What is 5.2b Outliers and Medians in NCERT Class 7 Mathematics Chapter 5?

5.2b Outliers and Medians is the topic of Part 2 of Chapter 5 (Ch 5 — Data Handling) in NCERT Class 7 Mathematics. This lesson builds intuition through examples, visuals, and graded practice.

TOPIC 14 OF 23

5.2b Outliers and Medians

Q: What is Part 2 u2014 Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool in NCERT Class 7 Mathematics?

Part 2 u2014 Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool is a key concept covered in NCERT Class 7 Mathematics, Chapter 5: Chapter 5. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

Q: How do I solve problems on Part 2 u2014 Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool step by step?

To solve problems on Part 2 u2014 Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 7 exercises gradually increase in difficulty u2014 start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

Q: Is Part 2 u2014 Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool important for the Class 7 board exam?

Part 2 u2014 Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool is part of the NCERT Class 7 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

Q: What mistakes should students avoid in Part 2 u2014 Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool?

Common mistakes in Part 2 u2014 Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Q: Where can I find more NCERT practice questions on Part 2 u2014 Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool?

End-of-chapter NCERT exercises for Part 2 u2014 Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 5, and solve at least one previous-year board paper to consolidate your understanding.

🎓 Class 7 Mathematics CBSE Theory Ch 5 — Data Handling ⏱ ~21 min

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5.2b Outliers and Medians

Does the average always give a reasonable summary of the values in a collection? If not, what is an alternative? Let us find out.

Height of a Family

The heights of the family members of Yaangba and Poovizhi are as follows:

Yaangba's family: 169 cm, 173 cm, 155 cm, 165 cm, 160 cm, 164 cm.
Poovizhi's family: 170 cm, 173 cm, 165 cm, 178 cm, 118 cm.

🔵 Find the average height of each family. Can we say that Yaangba's family is taller than Poovizhi's family?

Yaangba's mean = \((169+173+155+165+160+164) \div 6 = 986 \div 6 \approx 164.3\) cm.
Poovizhi's mean = \((170+173+165+178+118) \div 5 = 804 \div 5 = 160.8\) cm.

The average height of Poovizhi's family (160.8 cm) is less than that of Yaangba's family (164.3 cm). Although most members in Poovizhi's family are taller, their family's average height is less because one child is much younger and not as tall as the rest of the family. The average height, 160.8 cm, is less than the heights of 4 out of 5 members. Here, the average doesn't seem to represent the data very well.

Dot plot of family heights — Poovizhi's family has one outlier pulling the mean down.

The Median

Can you think of any other number that can represent the data better? One way is to sort the data and pick the number in the middle. This number is called the Median^?.

To find the median of Poovizhi's family, we first sort the heights: 118, 165, 170, 173, 178. The middle number is 170. Therefore, the median height is 170 cm.

Let us find the median height of Yaangba's family. Sorting the heights, we get: 155, 160, 164, 165, 169, 173. Since the number of values is even, we have two middle numbers — 164 and 165. We take the average of these two middle values to find the median. So, the median is \((164 + 165) \div 2 = 164.5\) cm.

Definition — Median

The median is the middle value of a sorted dataset.

If odd number of values: the middle one.
If even number of values: the average of the two middle ones.

🔵 In this case, does the median represent the heights of the families better than the average?
Yes, for Poovizhi's family — the median (170 cm) is close to 4 of the 5 members' heights, while the mean (160.8 cm) is pulled down by the young child. For Yaangba's family, mean and median are almost the same.

Outliers

In Poovizhi's family, the height of the youngest child is different from the heights of the rest of the family. We call such a value an outlier^?. Outliers are values which significantly deviate from the rest of the values in the data. Notice how the mean and the median are close to each other in Yaangba's data, in the absence of any outlier. In Poovizhi's data, because of the outlier, the mean is much lower than the median.

Find the median in the absence of the outlier

Remove 118 from Poovizhi's data. The sorted data becomes: 165, 170, 173, 178. Two middle values are 170 and 173. Median = \((170+173) \div 2 = 171.5\) cm. Compare: with outlier median = 170, without = 171.5 — very close! The median is robust to outliers.

