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ICSE • Class 9 • Mathematics • Ch 18
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Mean and Median of Ungrouped Data

In ICSE Class 9 Mathematics, "Mean and Median of Ungrouped Data" covers the fundamental measures of central tendency, which serve as representative single values describing the center of a quantitative data distribution. The Arithmetic Mean ($\bar{x}$) represents the mathematical average, calculated as the sum of all observations divided by the total number of observations: for raw data, $\bar{x} = \frac{\sum x_i}{n}$; and for ungrouped frequency distributions, $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$. The chapter explores the algebraic properties of the mean: if every observation is increased, decreased, multiplied, or divided by a non-zero constant $c$, the new mean is similarly transformed ($\bar{x}_{\text{new}} = \bar{x} \pm c$ or $c\bar{x}$); and the algebraic sum of deviations from the mean is always zero ($\sum (x_i - \bar{x}) = 0$). The Median ($M$) is defined as the positional middle value when data is arranged in ascending or descending order of magnitude: if $n$ is odd, $M = \left(\frac{n+1}{2}\right)^{\text{th}}$ observation; if $n$ is even, $M = \frac{1}{2}\left[\left(\frac{n}{2}\right)^{\text{th}} + \left(\frac{n}{2} + 1\right)^{\text{th}}\right]$ observation. The chapter rigorously compares mean vs median, highlighting that while the mean is highly sensitive to extreme outliers, the median remains robust and resistant.

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The Bill Gates in the Coffee Shop Paradox: Why the Average Income Can Lie, But the Median Never Does

Imagine a small coffee shop in Mumbai where nine ordinary citizens—teachers, engineers, and shopkeepers—are sitting, each earning an average salary of $\text{₹}50,000$ per month. Now imagine billionaire Mukesh Ambani walks through the door and sits down at a table! If you calculate the Arithmetic Mean of the ten people in the room, the average monthly income suddenly skyrockets to over $\text{₹}100,000,000$ (Ten Crore Rupees)! On paper, the "average" person in that coffee shop is an ultra-wealthy tycoon, even though nine of them cannot even afford a new scooter! This is the fundamental danger of the Mean—it is violently distorted by extreme values (outliers). But if you calculate the Median (the exact middle person when lined up by income), Ambani's entrance only shifts the median slightly from $\text{₹}50,000$ to $\text{₹}52,000$, painting a completely truthful, robust picture of the room! When should a scientist use the Mean, and when is the Median superior? Let us explore central tendency!

Why This Chapter Matters

Measures of central tendency are ubiquitous in economic poverty lines, national per capita income, school grade evaluations, laboratory measurement calibrations, and machine learning feature normalization.

Before You Begin (Prerequisites)

  • Summation notation ($\sum$) and basic algebraic manipulation.
  • Sorting numerical data in ascending order.

What You Will Learn (Core Objectives)

  • Calculate the arithmetic mean of raw data and ungrouped frequency distribution tables.
  • Apply algebraic transformation rules to determine the effect of arithmetic operations on the mean.
  • Determine the median of a raw dataset for both odd and even numbers of observations.
  • Find the median from an ungrouped frequency distribution using cumulative frequencies.
  • Evaluate missing frequencies or observations given the mean or median of a distribution.

Chapter Roadmap & Progression

1 1. The Arithmetic Mean: Formulas &...
2 2. The Median: Positional Measure o...
3 3. Worked ICSE Problem Archetypes

Complete Concept Guide (100% Curriculum Coverage)

1. The Arithmetic Mean: Formulas & Properties

Arithmetic Mean
A. Formulations:
  • For Raw Data ($x_1, x_2, ..., x_n$): $$\mathbf{\bar{x} = \frac{x_1 + x_2 + ... + x_n}{n} = \frac{\sum x_i}{n}}$$
  • For Ungrouped Frequency Distribution: $$\mathbf{\bar{x} = \frac{f_1x_1 + f_2x_2 + ... + f_kx_k}{f_1 + f_2 + ... + f_k} = \frac{\sum f_i x_i}{\sum f_i} = \frac{\sum f_i x_i}{N}}$$
B. Key Algebraic Properties:
  1. Sum of Deviations is Zero: $\sum (x_i - \bar{x}) = 0$.
  2. Linear Transformation Rule:
    • If a constant $c$ is added to/subtracted from each observation: $\bar{x}_{\text{new}} = \bar{x} \pm c$.
    • If each observation is multiplied by a non-zero constant $c$: $\bar{x}_{\text{new}} = c \cdot \bar{x}$.
  3. Combined Mean of Two Groups: If group 1 has $n_1$ items with mean $\bar{x}_1$ and group 2 has $n_2$ items with mean $\bar{x}_2$: $$\bar{x}_{\text{combined}} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$$

2. The Median: Positional Measure of Center

The Median

To find the median, first arrange the raw data in ascending order.

Case 1: When $n$ is ODD:
$$\mathbf{M = \left(\frac{n + 1}{2}\right)^{\text{th}} \text{ observation}}$$

There is a unique single observation lying exactly in the middle.

