MathsCBSE Class 10 · Chapter 13

Statistics Class 10: Mean, Median and Mode of Grouped Data

Q: Which method should I use to find the mean if the question does not say?

Any method gets full marks if it is correct and shown properly. Use the direct method for small class marks and the step deviation method when class marks are large and widths are equal. If the question names a method, you must use that one.

Q: Do I need to make the classes continuous to find the mean?

No. The class marks stay the same after conversion, so the mean is unaffected. You must convert to continuous classes before finding the median or mode, because the lower limit is used in those formulas.

Q: Are ogives (cumulative frequency graphs) in the 2026-27 syllabus?

No. Graphical representation of cumulative frequency distribution, including ogives, is not in the 2026-27 syllabus.

Q: Is the relation 3 Median = Mode + 2 Mean always exact?

No, it is an approximate relation that works well for moderately skewed data. In board questions you can use it as exact whenever two averages are given and the third is asked.

Q: What if two classes have the same highest frequency?

That is a bimodal situation, and the syllabus says such cases are to be avoided. Board questions will give a single class with the highest frequency, so you will always have one clear modal class.

Updated 13 October 2026 · Fact-checked

Statistics in Class 10 is about finding the mean, median and mode of grouped data. The board asks you to find the mean by the direct, assumed mean or step deviation method, find the median and mode using their formulas, work out a missing frequency, and use the relation 3 Median = Mode + 2 Mean.

What Statistics is about

This chapter teaches you to find the three averages of grouped data: the mean, the median and the mode. In grouped data you no longer see each observation. You only see classes such as 20–30 and how many observations fall in each. So every method rests on a sensible assumption about how the observations sit inside a class. Once you understand that assumption, the formulas stop looking like magic.

The chapter builds on the mean, median and mode of ungrouped data from Class 9. It needs careful arithmetic with fractions and decimals, and some linear equations when a frequency is missing. Probability also uses frequency tables, so the same table-handling skills carry over.

Board questions come in a few clear shapes. You may be asked to find the mean by a named method, find the mode or median of a table, find one or two missing frequencies when the mean or median is given, convert a cumulative frequency table back into an ordinary one, or use the empirical relation between mean, median and mode. Section E case studies can give real-life data, such as wages, marks or travel times, and ask you to compute and interpret an average.

Weightage: This chapter is part of Unit VII, Statistics and Probability, which carries 11 of the 80 theory marks together with Probability.Source: CBSE Mathematics syllabus 2026-27, Unit VII Statistics and Probability

In real board papers: Standard, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.8 marks per paper; Basic, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6.1 marks per paper. See the questions

Key concepts

Grouped data, class marks and continuous classes

NCERT section 13.2 Mean of Grouped Data

In grouped data, observations are placed in classes like 10–20, with a frequency telling you how many observations fall in each. Since you cannot see the actual values, you represent each class by its class mark (midpoint): $x_{i} = \frac{lower limit + upper limit}{2}$ . This works because the values in a class are assumed to be spread around its centre.

For the median and mode, the classes must be continuous (no gaps). If the classes are written as 1–10, 11–20, and so on, there is a gap of 1. You close it by subtracting half the gap ( $0.5$ ) from every lower limit and adding $0.5$ to every upper limit. The class marks do not change, so for the mean you can skip this step. For the median and mode, the lower limit $l$ must come from the continuous classes, or your answer will be wrong.

Worked example

The marks scored by 45 students in a 50-mark test are grouped as follows: 1–10: 4 students, 11–20: 9 students, 21–30: 15 students, 31–40: 11 students, 41–50: 6 students.
(i) Write the class mark of each class.
(ii) Find the mode of the marks.