Are You a Bookworm?

After the summer vacation, a class teacher asked his class how many short stories they had read. Each student answered the number of stories read on a piece of paper, as shown below. Find the mean and median of this data. Before calculating them, can you guess whether the mean will be less than or greater than the median?

6 8 5 15 30 2 7 12 40 10 5 1

Sorted: 1, 2, 5, 5, 6, 7, 8, 10, 12, 15, 30, 40. There are 12 values. Median = (7+8) ÷ 2 = 7.5. Mean = (1+2+5+5+6+7+8+10+12+15+30+40) ÷ 12 = 141 ÷ 12 ≈ 11.75.

Dot plot: outliers at 30 and 40 pull the mean above the median.

The outliers (30 and 40) are much larger than most other values; they pull the mean higher than the median. The mean of 11.75 is larger than half of the class members who have read 6 or more stories.

🔵 Which of the values would you consider an outlier? 30 and 40 — they are much larger than the rest (most values lie between 1 and 15).

🔵 Find the mean and median in the absence of the outliers. Without 30 and 40: sum = 141−70 = 71, values = 10, mean = 7.1. Median of {1,2,5,5,6,7,8,10,12,15} = (6+7)/2 = 6.5. Much closer to each other.

Are We on the Same Page?

Do you read newspapers? Have you noticed how many pages a newspaper has on different days of the week — is it the same or different? The list below shows the number of pages for a particular newspaper from Monday to Sunday: 16, 20, 22, 26, 16, 10. Mark the data, the mean, and the median on the dot plot below.

Sum = 16+20+22+26+16+10 = 110. Number of values = 6. Mean = 110 ÷ 6 ≈ 18.33. Sorted: 10, 16, 16, 20, 22, 26. Median = (16+20)/2 = 18.

Three Scenarios for Mean and Median

In the three examples we have considered — the heights, short-stories, and newspaper pages — we see the variability in data when:

(a) the mean and median are close to each other
(b) the mean and median are comparatively far apart, with mean < median
(c) the mean and median are comparatively far apart, with mean > median

Key Idea

When the data is more balanced or uniformly spread out about the mean, the median appears close to each other. When the outlier is on the lower end, the mean appears to shift in that direction, i.e., mean < median. When the outlier is on the higher end, the mean appears to shift in that direction, i.e., mean > median.

Of Ends and the Essence

As we have just seen, the mean and the median can give different perspectives on the data. As part of analysing data, it can also be valuable to look at the variability in the given data, i.e., its extremes (minimum and maximum values).

How Tall is Your Class?

Suppose you are asked the question, "How tall is your class? What would you say?" The table below shows the heights of students in a Grade 5 class in centimetres:

Boys	147, 135, 130, 154, 128, 135, 134, 158, 155, 146, 146, 143, 140, 141, 144, 145, 150
Girls	143, 136, 150, 144, 154, 140, 145, 148, 156, 150, 150

We can visualise the data using a dot plot, identify the ends and patterns, and look at the variability. We can also find the measures of central tendency. The dot plot for the whole class, followed by the dot plots for boys and girls, respectively, are shown. The mean and the median are also shown for each collection.

Whole class: Mean = 144.4, Median = 145
Boys: Mean = 142.94, Median = 143
Girls: Mean = 146.9, Median = 148

Some observations:

The boys' heights are more spread out and are between 128 and 158. The girls' heights lie between 136 and 156. Both the tallest and shortest in the class are boys.
Yet, the boys' average height is less than the whole class average, and also less than the girls' average height. We can say girls are taller than boys in this class. Of course, this doesn't mean every girl is taller than every boy!
For boys' heights, mean < median (142.94 < 144) indicating a small influence of values on the lower side. For girls' heights too, mean < median (146.9 < 148) indicating a small influence of values on the lower side.

🔵 How many students are taller than the class' average height? By counting students whose height is above 144.4 cm — answer depends on the dataset; typically more than half.

🔵 How many boys are taller than the class' average height? Boys whose height exceeds 144.4 cm.