Case 2: When $n$ is EVEN:
$$\mathbf{M = \frac{1}{2} \left[ \left(\frac{n}{2}\right)^{\text{th}} \text{ observation} + \left(\frac{n}{2} + 1\right)^{\text{th}} \text{ observation} \right]}$$

The median is the arithmetic mean of the two middle observations.

3. Worked ICSE Problem Archetypes

Exemplary Solutions
Problem 1: The mean of $5$ observations is $15$. If a new observation is included, the mean becomes $17$. Find the value of the included observation.

Solution:

$$\text{Sum of original 5 observations} = 5 \times \bar{x} = 5 \times 15 = 75$$

Let the new observation be $x$. The new number of observations is $5 + 1 = 6$.

$$\text{New Sum} = 6 \times \text{New Mean} = 6 \times 17 = 102$$ $$\text{Value of included observation} = 102 - 75 = \mathbf{27}$$
Problem 2: The following observations are arranged in ascending order: $11, 13, 15, 19, x + 2, x + 4, 30, 35, 39, 43$. If the median of the data is $25$, find the value of $x$.

Solution:

Count the number of observations: $n = 10$ (which is an even number).

The median is the average of the $\left(\frac{10}{2}\right)^{\text{th}} = 5^{\text{th}}$ and $\left(\frac{10}{2} + 1\right)^{\text{th}} = 6^{\text{th}}$ observations:

$$5^{\text{th}} \text{ term} = x + 2, \qquad 6^{\text{th}} \text{ term} = x + 4$$ $$\text{Median} = \frac{(x + 2) + (x + 4)}{2} = \frac{2x + 6}{2} = x + 3$$

Equating this to the given median ($25$):

$$x + 3 = 25 \implies \mathbf{x = 22}$$

Key Formulas, Identities & Theorems

Raw Data Mean
$$\bar{x} = \frac{\sum x_i}{n}$$
Sum of items divided by count.
Frequency Table Mean
$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$
Weighted sum divided by total frequency.
Median (n odd)
M = x_{(n+1)/2}
Single exact middle item in sorted array.
Median (n even)
$$M = \frac{1}{2}\left(x_{n/2} + x_{n/2 + 1}\right)$$
Average of the two middle sorted items.

Mathematics: Mean vs Median in Symmetric and Skewed Distributions

Central Tendency: Mean vs. Median & The Outlier Effect Right-Skewed Distribution (Outlier Effect) Mode Median Mean Order: Mode < Median < Mean Comparison of Key Formulas Arithmetic Mean (¯x): • Raw: ¯x = (Σx) / n • Frequency Table: ¯x = (Σfx) / (Σf) Median (M) when n is ODD: • Sort data ascending first! • M = [(n + 1) / 2]ᵗʰ observation Median (M) when n is EVEN: • Average of two central items: • M = ½ [ (n/2)ᵗʰ + (n/2 + 1)ᵗʰ ] Σ(xᵢ - ¯x) = 0  |  Median is robust against outliers!

Chapter Summary & 10 Key Takeaways

Takeaway 1
Central tendency represents the central clustering point of a quantitative data distribution.
Takeaway 2
Arithmetic mean of raw data is sum of observations divided by total count: x̄ = (∑ x) / n.
Takeaway 3
Arithmetic mean of frequency distribution is x̄ = (∑ fx) / (∑ f).
Takeaway 4
If every observation is altered by adding, subtracting, or multiplying by constant c, the mean undergoes the identical operation.
Takeaway 5
The sum of deviations of all observations from their arithmetic mean is strictly zero: ∑ (xi - x̄) = 0.
Takeaway 6
To find the median, data MUST strictly be arranged in ascending or descending order first.
Takeaway 7
When n is odd, the median is the single central term: ((n + 1) / 2)th term.
Takeaway 8
When n is even, the median is the average of the two middle terms: 1/2 * [(n/2)th + (n/2 + 1)th].
Takeaway 9
The mean is sensitive to extreme outliers; the median is robust and unaffected by extreme values.
Takeaway 10
Combined mean formula for two groups: x̄_combined = (n1*x̄1 + n2*x̄2) / (n1 + n2).

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
The mean of six numbers is $30$. If one number is excluded, their mean becomes $28$. Find the excluded number.
Reveal Answer & Explanation
Answer: • Original total sum of $6$ numbers:
$$\text{Sum}_6 = 6 \times 30 = 180$$
• When one number is excluded, $5$ numbers remain with mean $28$:
$$\text{Sum}_5 = 5 \times 28 = 140$$
• The excluded number is the difference between the two sums:
$$\text{Excluded Number} = 180 - 140 = \mathbf{40}$$
Old sum = 6 * 30 = 180. New sum = 5 * 28 = 140. Excluded number = 180 - 140 = 40.
2
Find the median of the following dataset: $25, 14, 18, 30, 19, 27, 21, 22$.
Reveal Answer & Explanation
Answer:

