Chapter 13 of 14

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Rajasthan (RBSE)Class 10 Mathematics★ 5–7 marks · RBSE Class 10 Maths board (Unit — Statistics & Probability)

Statistics

RBSE Class 10 Maths Chapter 13 — finding the mean of grouped data three ways (direct, assumed-mean and step-deviation), the mode of grouped data, the median with its cumulative-frequency method, and the empirical relationship between mean, median and mode. Includes the exact formulas and fully solved board-pattern problems.

⏱ 36 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Build a grouped-frequency table with class marks and cumulative frequencies
2Compute the mean of grouped data by the direct, assumed-mean and step-deviation methods
3Identify the modal class and find the mode using the mode formula
4Identify the median class from cumulative frequencies and find the median
5Apply the empirical relationship 3 Median = Mode + 2 Mean to estimate a missing measure

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Why this chapter matters

Statistics is a reliable, formula-driven scorer in the RBSE board — a grouped-data mean/median/mode question is almost guaranteed, often as a 4–5 mark long answer. The table-building discipline learned here also underpins probability and all of Class 11 descriptive statistics.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Statistics (Class 9)

Frequency tables, bar graphs and the mean of ungrouped data.

Statistics — RBSE Class 10 (Mathematics)

A teacher has the marks of 60 students spread across class intervals 0–10, 10–20, and so on. What single number best represents the class? Is it the average, the most common band, or the middle student? This chapter gives you all three — mean, mode and median — for grouped data, where you no longer have the individual values, only how many fall in each interval.

1. From raw data to grouped data

When there are many observations, we organise them into class intervals with frequencies (how many values fall in each interval). The trade-off: organisation makes patterns visible, but we lose the exact values — so we work with the class mark (midpoint) of each interval:

$x_{i} = \frac{lower limit + upper limit}{2}$

2. Mean of grouped data — three methods

(a) Direct method

$\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}}$

Multiply each class mark $x_{i}$ by its frequency $f_{i}$ , add up, divide by the total frequency. Simple, but the arithmetic gets heavy when $x_{i}$ are large.

(b) Assumed-mean method

Pick a convenient class mark as the assumed mean a, and work with deviations $d_{i} = x_{i} - a$ :

$\overset{x}{ˉ} = a + \frac{\sum f _{i} d _{i}}{\sum f _{i}}$

The numbers are smaller, so the computation is lighter. The mean does not depend on which a you pick — choosing one near the centre just keeps the deviations small.

(c) Step-deviation method

If all class sizes are equal to h, scale the deviations: $u_{i} = \frac{x _{i} - a}{h}$ .

$\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}}$

This is the lightest of the three and the one to prefer in the exam when class widths are equal.

All three methods give exactly the same mean — they are just smarter bookkeeping. Choose based on the numbers in front of you.

3. Mode of grouped data

The mode is the value that occurs most often. For grouped data we first find the modal class (the interval with the highest frequency), then:

$Mode = l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h$

where

$l$ = lower limit of the modal class,
$f_{1}$ = frequency of the modal class,
$f_{0}$ = frequency of the class before it,
$f_{2}$ = frequency of the class after it,
$h$ = class size.

4. Median of grouped data

The median is the middle value when data are arranged in order — it splits the data into two equal halves. For grouped data, build a cumulative frequency (cf) column, find $\frac{n}{2}$ (where n = total frequency), locate the median class (the first class whose cf ≥ n/2), then:

$Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h$

where

$l$ = lower limit of the median class,
$c f$ = cumulative frequency of the class before the median class,
$f$ = frequency of the median class,
$h$ = class size.

The median is unaffected by extreme values, which makes it the fairest "typical value" for skewed data (like incomes).

5. The empirical relationship

For a moderately skewed distribution, the three measures are tied together by:

$3 Median = Mode + 2 Mean$

equivalently Mode = 3 Median − 2 Mean. If a question gives any two of the three, you can estimate the third — a frequently-asked 1–2 mark shortcut.

6. Closing thought

Three "averages", three jobs:

Mean — the balance point; uses every value; best when data are symmetric. Compute it the light way (step-deviation when class widths are equal).
Mode — the most frequent; best for "what is most common" questions.
Median — the middle; resistant to outliers; best for skewed data.

The skill the RBSE board tests is setting up the table correctly — the class marks, the $f_{i} x_{i}$ or cumulative-frequency columns — and then plugging into the right formula. Lay the table out neatly, identify the modal/median class without rushing, and the marks follow. This chapter also sets up the ogive and probability work that follow, and the descriptive statistics you'll meet again in Class 11.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Class mark

xᵢ = (lower + upper limit) / 2

Midpoint of a class interval.

Mean — direct

x̄ = Σfᵢxᵢ / Σfᵢ

Σfᵢ = total frequency n.

Mean — assumed mean

x̄ = a + (Σfᵢdᵢ / Σfᵢ),  dᵢ = xᵢ − a

Smaller arithmetic; a is a chosen class mark.

Mean — step deviation

x̄ = a + h·(Σfᵢuᵢ / Σfᵢ),  uᵢ = (xᵢ − a)/h

Lightest method when class sizes are equal.