Try it first, then show the solutionSolution

Class marks: $\frac{1 + 10}{2} = 5.5$ , and similarly $15.5, 25.5, 35.5, 45.5$ .
The classes have a gap of 1 (10 to 11), so subtract $0.5$ from each lower limit and add $0.5$ to each upper limit.
Continuous classes: 0.5–10.5, 10.5–20.5, 20.5–30.5, 30.5–40.5, 40.5–50.5. The class marks stay the same.
Highest frequency is 15, so the modal class is 20.5–30.5.
$l = 20.5$ , $h = 10$ , $f_{1} = 15$ , $f_{0} = 9$ , $f_{2} = 11$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 20.5 + \frac{15 - 9}{30 - 9 - 11} \times 10$
$= 20.5 + \frac{6}{10} \times 10 = 20.5 + 6 = 26.5$
Check: $26.5$ lies inside the modal class 20.5–30.5.

Answer: (i) Class marks are 5.5, 15.5, 25.5, 35.5 and 45.5. (ii) Mode = 26.5 marks.

Mean by the direct method

NCERT section 13.2 Mean of Grouped Data

The mean of grouped data is $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}}$ , where $x_{i}$ is the class mark and $f_{i}$ the frequency. The idea: assume all $f_{i}$ observations of a class are equal to its class mark. Then the class contributes $f_{i} x_{i}$ to the total, and dividing the grand total by the number of observations gives the mean.

Use the direct method when the class marks and frequencies are small numbers, so the products $f_{i} x_{i}$ are easy. It is also the cleanest method when a frequency is missing and the mean is given: you write $\sum f_{i} x_{i}$ and $\sum f_{i}$ in terms of the unknown, set up an equation, and solve it. Always make a table with columns for class, $f_{i}$ , $x_{i}$ and $f_{i} x_{i}$ ; examiners give marks for this table.

Worked example

The mean of the following frequency distribution is 27. Find the missing frequency $p$ .
Classes: 0–10, 10–20, 20–30, 30–40, 40–50
Frequencies: 5, 8, $p$ , 12, 7

Try it first, then show the solutionSolution

Class marks $x_{i}$ : $5, 15, 25, 35, 45$ .
Products $f_{i} x_{i}$ : $25, 120, 25 p, 420, 315$ .
$\sum f_{i} x_{i} = 880 + 25 p$ and $\sum f_{i} = 32 + p$ .
Mean $= \frac{880 + 25 p}{32 + p} = 27$
$880 + 25 p = 864 + 27 p$
$16 = 2 p$ , so $p = 8$ .
Check: $\sum f_{i} = 40$ , $\sum f_{i} x_{i} = 880 + 200 = 1080$ , and $\frac{1080}{40} = 27$ . Correct.

Answer: $p = 8$

Mean by the assumed mean method

NCERT section 13.2 Mean of Grouped Data

When class marks are large, $f_{i} x_{i}$ becomes heavy to multiply. So you pick an assumed mean $a$ , usually a class mark near the middle of the table, and work with deviations $d_{i} = x_{i} - a$ . Then $\overset{x}{ˉ} = a + \frac{\sum f _{i} d _{i}}{\sum f _{i}}$ .

Why it works: each $x_{i} = a + d_{i}$ , so $\sum f_{i} x_{i} = a \sum f_{i} + \sum f_{i} d_{i}$ . Divide by $\sum f_{i}$ and you get the formula. The deviations are small numbers, some negative, so the arithmetic is lighter. The answer does not depend on which $a$ you choose; choosing a middle class mark just makes the numbers smallest. Watch the signs: classes below $a$ give negative $d_{i}$ , and $\sum f_{i} d_{i}$ can be negative, which makes the mean less than $a$ .

Worked example

The daily wages (in Rs) of 50 workers in a factory are: 100–120: 12 workers, 120–140: 14 workers, 140–160: 8 workers, 160–180: 6 workers, 180–200: 10 workers. Find the mean daily wage using the assumed mean method.

