Statistics — Mean, Median, Mode for Grouped Data (Class 9–10)

What Statistics Is Actually About

Most students treat this chapter as “apply formula, get answer.” That works for board exams, but you’ll understand it much better if you see the real picture first.

Statistics is about summarising a large dataset into a single representative number. When we have 30 students’ marks and the principal asks “how did the class do?”, we can’t list all 30 numbers. We pick one number that represents the group — that’s what mean, median, and mode do.

The catch is that Class 10 NCERT moves from individual data (Class 9) to grouped data — data organised into class intervals like 10–20, 20–30, etc. This requires different formulas, and that’s where most students lose marks.

We’ll cover all three methods for mean, both types of median calculation (formula + ogive), the mode formula, and the exam patterns for CBSE Class 9 and 10.

Key Terms and Definitions

Class Interval: A range of values grouped together. In 10–20, the lower boundary is 10 and upper is 20.

Class Width (h): The difference between upper and lower limits. For 10–20, $h = 10$ .

Midpoint / Class Mark ( $x_i$ ): The middle value of a class interval.

x_i = \frac{\text{lower limit} + \text{upper limit}}{2}

Frequency ( $f_i$ ): Number of observations falling in that class.

Cumulative Frequency: Running total of frequencies. If frequencies are 5, 8, 12, then cumulative frequencies are 5, 13, 25.

Assumed Mean ( $a$ ): A value we choose to simplify calculations — always the midpoint of the middle class interval.

Deviation ( $d_i$ ): $d_i = x_i - a$

Step Deviation ( $u_i$ ): $u_i = \dfrac{x_i - a}{h}$

Mean of Grouped Data

There are three methods. CBSE Class 10 expects you to know all three — they specify which one to use in each question.

Method 1: Direct Method

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

Use when: The midpoints ( $x_i$ ) and frequencies are small numbers. If you’re getting three-digit products everywhere, switch to assumed mean.

Steps:

Find midpoint $x_i$ of each class.
Multiply: $f_i \times x_i$ for each row.
Sum the $f_i$ column and the $f_i x_i$ column separately.
Divide.

Method 2: Assumed Mean Method

\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}

where $d_i = x_i - a$

Why this works: We’re essentially calculating how far the actual mean is from our assumed mean, then adding that correction. The algebra is identical to the direct method — this is just a computational shortcut.

Use when: Midpoints are large (like 150, 160, 170…) and direct method means multiplying large numbers.

Method 3: Step Deviation Method

\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h

where $u_i = \dfrac{x_i - a}{h}$

Use when: All class widths are equal (which they almost always are in CBSE problems). This gives the smallest numbers to work with.

In CBSE exams, step deviation method saves the most time. If the question says “find mean using step deviation method,” you must use this — switching methods costs marks.

Median of Grouped Data

Formula Method

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

$l$ = lower limit of median class
$n$ = total frequency
$cf$ = cumulative frequency of class before the median class
$f$ = frequency of the median class
$h$ = class width

Finding the median class: Calculate $\frac{n}{2}$ . The class whose cumulative frequency first exceeds this value is the median class.

Example: If $n = 50$ , we look for $\frac{50}{2} = 25$ . Find which class has the cumulative frequency that crosses 25.

The most common mistake: using $cf$ of the median class itself instead of the class before it. The formula needs the cumulative frequency that came before we entered the median class.

Ogive Method (Graphical)

An ogive (pronounced “oh-jive”) is a cumulative frequency curve. CBSE Class 10 asks you to draw it and find the median graphically.

Steps for “Less Than” Ogive:

Write upper limits of each class.
Write cumulative frequencies against each upper limit.
Plot points: (upper limit, cumulative frequency).
Join with a smooth curve — never straight lines between points.
Draw a horizontal line at $\frac{n}{2}$ on the y-axis.
Where it meets the ogive, drop a vertical line to the x-axis.
That x-value is the median.

For “More Than” Ogive: Use lower limits and “more than” cumulative frequencies. The intersection of both ogives also gives the median.

In CBSE 10 board exams, the ogive question is worth 3–4 marks. You must: correctly plot all points, join them as a smooth curve (not straight lines), and clearly mark where you read off the median.

Mode of Grouped Data

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

$l$ = lower limit 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
$h$ = class width

Modal class: The class with the highest frequency.

Why this formula?: Mode in grouped data can’t be read directly — we don’t know exact values. This formula interpolates within the modal class, assuming the mode is closer to whichever neighbouring class has a larger frequency.

If you’re unsure which is $f_0$ and which is $f_2$ : $f_0$ is always the class before (lower class interval), $f_2$ is always the class after. Don’t swap them — it changes the answer.

Solved Examples

Example 1 — Easy (CBSE Class 10)

Find the mean of the following data using the direct method:

Marks	0–10	10–20	20–30	30–40	40–50
Students	5	10	25	30	10

Solution:

Class	$x_i$	$f_i$	$f_i x_i$
0–10	5	5	25
10–20	15	10	150
20–30	25	25	625
30–40	35	30	1050
40–50	45	10	450
Total		80	2300

\bar{x} = \frac{2300}{80} = 28.75

Example 2 — Medium (CBSE Class 10 Board Pattern)

Find the mean marks using step deviation method:

Marks	10–25	25–40	40–55	55–70	70–85	85–100
Students	2	3	7	6	6	6

Solution: Take $a = 62.5$ (middle class midpoint), $h = 15$ .

