left(\frac{\frac{n}{2} - CF}{F}\right)\times h \\ = \amp 50.5 + \left(\frac{\frac{40}{2} - 19}{9}\right) \times 5\\ = \amp 50.0 + \frac{5}{9}\\ \text{Median} = \amp 51.05555 \end{align*}
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    Median mark\(\approx 51.06\)
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Exercises Exercises

1.
A group of \(40\) athletes recorded their finishing times (in minutes) during a practice run at Karura Forest.
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Table 3.1.70. Finishing Times
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Time (min) Frequency
\(120 - 129\) \(4\)
\(130 - 139\) \(12\)
\(140 - 149\) \(15\)
\(150 - 159\) \(9\)
Calculate the median finishing time.
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Answer.
Total frequency (\(n\)) \(= 40\text{.}\) Median position \(= \frac{40}{2} = 20\text{.}\)
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CF values: \(4, 16, 31, 40\text{.}\)
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The \(20\)th value lies in the \(140 - 149\) class.
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\(L = 140\text{,}\) \(\frac{n}{2} = 20\text{,}\) \(CF = 16\text{,}\) \(F = 15\text{,}\) \(C = 10\text{.}\)
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Median \(= 140 + (\frac{20 - 16}{15}) \times 10 = 140 + 2.67 = 142.67\) minutes.
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2.
A county hospital in Machakos surveyed the waiting time (in minutes) for \(60\) patients at the outpatient department.
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Table 3.1.71. Patient Waiting Times
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Time (min) No. of Patients
\(0 - 19\) \(10\)
\(20 - 39\) \(25\)
\(40 - 59\) \(15\)
\(60 - 79\) \(10\)
Estimate the median waiting time for a patient.
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Answer.
Total frequency (\(n\)) \(= 60\text{.}\) Median position \(= 30\text{.}\)
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CF values: \(10, 35, 50, 60\text{.}\)
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The \(30\)th value lies in the \(20 - 39\) class.
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\(L = 20\text{,}\) \(CF = 10\text{,}\) \(F = 25\text{,}\) \(C = 20\text{.}\)
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Median \(= 20 + (\frac{30 - 10}{25}) \times 20 = 20 + 16 = 36\) minutes.
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3.
An agricultural officer in Uasin Gishu measured the height (in cm) of \(50\) maize stalks in a test farm.
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Table 3.1.72. Maize Stalk Heights
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Height (cm) Frequency
\(100 - 119\) \(8\)
\(120 - 139\) \(18\)
\(140 - 159\) \(14\)
\(160 - 179\) \(10\)
Determine the median height of the maize stalks.
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Answer.
Total frequency (\(n\)) \(= 50\text{.}\) Median position \(= 25\text{.}\)
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CF values: \(8, 26, 40, 50\text{.}\)
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The \(25\)th value lies in the \(120 - 139\) class.
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\(L = 120\text{,}\) \(CF = 8\text{,}\) \(F = 18\text{,}\) \(C = 20\text{.}\)
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Median \(= 120 + (\frac{25 - 8}{18}) \times 20 = 120 + 18.89 = 138.89\) cm.
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4.
A survey was conducted on the monthly rent (in thousands of KES) for one-bedroom houses in a Nairobi estate.
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Table 3.1.73. Monthly Rent
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Rent (Ksh ’000) No. of Houses
\(5 - 9\) \(20\)
\(10 - 14\) \(40\)
\(15 - 19\) \(30\)
\(20 - 24\) \(10\)
Calculate the median rent.
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Answer.
Total frequency (\(n\)) \(= 100\text{.}\) Median position \(= 50\text{.}\)
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CF values: \(20, 60, 90, 100\text{.}\)
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The \(50\)th value lies in the \(10 - 14\) class.
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\(L = 10\text{,}\) \(CF = 20\text{,}\) \(F = 40\text{,}\) \(C = 5\text{.}\)
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Median \(= 10 + (\frac{50 - 20}{40}) \times 5 = 10 + 3.75 = 13.75\) (Ksh \(13,750\)).
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Subsubsection 3.1.3.4 Mode (Grouped)

In grouped data, we cannot easily identify the specific number that appears most often. Instead, we identify the group that contains the most data points.
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Learner Experience 3.1.10.