Try it first, then show the solutionSolution

Class marks $x_{i}$ : $110, 130, 150, 170, 190$ . Take $a = 150$ .
Deviations $d_{i} = x_{i} - 150$ : $- 40, - 20, 0, 20, 40$ .
$f_{i} d_{i}$ : $- 480, - 280, 0, 120, 400$ .
$\sum f_{i} d_{i} = - 760 + 520 = - 240$ , $\sum f_{i} = 50$ .
$\overset{x}{ˉ} = a + \frac{\sum f _{i} d _{i}}{\sum f _{i}} = 150 + \frac{- 240}{50}$
$= 150 - 4.8 = 145.2$
Check by direct method: $\sum f_{i} x_{i} = 1320 + 1820 + 1200 + 1020 + 1900 = 7260$ , and $\frac{7260}{50} = 145.2$ .

Answer: Mean daily wage = Rs 145.20

Mean by the step deviation method

NCERT section 13.2 Mean of Grouped Data

If the classes have equal width $h$ , every deviation $d_{i}$ is a multiple of $h$ . So you divide once more: $u_{i} = \frac{x _{i} - a}{h}$ . Now the $u_{i}$ are small whole numbers like $- 2, - 1, 0, 1, 2$ , and the mean is $\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}}$ .

Why it works: $d_{i} = h u_{i}$ , so $\frac{\sum f _{i} d _{i}}{\sum f _{i}} = h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}}$ . You shrank every deviation by $h$ , so you must multiply back by $h$ at the end. Forgetting this $h$ is the most common error here.

Use this method when the question names it, or when class marks are large and widths are equal. All three methods give the same mean, which is a good way to check your answer.

Worked example

The marks obtained by 50 students in an examination are: 0–20: 6, 20–40: 10, 40–60: 16, 60–80: 12, 80–100: 6. Find the mean marks using the step deviation method.

Try it first, then show the solutionSolution

Class marks $x_{i}$ : $10, 30, 50, 70, 90$ . Take $a = 50$ , $h = 20$ .
$u_{i} = \frac{x _{i} - 50}{20}$ : $- 2, - 1, 0, 1, 2$ .
$f_{i} u_{i}$ : $- 12, - 10, 0, 12, 12$ .
$\sum f_{i} u_{i} = 2$ , $\sum f_{i} = 50$ .
$\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}} = 50 + 20 \times \frac{2}{50}$
$= 50 + 0.8 = 50.8$
Check by direct method: $\sum f_{i} x_{i} = 60 + 300 + 800 + 840 + 540 = 2540$ , and $\frac{2540}{50} = 50.8$ .

Answer: Mean = 50.8 marks

Mode of grouped data

NCERT section 13.3 Mode of Grouped Data

The mode is the most frequent value. In grouped data the modal class is the class with the highest frequency, and the mode lies inside it. The formula is $Mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$ , where $l$ is the lower limit of the modal class, $h$ its width, $f_{1}$ its frequency, $f_{0}$ the frequency of the class before it and $f_{2}$ of the class after it.

Why this shape? The mode is pulled towards the neighbour with the bigger frequency. The denominator $2 f_{1} - f_{0} - f_{2} = (f_{1} - f_{0}) + (f_{1} - f_{2})$ splits the class in the ratio of the two drops. If $f_{0} = f_{2}$ , the mode sits exactly at the class mark. Use frequencies, never cumulative frequencies. The syllabus avoids bimodal data, so there will be one clear highest frequency.

Worked example

The ages (in years) of 58 visitors to a museum are: 10–20: 6, 20–30: 10, 30–40: 20, 40–50: 14, 50–60: 8. Find the modal age and comment on where it lies in the modal class.

Try it first, then show the solutionSolution

Highest frequency is 20, so the modal class is 30–40.
$l = 30$ , $h = 10$ , $f_{1} = 20$ , $f_{0} = 10$ , $f_{2} = 14$ .
Mode $= 30 + \frac{20 - 10}{2 ( 20 ) - 10 - 14} \times 10$
$= 30 + \frac{10}{16} \times 10 = 30 + 6.25 = 36.25$
Comment: $f_{2} = 14$ is greater than $f_{0} = 10$ , so the mode is pulled towards the upper class and lies above the class mark $35$ .

