Learning Resources

Data Interpretation (Grouped Data)

Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.

Grade 09 Pathway: N/A

First Principles

Objective

Learn how to organise large sets of continuous or discrete numerical data into class intervals and calculate the estimated mean using class midpoints and frequency tables.

Analogy: Sorting Maize Bags at a Milling Depot

Imagine a cooperative in Eldoret receiving hundreds of bags of maize. Measuring and listing every single bag's exact mass (e.g., \(48.2\text{ kg}\), \(51.7\text{ kg}\), \(49.1\text{ kg}\)) one by one would be overwhelming and unhelpful. Instead, the warehouse clerk sorts them into standard weight categories or bins: 40–49 kg, 50–59 kg, and 60–69 kg. Each bin is a class interval, and the count of bags in each bin is its frequency.

(a) Class Intervals and Midpoints

When data is grouped, we lose the exact individual values. To represent all values in an interval, we calculate its centre — the class midpoint (or class mark) \(x_i\):

\[x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}\]

We make the reasonable assumption that the data values inside an interval are evenly distributed around its midpoint.

(b) The Grouped Mean Formula

To find the estimated mean \(\bar{x}\), we multiply each class midpoint \(x_i\) by its frequency \(f_i\) to find the subtotal for that group, sum these totals, and divide by the total number of items \(\sum f_i\):

\[\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\]

Interactive Grouped Mean Explorer

Adjust the frequency of each height band (in cm) for a class of Kenyan students and observe how the estimated mean updates in real time.

130 – 140 cm (Midpoint = 135 cm): 4 learners

140 – 150 cm (Midpoint = 145 cm): 10 learners

150 – 160 cm (Midpoint = 155 cm): 6 learners

Total Learners (\(\sum f_i\)): 20

Sum of Products (\(\sum f_i x_i\)): 2920

Estimated Mean Height (\(\bar{x}\)): 146.00 cm

Key Formulas

\[\text{Class Midpoint } (x_i) = \frac{\text{Lower Class Limit} + \text{Upper Class Limit}}{2}\] — The representative central value for every data entry falling inside a given class interval.

\[\text{Class Width } (w) = \text{Upper Class Limit} - \text{Lower Class Limit}\] — The span or size of an interval for continuous contiguous boundaries.

\[\text{Estimated Mean } (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} = \frac{\sum (f \cdot x)}{N}\] — Multiplies each group's midpoint by its frequency, sums the products, and divides by the total sample size \(N\).

\[F_i = \sum_{k=1}^{i} f_k\] — Cumulative Frequency: The running total of frequencies from the lowest interval up to the current interval.

Worked Examples

Example 1 (Easy): Basic Grouped Mean

A tea collector in Kericho records the masses of plucked tea baskets in 3 intervals: 0–10 kg (frequency 2), 10–20 kg (frequency 5), and 20–30 kg (frequency 3). Estimate the mean mass per basket.

Find midpoints \(x_i\): \[x_1 = \frac{0+10}{2} = 5,\quad x_2 = \frac{10+20}{2} = 15,\quad x_3 = \frac{20+30}{2} = 25\]
Compute \(f_i x_i\): \[\sum f_i x_i = (2 \times 5) + (5 \times 15) + (3 \times 25) = 10 + 75 + 75 = 160\]
Find total frequency: \[\sum f_i = 2 + 5 + 3 = 10\]
Calculate estimated mean: \[\bar{x} = \frac{160}{10} = 16\text{ kg}\]

Example 2 (Medium): Multi-Class Frequency Distribution

A biology class in Nakuru recorded the lengths of 20 leaves into intervals: 0–10 mm (freq 4), 10–20 mm (freq 8), 20–30 mm (freq 6), and 30–40 mm (freq 2). Estimate the mean leaf length.

Midpoints: \(5, 15, 25, 35\).
Products: \[\sum f_i x_i = 4(5) + 8(15) + 6(25) + 2(35) = 20 + 120 + 150 + 70 = 360\]
Total frequency: \(4 + 8 + 6 + 2 = 20\).
Estimated Mean: \[\bar{x} = \frac{360}{20} = 18\text{ mm}\]

Example 3 (Hard): Finding a Missing Frequency

The daily transport expense of employees at a tech firm in Nairobi is summarised below. If the estimated mean expense is KES 275, find the missing frequency \(k\).