Class	$x_i$	$f_i$	$u_i = \frac{x_i - 62.5}{15}$	$f_i u_i$
10–25	17.5	2	–3	–6
25–40	32.5	3	–2	–6
40–55	47.5	7	–1	–7
55–70	62.5	6	0	0
70–85	77.5	6	1	6
85–100	92.5	6	2	12
Total		30		–1

\bar{x} = 62.5 + \frac{-1}{30} \times 15 = 62.5 - 0.5 = 62

Example 3 — Hard (Finding Both Median and Mode)

The following table shows daily wages:

Wages (₹)	100–120	120–140	140–160	160–180	180–200
Workers	12	14	8	6	10

Find the median and mode.

Finding Median:

Class	Frequency	Cumulative Frequency
100–120	12	12
120–140	14	26
140–160	8	34
160–180	6	40
180–200	10	50

$n = 50$ , so $\frac{n}{2} = 25$ .

Cumulative frequency first exceeds 25 in the class 120–140. So median class is 120–140.

Here $l = 120$ , $cf = 12$ , $f = 14$ , $h = 20$ .

\text{Median} = 120 + \frac{25 - 12}{14} \times 20 = 120 + \frac{13}{14} \times 20 = 120 + 18.57 = 138.57

Finding Mode:

Highest frequency is 14 (class 120–140). Modal class = 120–140.

$l = 120$ , $f_1 = 14$ , $f_0 = 12$ , $f_2 = 8$ , $h = 20$ .

\text{Mode} = 120 + \frac{14 - 12}{2(14) - 12 - 8} \times 20 = 120 + \frac{2}{8} \times 20 = 120 + 5 = 125

Exam-Specific Tips

CBSE Class 10 Board Exams

Statistics typically carries 5–6 marks in CBSE Class 10 boards — one 3-mark and one or two 2-mark questions. The ogive question (draw + find median graphically) appears almost every year. PYQs from 2019, 2020, 2022, and 2023 all had it.

Show your complete table — even if the final answer is wrong, you get step marks for correct $f_i x_i$ columns.
For ogive: scale matters. Label both axes clearly. The examiner checks that your curve is smooth, not pointy.
The empirical relationship $\text{Mode} = 3\text{Median} - 2\text{Mean}$ appears as a 1-mark verification question. Memorise it.

CBSE Class 9

In Class 9, statistics covers ungrouped data and frequency polygons. The shift to grouped data formulas happens in Class 10. Class 9 board exams will ask for mean/median/mode of raw data and drawing bar graphs or frequency polygons.

Relationship Between Mean, Median, Mode

\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}

This is an approximate relationship that holds for moderately skewed distributions. CBSE uses it to verify answers or as a short 1-mark question.

Common Mistakes to Avoid

Mistake 1: Wrong cumulative frequency in median formula. $cf$ in the median formula is the cumulative frequency of the class before the median class, not the median class itself. Write the cumulative frequency column first, identify the median class, then pick $cf$ from the row above it.

Mistake 2: Confusing $f_0$ and $f_2$ in mode formula. $f_0$ is the frequency of the class with a smaller class interval (comes before), $f_2$ comes after. Swapping them gives a different (wrong) answer that you can’t recover from.

Mistake 3: Using class boundaries instead of midpoints. For the mean, use the midpoint $x_i = \frac{\text{lower} + \text{upper}}{2}$ , never the lower or upper limit alone.

Mistake 4: Drawing straight lines instead of a smooth curve for ogive. An ogive is a smooth S-shaped curve. Straight-line segments between points is a frequency polygon, not an ogive. This is an explicit marking criterion.

Mistake 5: Wrong median class identification. The median class is the one whose cumulative frequency first exceeds $\frac{n}{2}$ . If $\frac{n}{2} = 25$ and cumulative frequencies are 12, 25, 34…, the median class is the one with CF = 34 (the one that crosses 25), not the one with CF = 25.

Practice Questions

Q1. Find the mean of the following data using the assumed mean method:

Class	0–20	20–40	40–60	60–80	80–100
Frequency	6	8	10	12	4

Take $a = 50$ (midpoint of 40–60), $h = 20$ .

$x_i$	$f_i$	$d_i = x_i - 50$	$f_i d_i$
10	6	–40	–240
30	8	–20	–160
50	10	0	0
70	12	20	240
90	4	40	160
	40		0

$\bar{x} = 50 + \frac{0}{40} = 50$

Q2. Find the mode of: 10–20 (4 students), 20–30 (7), 30–40 (12), 40–50 (8), 50–60 (5).

Modal class = 30–40 (highest frequency 12).

$l = 30$ , $f_1 = 12$ , $f_0 = 7$ , $f_2 = 8$ , $h = 10$ .