Answer: Modal age = 36.25 years, which lies in 30–40 and above its midpoint 35, as expected.

Cumulative frequency and the median class

NCERT section 13.4 Median of Grouped Data

The cumulative frequency (cf) of a class is the total frequency of that class and all classes before it. A "less than" table gives totals below each upper limit; a "more than" table gives totals at or above each lower limit. To get back ordinary frequencies, subtract consecutive cumulative frequencies.

The median is the middle observation, the one at position $\frac{n}{2}$ where $n = \sum f_{i}$ . The median class is the first class whose cumulative frequency is greater than or equal to $\frac{n}{2}$ . That class contains the middle observation. Note that the median class is found from cumulative frequencies, not from the highest frequency, so it is often different from the modal class. Always add a cf column to your table before finding the median; it earns a mark and prevents errors.

Worked example

The time (in minutes) taken by 40 students to solve a puzzle is given as a cumulative frequency table: less than 10: 4, less than 20: 10, less than 30: 26, less than 40: 35, less than 50: 40. Write the frequency distribution, find the median class and then the median time.

Try it first, then show the solutionSolution

Frequencies by subtraction: 0–10: $4$ ; 10–20: $10 - 4 = 6$ ; 20–30: $26 - 10 = 16$ ; 30–40: $35 - 26 = 9$ ; 40–50: $40 - 35 = 5$ .
$n = 40$ , so $\frac{n}{2} = 20$ .
The first cf $\geq 20$ is 26, so the median class is 20–30.
$l = 20$ , $c f = 10$ (class before), $f = 16$ , $h = 10$ .
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 20 + \frac{20 - 10}{16} \times 10$
$= 20 + \frac{100}{16} = 20 + 6.25 = 26.25$

Answer: Frequencies are 4, 6, 16, 9, 5; median class is 20–30; median time = 26.25 minutes.

Median formula and missing frequencies

NCERT section 13.4 Median of Grouped Data

The median formula is $Median = l + \frac{\frac{n}{2} - c f}{f} \times h$ , where $l$ is the lower limit of the median class, $c f$ the cumulative frequency of the class before it, $f$ the frequency of the median class and $h$ its width.

Why it works: up to $l$ you already have $c f$ observations. You need $\frac{n}{2} - c f$ more to reach the middle. Assuming the $f$ observations of the median class are spread evenly across its width, you move that fraction $\frac{\frac{n}{2} - c f}{f}$ of the width $h$ beyond $l$ .

A common 5-mark question type gives the total and the median but hides two frequencies. Use two equations: one from the total frequency, one from the median formula. First decide the median class from the given median value, then write its $c f$ in terms of the unknown.

Worked example

The distances (in km) travelled by 50 delivery riders in a day are: 0–10: 8, 10–20: $p$ , 20–30: 12, 30–40: $q$ , 40–50: 6. The median distance is 25 km. Find $p$ and $q$ .

Try it first, then show the solutionSolution

Total: $8 + p + 12 + q + 6 = 50$ , so $p + q = 24$ .
The median 25 lies in 20–30, so the median class is 20–30.
$l = 20$ , $h = 10$ , $f = 12$ , $c f = 8 + p$ , $\frac{n}{2} = 25$ .
$25 = 20 + \frac{25 - ( 8 + p )}{12} \times 10$
$5 = \frac{( 17 - p ) \times 10}{12}$ , so $17 - p = 6$ .
$p = 11$ , and $q = 24 - 11 = 13$ .
Check: cf values are $8, 19, 31, \dots$ ; $25$ falls in 20–30 and Median $= 20 + \frac{25 - 19}{12} \times 10 = 25$ .

Answer: $p = 11$ and $q = 13$

Empirical relation between mean, median and mode

NCERT section 13.4 Median of Grouped Data

For distributions that are only moderately lopsided (skewed), the three averages are linked approximately by $3 Median = Mode + 2 Mean$ . This is an observed (empirical) result, not a theorem, but board questions accept it as exact.