Expense (KES)	Midpoint (\(x\))	Frequency (\(f\))	\(f \times x\)
100–200	150	5	750
200–300	250	8	2000
300–400	350	\(k\)	\(350k\)
400–500	450	3	1350

Set up sums in terms of \(k\): \[\sum f = 5 + 8 + k + 3 = 16 + k\] \[\sum fx = 750 + 2000 + 350k + 1350 = 4100 + 350k\]
Equate to known mean \(\bar{x} = 275\): \[\frac{4100 + 350k}{16 + k} = 275\]
Cross-multiply and solve for \(k\): \[4100 + 350k = 275(16 + k)\] \[4100 + 350k = 4400 + 275k\] \[350k - 275k = 4400 - 4100\] \[75k = 300 \implies k = 4\]

Common Mistakes

Mistake Using the lower or upper boundary of an interval directly instead of its midpoint when computing \(\sum f_i x_i\).

Why it feels right The boundary numbers (like 20 or 30 in 20–30) are printed right in front of you, making them easy to pick.

Correction Always compute the midpoint \(x_i = \frac{L + U}{2}\). Using a boundary skews the estimated mean artificially lower or higher.

Mistake Simply averaging the midpoints alone: \(\bar{x} = \frac{\sum x_i}{\text{number of classes}}\).

Why it feels right It looks like the standard arithmetic average of the column.

Correction Class intervals have different numbers of items! You must weight each midpoint by its corresponding frequency \(f_i\) before dividing by \(\sum f_i\).

Mistake Thinking the grouped mean is the exact true mean of the raw data.

Why it feels right The math formula gives a precise number with decimals.

Correction Grouped mean is strictly an estimate because the exact distribution of values inside each bin is simplified to the single midpoint.

Real World

Agriculture & Crop Yields: In Kenya's Rift Valley, grain marketing boards group harvest bag weights into bins (e.g., 40–50 kg, 50–60 kg) to estimate total tonnage and calculate fair payout averages for large farming cooperatives.

Retail and Matatu Transit: Matatu SACCOs in Nairobi group daily fare collections into intervals of KES 500 to analyse revenue trends across peak and off-peak hours without logging every individual passenger coin.

National Examination Analysis: The Kenya National Examinations Council (KNEC) groups raw test scores across thousands of schools into grade brackets (e.g., 70–79, 80–89) to evaluate overall national student performance curves.

Practice

What is the class midpoint of the interval 20 – 30? (Type only the number, e.g., 25)

Review the concepts above.

A tea cooperative in Kericho records the masses of bags. The class interval 40 – 50 kg has a frequency of 6. What is the value of \(f \times x\) (frequency multiplied by class midpoint) for this interval? (Type only the number, e.g., 270)

Review the concepts above.

A group of 20 learners at a school in Machakos have their heights recorded in a grouped frequency table: • 130 – 140 cm: frequency 4 • 140 – 150 cm: frequency 10 • 150 – 160 cm: frequency 6 Calculate the estimated mean height in cm. (Type only the number, e.g., 146)

Review the concepts above.

A fish vendor in Kisumu records the weights of 10 tilapia fish: • 0 – 200 g: frequency 3 • 200 – 400 g: frequency 5 • 400 – 600 g: frequency 2 What is the estimated mean mass of the fish in grams? (Type only the number, e.g., 280)

Review the concepts above.

A cooperative in Nyeri collected 40 bags of coffee. The table below shows the distribution of weights: • 10 – 30 kg: frequency 6 • 30 – 50 kg: frequency 14 • 50 – 70 kg: frequency 12 • 70 – 90 kg: frequency 8 Calculate the estimated mean weight of a bag of coffee in kg. (Type only the number, e.g., 51)

Review the concepts above.

The daily transport expenses (in KES) of staff at a Nairobi company are grouped as follows: • 100 – 200 KES: frequency 5 • 200 – 300 KES: frequency 8 • 300 – 400 KES: frequency k • 400 – 500 KES: frequency 3 If the estimated mean expense is KES 275, find the value of the missing frequency k. (Type only the number, e.g., 4)

Review the concepts above.