• Step 1: Count the number of observations: $n = 8$ (which is even).
• Step 2: Arrange the data in ascending order:

$$14, 18, 19, 21, 22, 25, 27, 30$$


• Step 3: Since $n = 8$ is even, median is the average of the $\left(\frac{8}{2}\right) = 4^{\text{th}}$ and $\left(\frac{8}{2} + 1\right) = 5^{\text{th}}$ observations:
- $4^{\text{th}} \text{ term} = 21$
- $5^{\text{th}} \text{ term} = 22$
• Calculation:

$$\text{Median} = \frac{21 + 22}{2} = \frac{43}{2} = \mathbf{21.5}$$


Sort: 14, 18, 19, 21, 22, 25, 27, 30. Median is (21 + 22)/2 = 21.5.
3
If the mean of numbers $x, x + 3, x + 5, x + 7, x + 10$ is $9$, find the value of $x$ and determine the median.
Reveal Answer & Explanation
Answer: • Number of observations $n = 5$.
• Mean equation:
$$\frac{x + (x + 3) + (x + 5) + (x + 7) + (x + 10)}{5} = 9$$
$$\frac{5x + 25}{5} = 9 \implies x + 5 = 9 \implies \mathbf{x = 4}$$
• Substitute $x = 4$ into the five terms:
$$4, 7, 9, 11, 14$$
• Since $n = 5$ is odd, the median is the $\left(\frac{5+1}{2}\right) = 3^{\text{rd}}$ observation:
$$\mathbf{\text{Median} = 9}$$
5x + 25 = 45 -> x = 4. Numbers are 4, 7, 9, 11, 14. 3rd term is 9.
4
Prove that the algebraic sum of deviations of a set of $n$ observations from their arithmetic mean is zero ($\sum_{i=1}^n (x_i - \bar{x}) = 0$).
Reveal Answer & Explanation
Answer: • Expand the summation:
$$\sum_{i=1}^n (x_i - \bar{x}) = (x_1 - \bar{x}) + (x_2 - \bar{x}) + ... + (x_n - \bar{x})$$
$$= (x_1 + x_2 + ... + x_n) - n\bar{x}$$
• By the definition of arithmetic mean, $\bar{x} = \frac{\sum x_i}{n} \implies \sum x_i = n\bar{x}$.
• Substitute $\sum x_i = n\bar{x}$:
$$= n\bar{x} - n\bar{x} = \mathbf{0} \quad \blacksquare$$
∑ (xi - x̄) = ∑ xi - n*x̄ = n*x̄ - n*x̄ = 0.
5
The mean weight of $150$ students in a class is $60\text{ kg}$. The mean weight of male students is $70\text{ kg}$ and that of female students is $55\text{ kg}$. Find the number of male and female students.
Reveal Answer & Explanation
Answer: • Let the number of male students be $m$. Then female students $= 150 - m$.
• Total weight of all students $= 150 \times 60 = 9000\text{ kg}$.
• Total weight from subgroups:
$$70m + 55(150 - m) = 9000$$
$$70m + 8250 - 55m = 9000$$
$$15m = 9000 - 8250 = 750 \implies m = \frac{750}{15} = 50$$
• Number of male students $= \mathbf{50}$.
• Number of female students $= 150 - 50 = \mathbf{100}$.
70m + 55(150 - m) = 150 * 60 = 9000. 15m = 750 -> m = 50.
6
If the median of an arranged dataset is $17$, what will be the new median if $3$ is added to every observation?
Reveal Answer & Explanation
Answer: • Adding a constant $c = 3$ to every observation shifts the entire ordered distribution upward along the number line by $3\text{ units}$.
• The relative order of the observations remains completely unchanged.
• Therefore, the central observation is also shifted upward by $3$:
$$\text{New Median} = 17 + 3 = \mathbf{20}$$
Adding a constant c to every observation increases the median by c: 17 + 3 = 20.
7
Why is the median preferred over the mean when reporting national household incomes or house prices?
Reveal Answer & Explanation
Answer:

• Income and real estate distributions are heavily skewed to the right (positive skew) by a tiny fraction of multi-billionaires and luxury mansions (extreme high outliers).
• The arithmetic mean is heavily inflated by these few massive values, presenting a misleadingly wealthy impression of the typical family.
• The median represents the exact middle ($50^{\text{th}}$ percentile) household, giving an accurate, robust measure of what a typical ordinary citizen earns, unaffected by outliers.


Income distributions have extreme high outliers that distort the mean. The median is resistant to outliers.
8
Find the mean of the first $10$ natural numbers.
Reveal Answer & Explanation
Answer: • The first $10$ natural numbers are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$.
• The sum of the first $n$ natural numbers is given by $\frac{n(n+1)}{2}$:
$$\text{Sum} = \frac{10(11)}{2} = 55$$
• Arithmetic Mean:
$$\bar{x} = \frac{\text{Sum}}{n} = \frac{55}{10} = \mathbf{5.5}$$
Sum = 55. Mean = 55 / 10 = 5.5.
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