You can rearrange it to find whichever value is missing: $Mode = 3 Median - 2 Mean$ , or $Mean = \frac{3 Median - Mode}{2}$ . A neat consequence is that the median always lies between the mean and the mode, about one-third of the way from the mean. Use the relation when two averages are given and the third is asked, or to check whether three computed values are consistent. If your three values are wildly off this relation, recheck your work.

Worked example

Assertion (A): If the mean and mode of a distribution are 45.5 and 47 respectively, then its median is 46.
Reason (R): For a moderately skewed distribution, Mode = 3 Median − 2 Mean.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

R is the empirical relation $3 Median = Mode + 2 Mean$ rearranged, so R is true.
Using it: $3 Median = 47 + 2 (45.5) = 47 + 91 = 138$ .
Median $= \frac{138}{3} = 46$ , so A is true.
A was obtained directly from R, so R explains A.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Maths in real life: Statistics

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Mean by the direct method

Suppose your school's opening batter keeps a record of runs scored in each innings this season, grouped into classes of 20 runs. The scorer has lost the count for one class, but the coach remembers that the batter's mean score was exactly 50 runs per innings.

The runs per innings are: 0–20: 3 innings, 20–40: $p$ innings, 40–60: 5 innings, 60–80: 4 innings, 80–100: 2 innings. The mean is 50 runs. Using the direct method, find $p$ and the total number of innings played.

Try it first, then show the solutionSolution

Class marks: $x_{i} = 10, 30, 50, 70, 90$ .
Products $f_{i} x_{i}$ : $3 \times 10 = 30$ , $30 p$ , $5 \times 50 = 250$ , $4 \times 70 = 280$ , $2 \times 90 = 180$ .
$\sum f_{i} x_{i} = 30 + 30 p + 250 + 280 + 180 = 740 + 30 p$ .
$\sum f_{i} = 3 + p + 5 + 4 + 2 = 14 + p$ .
Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}}$ , so $\frac{740 + 30 p}{14 + p} = 50$ .
$740 + 30 p = 700 + 50 p$ , so $20 p = 40$ and $p = 2$ .
Total innings $= 14 + 2 = 16$ .
Check: $\sum f_{i} x_{i} = 740 + 60 = 800$ and $\frac{800}{16} = 50$ . Correct.

Answer: $p = 2$ , and the batter played 16 innings.

Memory hook: Missing innings, known average: write totals with $p$ , set mean equal, solve.

Mean by the assumed mean method

Imagine your housing society's secretary wants the average monthly electricity bill of 50 flats before deciding on a solar panel plan. The bills run into hundreds of rupees, so the class marks are large and multiplying them directly is tiring. Suppose the solar plan would cost the society Rs 15,000 a month in instalments and is expected to cut every flat's bill by 40%.

Monthly bills (in Rs): 500–700: 6 flats, 700–900: 10 flats, 900–1100: 14 flats, 1100–1300: 12 flats, 1300–1500: 8 flats. Using the assumed mean method, find the mean bill and the total amount billed to all 50 flats. Then find the total monthly saving from the solar plan and decide whether the plan saves the society money after paying the instalment. If it does, by how much each month?

Try it first, then show the solutionSolution

Class marks: $x_{i} = 600, 800, 1000, 1200, 1400$ .
Take assumed mean $a = 1000$ , the middle class mark.
Deviations $d_{i} = x_{i} - 1000$ : $- 400, - 200, 0, 200, 400$ .
Products $f_{i} d_{i}$ : $- 2400, - 2000, 0, 2400, 3200$ .
$\sum f_{i} d_{i} = - 4400 + 5600 = 1200$ , and $\sum f_{i} = 50$ .
Mean $= a + \frac{\sum f _{i} d _{i}}{\sum f _{i}} = 1000 + \frac{1200}{50} = 1000 + 24 = 1024$ , so the mean bill is Rs 1024.
Total billed $= 50 \times 1024 = 51200$ , so Rs 51,200 a month.
Saving at 40% $= \frac{40}{100} \times 51200 = 20480$ , so Rs 20,480 a month.
Compare with the instalment: $20480 - 15000 = 5480$ , which is positive, so the plan saves money.
Check by direct method: $\sum f_{i} x_{i} = 3600 + 8000 + 14000 + 14400 + 11200 = 51200$ , and $\frac{51200}{50} = 1024$ . Correct.

