CBSE Class 10 Maths — Statistics

Chapter Overview & Weightage

Statistics in Class 10 CBSE is a high-weightage chapter — it consistently carries 10–12 marks in the board exam. The questions are formula-based and predictable, making this one of the easiest chapters to score full marks in with focused preparation.

Year	Marks	Topics Tested
2024	11	Mean (assumed mean), Median, Ogive
2023	10	Mean (direct method), Mode, Median
2022	11	Mean (step-deviation), Mode, Ogive-based median

Statistics is a scoring topic. You can expect 1–2 MCQs (1 mark each), one 3-mark question, and one 5-mark question. If you practise all three measures of central tendency with full table work, 10–12 marks are near-guaranteed.

Key Concepts You Must Know

Grouped Data: Data arranged in class intervals (e.g., 10–20, 20–30). You work with the class mark (midpoint of each interval) for mean calculations.

Three Measures of Central Tendency:

Mean: Average. Three methods — direct, assumed mean, step-deviation.
Median: Middle value when arranged in order. For grouped data, use the median formula with cumulative frequencies.
Mode: Most frequently occurring value. For grouped data, the modal class is the class with the highest frequency.

Cumulative Frequency: Running total of frequencies. Used to draw the ogive and to locate median.

Ogive (Cumulative Frequency Curve): A graph of cumulative frequency vs upper class boundary. The x-coordinate of the point where the ogive hits n/2 gives the median.

Important Formulas

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

Where $x_i$ = class mark, $f_i$ = frequency. Use when numbers are small and manageable.

\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}, \quad d_i = x_i - a

Choose $a$ = class mark near the middle. Reduces arithmetic in large-number problems.

\bar{x} = a + h \cdot \frac{\sum f_i u_i}{\sum f_i}, \quad u_i = \frac{x_i - a}{h}

Use when all class widths are equal (same $h$ ). Fastest method for board exams.

\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h

$l$ = lower boundary of median class, $n$ = total frequency, $cf$ = cumulative frequency before median class, $f$ = frequency of median class, $h$ = class width.

\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h

$l$ = lower boundary of modal class, $f_1$ = frequency of modal class, $f_0$ = frequency of class before modal class, $f_2$ = frequency of class after modal class.

Solved Previous Year Questions

PYQ 1 — Mean by Step-Deviation Method (5 marks)

Q: Find the mean of the following data:

Class	10–20	20–30	30–40	40–50	50–60
Frequency	4	8	14	10	4

Solution: Take $a = 35$ , $h = 10$ .

Class	$x_i$	$f_i$	$u_i = \frac{x_i-35}{10}$	$f_i u_i$
10–20	15	4	−2	−8
20–30	25	8	−1	−8
30–40	35	14	0	0
40–50	45	10	1	10
50–60	55	4	2	8
Total		40		2

\bar{x} = 35 + 10 \times \frac{2}{40} = 35 + 0.5 = \mathbf{35.5}

PYQ 2 — Median (5 marks)

Q: Find the median of the following distribution:

Class	0–10	10–20	20–30	30–40	40–50
Frequency	5	8	20	15	2

Solution: $n = 50$ , so $n/2 = 25$ .

Cumulative frequencies: 5, 13, 33, 48, 50.

The 25th value falls in class 20–30 (cf before = 13, this class brings total to 33).

Median class: 20–30. Here $l = 20$ , $cf = 13$ , $f = 20$ , $h = 10$ .

\text{Median} = 20 + \frac{25 - 13}{20} \times 10 = 20 + \frac{12}{20} \times 10 = 20 + 6 = \mathbf{26}

PYQ 3 — Mode (3 marks)

Q: From the above distribution, find the mode.

Solution: Highest frequency = 20 (class 20–30). Modal class = 20–30.

$l = 20$ , $f_1 = 20$ , $f_0 = 8$ , $f_2 = 15$ , $h = 10$ .

\text{Mode} = 20 + \frac{20 - 8}{2(20) - 8 - 15} \times 10 = 20 + \frac{12}{17} \times 10 = 20 + 7.06 \approx \mathbf{27.06}

Difficulty Distribution

Difficulty	% of Questions	Types
Easy	40%	Direct mean calculation, identifying modal class
Medium	40%	Median with cumulative frequency table, mode formula
Hard	20%	Missing frequency from given mean/median, ogive-based median

Expert Strategy

Always draw a frequency table even if the data is already given in table form — lay out all columns ( $x_i$ , $f_i$ , $f_i x_i$ or $u_i$ , etc.) systematically. CBSE awards step marks for each column, so a complete table guarantees marks even if your final answer has an arithmetic error.

For median, find $n/2$ first and locate the median class by checking cumulative frequencies. Students waste time recalculating — just build the cumulative frequency column from the start.

For mode, the modal class is the one with the maximum frequency — not the one with the maximum class mark or maximum cumulative frequency.

When the question gives missing frequencies and asks you to find them using mean, set up the mean formula with the unknown as $x$ or $k$ . Solve the linear equation. This appears every 2–3 years in CBSE boards and is worth 5 marks.

Common Traps

Trap 1 — Class mark confusion: Always use the midpoint of the class interval as $x_i$ , not the upper or lower boundary. For class 10–20, $x_i = 15$ , not 10 or 20.

Trap 2 — Wrong median class: Students often pick the class where cumulative frequency first exceeds $n/2$ . That is correct — but they must use the lower boundary of that class as $l$ , not the upper boundary.

Trap 3 — Mode formula sign errors: In the mode formula, the denominator is $2f_1 - f_0 - f_2$ . Students sometimes write $f_0 - 2f_1 + f_2$ or similar. Write it out fully each time — do not try to memorise sign patterns.

Trap 4 — Forgetting $h$ in step-deviation: The step-deviation method multiplies $\sum f_i u_i / \sum f_i$ by $h$ . Students who forget this multiplication will get an answer that is $h$ times too small.