Lesson Notes By Weeks and Term v3 - Junior Secondary 1
Data presentation
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Subject: General Mathematics
Class: Junior Secondary 1
Term: 2nd Term
Week: 4
Theme: Everyday Statistics
This page supports the lesson note with a companion video and a short classroom-ready summary.
For class groups and homework, share this lesson page so learners also get the summary, objectives, and full lesson context.
Students should be able to determine the median of a given set of data
This section provides a detailed explanation of the median, including definitions and step-by-step procedures for calculation. 2.1 Definition of Median The median is the middle value in a set of data that has been arranged in order of magnitude (either ascending from smallest to largest or descending from largest to smallest). It divides the data into two equal halves, meaning 50% of the data points are below the median and 50% are above it. 2.2 Steps to Determine the Median The process of finding the median depends on whether the number of data points (n) in the set is odd or even.
General Steps: Arrange the Data: The first and most critical step is to arrange all the data values in either ascending order (from smallest to largest) or descending order (from largest to smallest).
Count the Data Points (n): Determine the total number of values in the dataset.
Case 1: When the number of data points (n) is ODD. If the total number of data values is odd, there will be exactly one middle value.
Identify the Middle Position: The median will be the value at the $\frac{(n+1)}{2}$-th position.
Select the Value: The value at this identified position in the ordered list is the median. Worked Example 1 (Odd Number of Data Points): A teacher recorded the ages (in years) of 7 students in her JSS1 class: 11, 13, 10, 12, 11, 14,
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0. Determine the median age.
Solution: Arrange the data in ascending order: 10, 10, 11, 11, 12, 13, 14 Count the number of data points (n): n = 7 (which is an odd number)
Identify the middle position: Position = $\frac{(n+1)}{2} = \frac{(7+1)}{2} = \frac{8}{2} = 4$-th position.
Select the value at the 4th position: 10, 10, 11, 11, 12, 13, 14 The value at the 4th position is
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1. Therefore, the median age is 11 years.
Case 2: When the number of data points (n) is EVEN. If the total number of data values is even, there will be two middle values.
Identify the Two Middle Positions: The two middle values will be at the $\frac{n}{2}$-th position and the $(\frac{n}{2} + 1)$-th position.
Calculate the Average: The median is the average (mean) of these two middle values. Sum the two middle values and divide by
2. Worked Example 2 (Even Number of Data Points): A shopkeeper recorded the number of bags of rice sold each day for 6 days: 8, 12, 7, 15, 10,
9. Determine the median number of bags of rice sold.
Solution: Arrange the data in ascending order: 7, 8, 9, 10, 12, 15 Count the number of data points (n): n = 6 (which is an even number)
Identify the two middle positions: First middle position = $\frac{n}{2} = \frac{6}{2} = 3$-rd position. Second middle position = $(\frac{n}{2} + 1) = (3 + 1) = 4$-th position. Select the values at the 3rd and 4th positions: 7, 8, 9, 10, 12, 15 The values are 9 and
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0. Calculate the average of these two middle values: Median = $\frac{(9 + 10)}{2} = \frac{19}{2} = 9.5$ Therefore, the median number of bags of rice sold is 9.
5. Important
Note: Always ensure the data is ordered before attempting to find the median. Failure to order the data will lead to an incorrect median. This section outlines the activities for both the teacher and the students during the lesson.
Teacher Activities: Introduction (5 minutes): Begins by briefly revising the concept of data and its collection.
Introduces the lesson topic: "Median" as a measure of central tendency. States the learning objectives clearly for the students. Engages students by asking what "middle" means in everyday contexts (e.g., middle of a road, middle seat). Explanation and Demonstration (15 minutes): Clearly defines median and explains its significance (e.g., less affected by extremes). Demonstrates the step-by-step process of finding the median using a simple, small dataset with an odd number of values (e.g., student heights, scores). The teacher should use a flip chart or chalkboard, showing each step: Writing down raw data. Arranging data in ascending order. Counting data points (n). Identifying the middle position using the $\frac{(n+1)}{2}$ formula. Pointing to the median value. Repeats the demonstration with a dataset containing an even number of values. Again, meticulously showing: Writing down raw data. Arranging data in ascending order. Counting data points (n). Identifying the two middle positions using $\frac{n}{2}$ and $(\frac{n}{2} + 1)$. Calculating the average of the two middle values.
Guided Practice Facilitation (10 minutes): Provides a new set of data (one with odd 'n', one with even 'n') for students to work on in small groups or pairs. Circulates around the classroom, monitoring students' progress, offering guidance, and correcting misconceptions. Encourages students to explain their steps to each other.
Class Discussion and Feedback (5 minutes): Calls on different groups to present their solutions. Clarifies any remaining doubts and reinforces correct understanding of the steps involved in finding the median.
Assignment and Conclusion (5 minutes): Assigns independent practice questions for homework or further classwork. Summarizes the key learning points of the lesson.
Student Activities: Active Listening and Participation: Listens attentively to the teacher's explanations and demonstrations. Asks clarifying questions when concepts are unclear. Responds to the teacher's introductory questions.
Note-taking: Copies definitions, steps, and worked examples into their notebooks.
Group/Pair Work: Works collaboratively in small groups or pairs on guided practice questions. Discusses strategies for arranging data and identifying middle values. Calculates the median for given datasets.
Presentation: A representative from each group presents their solution and explains their working to the class.