Work in groups
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Review the pocket money distribution table below:
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Table 3.1.75. Pocket Money Distribution
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Pocket Money (Ksh) \(100 - 199\) \(200 - 299\) \(300 - 399\) \(400 - 499\) \(500 - 599\)
Frequency \(8\) \(15\) \(22\) \(20\) \(10\)
  1. Which group has the highest number of students?
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  2. What is the modal class of pocket money given to students?
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Key Takeaway 3.1.76.

Modal Class
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For grouped data, the Modal Class is simply the class interval with the highest frequency. It represents the range where the majority or the highest density of the data points lie.
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Example 3.1.77.

A company records the monthly salaries (in KES) of employees. Identify the modal class from the table below.
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Table 3.1.78. Employee Salaries
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Salary Range(KES) Number of Employees
\(20,000 - 29,999\) \(3\)
\(30,000 - 39,999\) \(5\)
\(40,000 - 49,999\) \(7\)
\(50,000 - 59,999\) \(10\)
\(60,000 - 69,999\) \(9\)
\(70,000 - 79,999\) \(6\)
\(80,000 - 89,999\) \(5\)
Solution.
To find the modal class, we look for the highest frequency in the "Number of Employees" column.
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The highest frequency is \(10\text{.}\) This corresponds to the salary range \(50,000 - 59,999\text{.}\)
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Therefore, the Modal Class is \(50,000 - 59,999\) KES.
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Example 3.1.79.

The scores of 50 students in a statistics test are summarized below:
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Class Interval Frequency (f)
\(0-9\) \(2\)
\(10-19\) \(5\)
\(20-29\) \(8\)
\(30-39\) \(12\)
\(40-49\) \(11\)
\(50-59\) \(7\)
\(60-69\) \(5\)
Find the modal score.
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Solution.
  1. Identify the modal class
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    The modal class is the class with the highest frequency. From the table, the highest frequency is \(12\text{,}\) which corresponds to \(30-39\text{.}\)So, modal class is \(30-39\text{.}\)
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  2. Apply the mode formula for grouped data
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    \begin{align*} Mode = \amp L + \left( \frac {f_1 - f_0}{2 \times f_1 - f_0 - f_2} \right) \times h \\ = \amp 29.5 + \frac{12 - 8}{2 \times 12 - 8 - 11} \times 10\\ = \amp 29.5 + \frac{4}{5} \times 10\\ = \amp 37.8 \end{align*}
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    Modal score \(= 37.5\) units
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Checkpoint 3.1.80.

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Checkpoint 3.1.81.

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Exercises Exercises

1.
A Bata shop in Mombasa recorded the sizes of shoes sold during a "Back to School" sale.
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Table 3.1.82. Shoe Sizes Sold
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Shoe Size Pairs Sold
\(2 - 4\) \(15\)
\(5 - 7\) \(45\)
\(8 - 10\) \(30\)
\(11 - 13\) \(10\)
Identify the modal class of the shoe sizes.
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Answer.
The modal class is \(5 - 7\text{.}\)
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This is because it has the highest frequency of \(45\) pairs sold.
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2.
The meteorological department recorded the daily rainfall (in mm) in Nyeri for the month of April.
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Table 3.1.83. Daily Rainfall
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Rainfall (mm) No. of Days
\(0 - 4\) \(10\)
\(5 - 9\) \(4\)
\(10 - 14\) \(12\)
\(15 - 19\) \(3\)
\(20 - 24\) \(1\)
State the modal class for the rainfall.
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Answer.
The modal class is \(10 - 14\) mm.
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Although \(0 - 4\) is high, \(10 - 14\) has the highest frequency of \(12\) days.
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3.
A speed camera on the Thika Superhighway recorded the speeds (in km/h) of \(100\) vehicles passing a specific point.
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Table 3.1.84. Vehicle Speeds
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Speed (km/h) No. of Vehicles
\(60 - 79\) \(15\)
\(80 - 99\) \(55\)
\(100 - 119\) \(25\)
\(120 - 139\) \(5\)
Find the modal class of the vehicle speeds.
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Answer.
The modal class is \(80 - 99\) km/h.
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This interval contains the highest number of vehicles (\(55\)).
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4.
A survey among teenagers in a Nairobi estate showed their weekly mobile data usage (in MBs).
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Table 3.1.85. Data Usage
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Data (MB) No. of Teenagers
\(0 - 499\) \(8\)
\(500 - 999\) \(22\)
\(1000 - 1499\) \(14\)
\(1500 - 1999\) \(6\)
What is the modal class for the data usage?
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Answer.
The modal class is \(500 - 999\) MB.
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This group has the highest frequency of \(22\) teenagers.
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