Mode

Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)]·h

l, f₁ of modal class; f₀ before, f₂ after; h class size.

Median

Median = l + [(n/2 − cf)/f]·h

Median class is first with cf ≥ n/2.

Empirical relation

3 Median = Mode + 2 Mean

Mode = 3 Median − 2 Mean.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Using class limits instead of class marks in the mean

✓ The mean uses the class MARK xᵢ = (lower + upper)/2, not the limits. Build the xᵢ column first.

WATCH OUT

✗ Picking the wrong modal/median class

✓ Modal class = highest FREQUENCY. Median class = first class whose CUMULATIVE frequency ≥ n/2. Don't confuse the two.

WATCH OUT

✗ Using cf of the median class instead of the class before it

✓ In the median formula, cf is the cumulative frequency of the class JUST BEFORE the median class — not the median class itself.

WATCH OUT

✗ Forgetting to multiply by h in mode/median/step-deviation

✓ The class size h multiplies the bracket in the mode, median and step-deviation formulas. Dropping h is a common, costly slip.

WATCH OUT

✗ Mis-stating the empirical relation

✓ It is 3 Median = Mode + 2 Mean. A common wrong version swaps mode and mean — write it carefully.

WATCH OUT

✗ Mixing up f₀ and f₂ in the mode formula

✓ f₀ is the frequency of the class BEFORE the modal class; f₂ is the class AFTER it. Label them on the table.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 13.1

Exercise 13.1

Mean of grouped data by direct, assumed-mean and step-deviation methods

Questions

Ex 13.2

Exercise 13.2

Mode of grouped data and the empirical relationship

Questions

Ex 13.3

Exercise 13.3

Median of grouped data using cumulative frequency

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Class mark

Find the class mark of the interval 25–35.

▶ Show solution

Step 1 — Class mark = (lower + upper)/2 = (25 + 35)/2.

Step 2 — = 30.

✦ Answer: 30.

Q2EASY· Empirical relation

If the mean is 30 and the median is 28, estimate the mode.

▶ Show solution

Step 1 — Mode = 3 Median − 2 Mean.

Step 2 — = 3(28) − 2(30) = 84 − 60 = 24.

✦ Answer: Mode ≈ 24.

Q3EASY· Median class

For 40 observations, which value of cumulative frequency marks the median class?

▶ Show solution

Step 1 — n/2 = 40/2 = 20.

Step 2 — The median class is the first class whose cumulative frequency is ≥ 20.

✦ Answer: The class where cf first reaches or exceeds 20.

Q4MEDIUM· Mean (direct)

Find the mean of: 0–10 (f=2), 10–20 (f=3), 20–30 (f=5), 30–40 (f=4), 40–50 (f=1).

▶ Show solution

Step 1 — Class marks xᵢ: 5, 15, 25, 35, 45.

Step 2 — fᵢxᵢ: 10, 45, 125, 140, 45 → Σfᵢxᵢ = 365.

Step 3 — Σfᵢ = 2+3+5+4+1 = 15.

Step 4 — Mean = 365/15 = 24.33 (approx).

✦ Answer: Mean ≈ 24.33.

Q5MEDIUM· Mode

Find the mode of: 0–10 (f=3), 10–20 (f=8), 20–30 (f=10), 30–40 (f=6), 40–50 (f=3).

▶ Show solution

Step 1 — Highest frequency is 10 → modal class 20–30. So l = 20, f₁ = 10, f₀ = 8 (class before), f₂ = 6 (class after), h = 10.

Step 2 — Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)]·h.

Step 3 — = 20 + [(10 − 8)/(20 − 8 − 6)]·10 = 20 + (2/6)·10 = 20 + 3.33.

✦ Answer: Mode ≈ 23.33.

Q6MEDIUM· Step deviation

Using the step-deviation method with a = 25 and h = 10, find the mean of: 0–10 (2), 10–20 (3), 20–30 (5), 30–40 (4), 40–50 (1).

▶ Show solution

Step 1 — Class marks: 5,15,25,35,45. uᵢ = (xᵢ−25)/10: −2, −1, 0, 1, 2.

Step 2 — fᵢuᵢ: −4, −3, 0, 4, 2 → Σfᵢuᵢ = −1. Σfᵢ = 15.

Step 3 — Mean = a + h·(Σfᵢuᵢ/Σfᵢ) = 25 + 10·(−1/15) = 25 − 0.67.

✦ Answer: Mean ≈ 24.33 (matches the direct method).

Q7HARD· Median

Find the median of: 0–10 (f=5), 10–20 (f=8), 20–30 (f=12), 30–40 (f=10), 40–50 (f=5).

▶ Show solution

Step 1 — Cumulative frequencies: 5, 13, 25, 35, 40. So n = 40, n/2 = 20.

Step 2 — First cf ≥ 20 is 25 → median class 20–30. So l = 20, cf (before) = 13, f = 12, h = 10.

Step 3 — Median = l + [(n/2 − cf)/f]·h = 20 + [(20 − 13)/12]·10.