Answer: Mean bill is Rs 1024 and the total billed is Rs 51,200 a month. The saving of Rs 20,480 is more than the Rs 15,000 instalment, so the plan is worth it, with a net saving of Rs 5,480 a month.

Memory hook: Big bills, small sums: measure everything from a middle bill of Rs 1000, then let the total decide.

Mean by the step deviation method

Suppose your class teacher surveys 50 students on how many minutes a day they spend on their phones outside study time. She wants to know whether the class average is within a suggested limit of 60 minutes a day.

Screen time (minutes per day): 0–30: 8 students, 30–60: 12, 60–90: 20, 90–120: 6, 120–150: 4. Using the step deviation method, find the mean screen time and state by how many minutes it is above or below the 60-minute limit.

Try it first, then show the solutionSolution

Class marks: $x_{i} = 15, 45, 75, 105, 135$ ; class width $h = 30$ .
Take $a = 75$ . Then $u_{i} = \frac{x _{i} - 75}{30} = - 2, - 1, 0, 1, 2$ .
Products $f_{i} u_{i}$ : $- 16, - 12, 0, 6, 8$ .
$\sum f_{i} u_{i} = - 28 + 14 = - 14$ , and $\sum f_{i} = 50$ .
Mean $= a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}} = 75 + 30 \times \frac{- 14}{50}$ .
$= 75 - \frac{420}{50} = 75 - 8.4 = 66.6$ .
$66.6 - 60 = 6.6$ , so the mean is above the limit.
Check by direct method: $\sum f_{i} x_{i} = 120 + 540 + 1500 + 630 + 540 = 3330$ , and $\frac{3330}{50} = 66.6$ . Correct.

Answer: Mean screen time is 66.6 minutes per day, which is 6.6 minutes above the 60-minute limit.

Memory hook: Shrink by $h$ to count steps, then multiply back by $h$ , like converting half-hours back to minutes.

Mode of grouped data

Imagine a stall selling water purifiers at a Diwali mela notes the ages (to the nearest year) of the visitors who stop to ask questions. The owner remembers that the modal age worked out to 38.5 years, but the count for the 45–54 age group has been smudged. He wants to know what share of visitors are older than the modal age group, so the next poster speaks to the right people.

Ages (years): 5–14: 14 visitors, 15–24: 10, 25–34: 16, 35–44: 24, 45–54: $x$ , 55–64: 4. The modal age is 38.5 years. Find $x$ . Then find the percentage of visitors aged 45 years or more, all of whom are older than the modal age.

Try it first, then show the solutionSolution

Ages are recorded to the nearest year, so there is a gap of 1 between classes. Subtract 0.5 from each lower limit and add 0.5 to each upper limit: 4.5–14.5, 14.5–24.5, 24.5–34.5, 34.5–44.5, 44.5–54.5, 54.5–64.5.
The mode 38.5 lies in 34.5–44.5, so this is the modal class.
$l = 34.5$ , $h = 10$ , $f_{1} = 24$ , $f_{0} = 16$ (class before), $f_{2} = x$ (class after).
$38.5 = 34.5 + \frac{24 - 16}{48 - 16 - x} \times 10$
$4 = \frac{80}{32 - x}$ , so $32 - x = 20$ and $x = 12$ .
Check: $34.5 + \frac{8}{20} \times 10 = 34.5 + 4 = 38.5$ , and $12 < 24$ , so 34.5–44.5 is still the modal class.
Total visitors $= 14 + 10 + 16 + 24 + 12 + 4 = 80$ .
Visitors aged 45 or more (classes 45–54 and 55–64) $= 12 + 4 = 16$ .
Percentage $= \frac{16}{80} \times 100 = 20%$ .