Independent Practice: Attempts the independent practice questions provided by the teacher. The following questions are designed for guided practice, where students can work on them with teacher supervision and immediate feedback.
Question 1: The scores of 5 students in a Mathematics quiz are: 8, 12, 6, 10,
9. Find the median score.
Solution: Arrange the data in ascending order: 6, 8, 9, 10, 12 Count the number of data points (n): n = 5 (odd number)
Identify the middle position: Position = $\frac{(5+1)}{2} = \frac{6}{2} = 3$-rd position.
Select the value at the 3rd position: 6, 8, 9, 10, 12 The value is
9. Answer: The median score is
9. Commentary: This demonstrates finding the median for an odd number of data points, reinforcing the ordering and direct selection of the middle value.
Question 2: A farmer recorded the number of bags of maize harvested from 8 different plots: 25, 30, 22, 28, 35, 20, 26,
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2. Determine the median number of bags harvested.
Solution: Arrange the data in ascending order: 20, 22, 25, 26, 28, 30, 32, 35 Count the number of data points (n): n = 8 (even number)
Identify the two middle positions: First middle position = $\frac{8}{2} = 4$-th position. Second middle position = $(\frac{8}{2} + 1) = (4 + 1) = 5$-th position. Select the values at the 4th and 5th positions: 20, 22, 25, 26, 28, 30, 32, 35 The values are 26 and
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8. Calculate the average of these two middle values: Median = $\frac{(26 + 28)}{2} = \frac{54}{2} = 27$ Answer: The median number of bags harvested is
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7. Commentary: This example addresses the calculation for an even number of data points, requiring students to find the average of the two central values.
Question 3: The amount of money (in Naira) spent by 7 friends on lunch are: N500, N450, N600, N550, N400, N500, N
7
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0. Find the median amount spent.
Solution: Arrange the data in ascending order: N400, N450, N500, N500, N550, N600, N700 Count the number of data points (n): n = 7 (odd number)
Identify the middle position: Position = $\frac{(7+1)}{2} = \frac{8}{2} = 4$-th position.
Select the value at the 4th position: N400, N450, N500, N500, N550, N600, N700 The value is N
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0. Answer: The median amount spent is N
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0. Commentary: This reinforces the concept with repeated values, showing that the process remains the same – ordering is key.
Differentiation Strategies: Group Work: Form mixed-ability groups where stronger students can support weaker ones during guided practice.
Visual Aids: Use number lines or physical objects (e.g., counters, small stones) to visually represent data ordering for kinesthetic learners.
Remediation (for struggling learners): Revisiting Ordering Numbers: Provide additional exercises focused solely on arranging sets of numbers (both small and large) in ascending and descending order. Use smaller datasets (3-5 numbers) to build confidence. Simplified
Examples: Work through simpler examples step-by-step, focusing on very small datasets (e.g., 3 numbers, then 4 numbers) before progressing to more complex ones.
One-on-One Support: Dedicate specific time during group work or after class to provide individual attention and explanation.
Colour-Coding: Encourage students to colour-code the middle value(s) in their ordered lists to make identification clearer.
Peer Tutoring: Pair a struggling learner with a more proficient peer to review concepts and practice problems.
Extension (for high-achieving learners): Comparison of Measures: Challenge students to calculate the mean, mode (if applicable), and median for the same dataset and discuss which measure best represents the data and why (e.g., when data has outliers, median is often preferred).
Larger Datasets: Introduce slightly larger datasets (e.g., 15-20 numbers) or even simple frequency tables (e.g., scores and their frequencies) and ask them to determine the median.
Caution: Keep frequency tables very basic, as formal calculation of median from grouped frequency tables is beyond JSS
1. Real-world Data Collection: Task students to collect a small dataset from their environment (e.g., number of windows in houses on their street, prices of a specific item in different shops) and then calculate and interpret the median.
Justification: Ask them to provide situations where the median would be a more appropriate measure of central tendency than the mean, and vice versa, providing justifications for their choices.
Market Price Stability Analysis: Application: A local government official wants to understand the typical price of a basket of local rice in different markets within their jurisdiction to ensure fair pricing.
They collect prices from 9 markets: NGN 15,000, NGN 14,500, NGN 16,000, NGN 15,500, NGN 14,800, NGN 25,000 (an outlier due to scarcity in one remote market), NGN 15,200, NGN 14,700, NGN 15,
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0. Relevance: Calculating the median price (NGN 15,200 after ordering) would give a better representation of the 'typical' price than the mean, which would be skewed by the NGN 25,000 outlier. This helps in policy-making regarding market regulation or consumer protection in Nigeria.
Community Health Data: Application: Health workers conducting a survey in a rural Nigerian community record the number of children per household for 10 households: 4, 2, 5, 3, 7, 1, 6, 4, 3,
5. They need to find the median number of children to understand the typical family size for planning vaccination campaigns or food distribution.
Relevance: The median (3.5 children) provides insights into family structures, helping health and social services plan resource allocation more effectively, e.g., for child welfare programs or reproductive health education.
Educational Performance Trends: Application: A school principal wants to assess the central performance of students in a JSS1 Mathematics external examination. The scores (out of 100) for 12 students are: 55, 60, 78, 45, 62, 70, 50, 85, 65, 72, 58,
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8. Relevance: Finding the median score (63.5) gives a clear indication of the typical student performance, which can be compared across different years or schools. This information is valuable for curriculum review, teacher training, and identifying areas for academic support. Unlike the mean, it won't be overly influenced by a few exceptionally high or low scores.