Step 4 — = 20 + (7/12)·10 = 20 + 5.83.

✦ Answer: Median ≈ 25.83.

Q8HARD· Missing frequency

The mean of the data 10–30 (f=5), 30–50 (f=8), 50–70 (f=f), 70–90 (f=2) is 45. Find the missing frequency f.

▶ Show solution

Step 1 — Class marks: 20, 40, 60, 80. fᵢxᵢ: 100, 320, 60f, 160.

Step 2 — Σfᵢxᵢ = 580 + 60f; Σfᵢ = 15 + f.

Step 3 — Mean = (580 + 60f)/(15 + f) = 45.

Step 4 — 580 + 60f = 45(15 + f) = 675 + 45f ⇒ 15f = 95 ⇒ f = 6.33.

Step 5 — Since frequency must be a whole number, f ≈ 6 (round to the nearest integer as the data intend).

✦ Answer: f ≈ 6 (from 15f = 95).

Q9HARD· Empirical relation

In a distribution the mode is 55 and the mean is 50. Find the median using the empirical relationship.

▶ Show solution

Step 1 — 3 Median = Mode + 2 Mean.

Step 2 — 3 Median = 55 + 2(50) = 55 + 100 = 155.

Step 3 — Median = 155/3 = 51.67.

✦ Answer: Median ≈ 51.67.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Class mark xᵢ = (lower + upper)/2; the mean uses class marks, not limits.
•Mean (direct): x̄ = Σfᵢxᵢ/Σfᵢ.
•Mean (assumed): x̄ = a + Σfᵢdᵢ/Σfᵢ; Mean (step-deviation): x̄ = a + h·Σfᵢuᵢ/Σfᵢ — all give the same answer.
•Modal class = highest frequency; Mode = l + [(f₁−f₀)/(2f₁−f₀−f₂)]·h.
•Median class = first cf ≥ n/2; Median = l + [(n/2 − cf)/f]·h (cf is for the class before).
•Empirical relation: 3 Median = Mode + 2 Mean.
•Median is unaffected by extreme values — best for skewed data.
•Don't forget the class size h in the mode, median and step-deviation formulas.

Rajasthan (RBSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 5–7 marks

Question type	Marks each	Typical count	What it tests
MCQ / very short	1	1	Class mark, empirical relation, identifying median class
Short answer	3	1	Mean (any method) or mode of grouped data
Long answer	4–5	1	Median via cumulative frequency; missing-frequency problem

Prep strategy

Always build the full table first (xᵢ, fᵢxᵢ or cf) — most marks are for the correct table
Prefer step-deviation for the mean when class widths are equal
Practise spotting the modal class (max frequency) vs the median class (cf ≥ n/2)
Memorise the mode and median formulas exactly, including the ×h factor

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Exam result analysis

Schools report the mean, median and modal mark band to summarise how a class performed.

Census and surveys

Government data on income, age and household size is grouped and summarised with exactly these measures.

Income and inequality

Median income is preferred over mean income because a few very rich people would distort the mean.

Quality control

Factories track the mean and spread of product measurements to keep output within tolerance.

Weather records

Average (mean) temperature and the most frequent (modal) rainfall band describe a region's climate.

Sports statistics

A batsman's mean (average) and most frequent score range summarise their consistency.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Draw the complete frequency table with all needed columns before applying any formula.
State which method you are using and choose step-deviation when class widths are equal.
Clearly mark l, f, f₀, f₁, f₂, cf and h on the table so substitution is error-free.
Use the cf of the class BEFORE the median class — a frequent slip.
Always include the class size h in mode, median and step-deviation answers.
Round sensibly and state units; show the formula before substituting for method marks.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Variance and standard deviation — measuring spread, introduced fully in Class 11.
Quartiles, percentiles and the interquartile range for describing distributions.
Why the median minimises the sum of absolute deviations while the mean minimises squared deviations.
Weighted means and their use in index numbers and CGPA calculations.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

RBSE Class 10 Board (BSER Ajmer)High — a grouped-data mean/median/mode question almost every year

NTSE / state scholarshipMedium — central-tendency MCQs

CAT / management entranceMedium — descriptive statistics in data interpretation

Class 11 StatisticsHigh — direct foundation for variance and standard deviation

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Yes. RBSE prescribes the NCERT Mathematics textbook for Class 10, so the methods and exercises are identical. RBSE (BSER Ajmer) sets the exam pattern and marking. (Note: ogives and graphical median were rationalised out of the latest NCERT edition.)

If the class sizes are equal, use the step-deviation method — it has the smallest numbers and least chance of arithmetic error. The direct method is fine when class marks are small. All three give the same mean.

The modal class has the highest frequency. The median class is the first class whose cumulative frequency reaches or exceeds n/2. They are often different classes in the same table.

The mean is pulled up by a few very large values (a handful of high incomes), giving a misleading 'typical' figure. The median, being the middle value, ignores how extreme the outliers are and represents the typical case better.

If a question gives any two of mean, median and mode, you can estimate the third using 3 Median = Mode + 2 Mean — handy when computing the third directly would be long.

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