How Long is a Minute?

Two groups of children were asked to estimate the length of 1 minute. They kept their eyes closed and then opened them when they think 1 minute has passed. Of course, they are not supposed to count while their eyes are closed. The dot plots below show after how many seconds the children opened their eyes.

Group A: Mean = 58.23, Median = 48
Group B: Mean = 59.28, Median = 58.5

🔵 Discuss how well the groups fared at this activity. Group B's median (58.5) is very close to the true 60 seconds; Group A's median (48) is lower — Group B was more accurate. Means are similar, but Group A has more spread.

Zero Median Runs Scored!

In a cricket match, can a team's median runs scored by a player be 0 but the team's total score be 407/10? Let's think.

If at least 6 players (out of 11) scored 0 runs, then the sorted runs would have 0 at the middle position — giving a median of 0. Meanwhile, a few players could still combine to reach 407 runs. For example, scores like {0, 0, 0, 0, 0, 0, 50, 80, 87, 90, 100, + extras (11)} sum to 407.

Accounting for the extras, the average runs scored by a player in this innings is \((407 − 11) \div 11 \approx 36\). So mean ≈ 36 while median = 0 — a huge gap, driven by many players scoring 0.

Zero Vs. No value

Suppose a player scored 57, 13, 0, 84, —, 51, 27 in a series. The player played Match 3 and scored 0 runs whereas the player did not play Match 5. So, we consider the total number of matches to be 6, not 7. We calculate their average runs scored per match as \((57 + 13 + 0 + 84 + 51 + 27) \div 6 = 38.67\).

Sita has a mango tree in her backyard. The number of mangoes the tree gave every month over the last year — from January to December — is 0, 8, 0, 24, 41, 16, 5, 0, 0, 0, 0, respectively. If we want to find the mean or median number of mangoes per month, it would be appropriate to consider only the (summer) months when mangoes are expected to grow.

A Mean Foot

In the early 1500s in Europe, the basic unit of land measurement was the rod, defined as 16 feet long. At that time, a foot length of 8 human feet — but foot sizes vary, so whose foot could they measure? A 16th century mathematician, Jacob Köbel, suggested the 16-foot rod. After the rod was determined, it was split into 16 equal sections, each representing the measure of a single foot. In essence, this was the arithmetic mean of the 16 individual feet, even though the term 'mean' was not mentioned anywhere.

Figure it Out

Q1. Find the median of onion prices in Yahapur and Wahapur.

Yahapur sorted: 24, 25, 26, 28, 28, 30, 35, 39, 40, 43, 49, 56. Median = (30+35)/2 = 32.5 ₹/kg.
Wahapur sorted: 17, 19, 23, 30, 30, 35, 39, 42, 42, 52, 53, 60. Median = (35+39)/2 = 37 ₹/kg.

Q2. Sanskruti asked her class how many domestic animals and pets each had at home. Some of the students were absent. The data values were: 0, 1, 0, 4, 8, 0, 2, 1, 1, 5, 3, 4, 0, ..., 10, 25, 2, ..., 2, 4. Find the mean and median. How would you describe this data?

The value 25 appears to be an outlier (perhaps a student from a farming family). With the outlier, the mean gets pulled higher; the median is more representative. Compute with actual dataset: arrange values, pick middle. For class-reported data, median (often 2-3 pets) describes the typical family better than the mean.

Q3. Rinzu takes care of a date-palm tree farm in Bhabra. The heights of the trees (in feet) are: 45, 43, 52, 61, 43, 46, 55, 46, 65, 66, 33, 44, 58, 60, 54, 60, 52, 37, 61, 62. Find the mean and median. How would you describe the heights of these palm trees? Can you think of a question that makes the mean more informative here? How many trees are taller than the average height?

Sum ≈ 1043. Count = 20. Mean ≈ 52.15 ft. Sorted values: median ≈ (52+52)/2 = 52 ft. Roughly half the trees are above 52 ft.

Q4. The weights of a few newborn babies are given. Fill the dot plot provided below. Analyse and compare this data.