\text{Mode} = 30 + \frac{12 - 7}{2(12) - 7 - 8} \times 10 = 30 + \frac{5}{9} \times 10 = 30 + 5.56 = 35.56

Q3. The mean of a distribution is 28 and mode is 16. Find the median using the empirical formula.

Mode = 3 Median – 2 Mean

$16 = 3 \times \text{Median} - 2 \times 28$

$16 = 3 \times \text{Median} - 56$

$3 \times \text{Median} = 72$

Median = 24

Q4. Find median from: 0–10 (3), 10–20 (5), 20–30 (9), 30–40 (5), 40–50 (3).

$n = 25$ , $\frac{n}{2} = 12.5$ .

Cumulative frequencies: 3, 8, 17, 22, 25.

CF first exceeds 12.5 at 17. Median class = 20–30.

$l = 20$ , $cf = 8$ , $f = 9$ , $h = 10$ .

\text{Median} = 20 + \frac{12.5 - 8}{9} \times 10 = 20 + 5 = 25

Q5. Ages of 40 patients: 5–14 (6), 15–24 (11), 25–34 (21), 35–44 (23), 45–54 (14), 55–64 (5). Find mean using step deviation.

Note: classes are not of equal upper–lower difference if written as above. Treat as continuous: 4.5–14.5, etc. Mid-values: 9.5, 19.5, 29.5, 39.5, 49.5, 59.5. Take $a = 29.5$ , $h = 10$ .

Wait — re-reading, this is a standard NCERT-type problem. Take midpoints as 9.5, 19.5, 29.5, 39.5, 49.5, 59.5. Choose $a = 29.5$ .

$x_i$	$f_i$	$u_i$	$f_i u_i$
9.5	6	–2	–12
19.5	11	–1	–11
29.5	21	0	0
39.5	23	1	23
49.5	14	2	28
59.5	5	3	15
	80		43

$\bar{x} = 29.5 + \frac{43}{80} \times 10 = 29.5 + 5.375 = 34.875 \approx 34.9$

Q6. If $\sum f_i = 50$ and $\sum f_i u_i = -7$ with $a = 25$ and $h = 5$ , find the mean.

\bar{x} = 25 + \frac{-7}{50} \times 5 = 25 - 0.7 = 24.3

Q7. A dataset has mean 35 and median 33. Estimate the mode.

Mode = 3 × Median – 2 × Mean = 3(33) – 2(35) = 99 – 70 = 29

Q8. Draw a less-than ogive for the data below and find the median:

Height (cm)	140–145	145–150	150–155	155–160	160–165
Girls	5	11	14	10	10

Less-than cumulative frequencies:

Less than	CF
145	5
150	16
155	30
160	40
165	50

Plot these 5 points: (145, 5), (150, 16), (155, 30), (160, 40), (165, 50). Draw smooth curve.

$\frac{n}{2} = 25$ . Draw horizontal line at CF = 25. It meets the curve above approximately 152.

Verify with formula: $n = 50$ , $\frac{n}{2} = 25$ . Median class = 150–155 (CF crosses 25 at 30).

$\text{Median} = 150 + \frac{25 - 16}{14} \times 5 = 150 + \frac{45}{14} = 150 + 3.21 = 153.21$ cm.

Frequently Asked Questions

Which mean method should I use when the question doesn’t specify?

Direct method is fine for simple numbers. For large midpoints or when class width is uniform, step deviation is faster and less error-prone. In a board exam, use what makes the arithmetic cleanest — and always show your table in full.

Can mode have two values?

Yes — if two classes tie for highest frequency, the data is bimodal. The formula applies to each modal class separately. CBSE rarely tests this, but know the term.

Why do we use midpoints instead of actual values?

In grouped data, we don’t have individual values — only that 10 students scored between 30 and 40. The midpoint (35) is our best estimate of what those 10 students actually scored on average.

How is the ogive different from a frequency polygon?

A frequency polygon plots frequency against midpoints — it shows where data is concentrated. An ogive plots cumulative frequency against upper (or lower) limits — it tells you “how many observations fall below this value.” The shape is different: ogive is S-shaped, frequency polygon is wave-shaped.

The empirical formula gives an approximation, not exact values. When is it exact?

The formula Mode = 3 Median – 2 Mean holds exactly for a perfectly symmetric or normally distributed dataset. For real data, it’s an approximation. CBSE uses it as an exact relationship for problem-solving — just apply it without worrying about this.

What’s the difference between less-than and more-than ogive?

Less-than ogive: plot upper class limits vs cumulative frequency (increasing curve). More-than ogive: plot lower class limits vs “more than” cumulative frequency (decreasing curve). Their intersection gives the median. Both are ascending/descending S-curves.

How many marks is Statistics in Class 10 boards?

Statistics falls under the “Statistics and Probability” unit, which carries 11 marks in the CBSE Class 10 board exam. Statistics alone (Chapter 14) typically accounts for 6–7 marks.

Is step deviation method only for equal class widths?

Technically yes — the formula divides by $h$ , which assumes uniform class width. If class widths vary, use the assumed mean method instead. CBSE problems almost always have equal class widths, so this rarely comes up in practice.