Plot weights on a number line; identify minimum, maximum, mean, median. Newborn weights cluster around 2.5–4 kg; outliers may indicate premature or exceptionally large babies.

Activity: Shoe-Size Detective

L4 Analyse

Materials: Notebook, pencil, class survey.

Predict: Will the mean shoe size of your class be the same as the median? Why or why not?

Ask each classmate their shoe size. Record values.
Sort the data and find the median.
Compute the mean.
Identify any outliers (very small or very large sizes).
Compare mean and median; discuss which better describes your class.

Tip: In most classrooms with no outliers, mean ≈ median. Outliers (e.g., one student with a very large size) can make mean > median.

Competency-Based Questions

Scenario: A small company has 9 employees. Their monthly salaries (in ₹ thousands) are: 15, 18, 18, 20, 22, 22, 24, 25, 200. The owner (earning ₹2,00,000) claims, "The average salary here is ₹40,000!"

Q1. Verify the owner's claim. Compute the mean salary.

L3 Apply

(a) ₹22,000
(b) ₹40,444
(c) ₹25,000
(d) ₹200,000

Answer: (b) ₹40,444. Sum = 15+18+18+20+22+22+24+25+200 = 364. Mean = 364 ÷ 9 ≈ 40.44 thousand = ₹40,444. So the owner's claim is arithmetically correct.

Q2. Find the median salary and analyse: does the mean or the median better represent a 'typical' employee's pay?

L4 Analyse

Answer: Sorted: 15, 18, 18, 20, 22, 22, 24, 25, 200. Median = ₹22,000 (5th value). The median is much closer to what most employees actually earn. Median better represents a typical employee because the ₹2,00,000 is an outlier that distorts the mean.

Q3. Evaluate: is the owner's claim of "average salary ₹40,000" honest or misleading? Justify with reasoning.

L5 Evaluate

Answer: Arithmetically honest but misleading. Only 1 of 9 employees earns above ₹40,000 — the outlier. A potential new employee would expect ₹40,000 but most employees earn ₹15,000–25,000. Ethically, the median (₹22,000) should be quoted for transparency.

Q4. Design a fairer 9-employee salary structure such that the mean equals the median. Show your data.

L6 Create

Example: 18, 20, 22, 24, 25, 26, 28, 30, 32. Sum = 225, Mean = 25. Median = 25. ✓ When data is symmetric with no outliers, mean = median.

Assertion–Reason Questions

Assertion (A): The median is a better measure of central tendency than the mean when data contains extreme outliers.
Reason (R): The median only depends on the position of the middle value, not on the magnitude of all values.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a) — R directly explains A: since median ignores extreme magnitudes, it is robust to outliers.

Assertion (A): For the dataset {1, 2, 3, 4, 100}, the mean and median are equal.
Reason (R): The middle value of a sorted 5-element dataset is the 3rd value.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (d) — Mean = 110/5 = 22, Median = 3. So A is false (mean ≠ median). R is true — it correctly identifies the median position.

Assertion (A): In a symmetric dataset with no outliers, mean = median.
Reason (R): Outliers always increase the mean.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (c) — A is true (symmetric data has mean = median). R is false: a very low outlier (e.g., 0 in a set otherwise around 100) decreases the mean, not increases it.

Frequently Asked Questions — Chapter 5

What is Part 2 — Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool in NCERT Class 7 Mathematics?

Part 2 — Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool is a key concept covered in NCERT Class 7 Mathematics, Chapter 5: Chapter 5. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Part 2 — Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool step by step?

To solve problems on Part 2 — Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 7 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 5: Chapter 5?

The essential formulas of Chapter 5 (Chapter 5) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Part 2 — Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool important for the Class 7 board exam?

Part 2 — Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool is part of the NCERT Class 7 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Part 2 — Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool?

Common mistakes in Part 2 — Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Part 2 — Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool?

End-of-chapter NCERT exercises for Part 2 — Median, Outliers & Comparing Data | Class 7 Maths Ch 5 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 5, and solve at least one previous-year board paper to consolidate your understanding.

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