Answer: $x = 12$ ; there were 80 visitors, and 20% of them were aged 45 years or more.

Memory hook: Close the age gaps first, then the busiest age group at the stall holds the mode.

Cumulative frequency and the median class

Suppose your colony's water committee records the daily water use of 40 households to size a new overhead tank. Their register shows only running totals: how many homes use less than 150 litres, less than 200 litres, and so on. They want the median use, so that half the homes use less and half use more, and they plan to give a water-saving tip card to every home using more than 250 litres.

Every household uses at least 100 litres a day. Households using less than 150 litres: 5; less than 200: 14; less than 250: 26; less than 300: 34; less than 350: 40. (i) Write the frequency distribution. (ii) Find the median daily water use. (iii) About how many households use more than the median amount, and how many tip cards are needed?

Try it first, then show the solutionSolution

Frequencies by subtraction: 100–150: $5$ ; 150–200: $14 - 5 = 9$ ; 200–250: $26 - 14 = 12$ ; 250–300: $34 - 26 = 8$ ; 300–350: $40 - 34 = 6$ .
$n = 40$ , so $\frac{n}{2} = 20$ .
The first cumulative frequency that is at least 20 is 26, so the median class is 200–250.
$l = 200$ , $c f = 14$ (class before), $f = 12$ , $h = 50$ .
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 200 + \frac{20 - 14}{12} \times 50$ .
$= 200 + \frac{6}{12} \times 50 = 200 + 25 = 225$ .
Check: 225 lies inside 200–250, the median class. Correct.
The median splits the households into two equal halves, so about $\frac{40}{2} = 20$ households use more than 225 litres.
Households using more than 250 litres $= 40 - 26 = 14$ , so 14 tip cards are needed.

Answer: Frequencies are 5, 9, 12, 8, 6; the median daily water use is 225 litres; about 20 households use more than the median, and 14 tip cards are needed.

Memory hook: Keep adding households till you reach the 20th; that class holds the median tank need.

Empirical relation between mean, median and mode

Imagine your maths teacher shares the grouped marks (out of 100) of 40 students in a class test. You want to know the most common mark, and you also want to see whether the empirical relation between the three averages gives a sensible answer for real class data.

Marks: 30–40: 4 students, 40–50: 8, 50–60: 16, 60–70: 10, 70–80: 2. Find the mean mark by the step deviation method and the median mark. Then use the empirical relation to estimate the modal mark, and check that it lies in the modal class.

Try it first, then show the solutionSolution

Class marks: $x_{i} = 35, 45, 55, 65, 75$ . Take $a = 55$ , $h = 10$ , so $u_{i} = - 2, - 1, 0, 1, 2$ .
Products $f_{i} u_{i}$ : $- 8, - 8, 0, 10, 4$ , so $\sum f_{i} u_{i} = - 2$ and $\sum f_{i} = 40$ .
Mean $= 55 + 10 \times \frac{- 2}{40} = 55 - 0.5 = 54.5$ .
Cumulative frequencies: 4, 12, 28, 38, 40. Since $\frac{n}{2} = 20$ , the median class is 50–60.
Median $= 50 + \frac{20 - 12}{16} \times 10 = 50 + 5 = 55$ .
Empirical relation: Mode $= 3 Median - 2 Mean = 165 - 109 = 56$ .
Check: the highest frequency is 16, so the modal class is 50–60, and 56 lies inside it. The mode formula gives $50 + \frac{16 - 8}{32 - 8 - 10} \times 10 = 50 + \frac{40}{7} \approx 55.7$ , very close to 56.

Answer: Mean = 54.5 marks, median = 55 marks, and the estimated mode = 56 marks, which lies in the modal class 50–60.

Memory hook: Three times the middle equals mode plus twice the mean: median always sits between the other two.

Topics in this chapter

Statistics formula sheet

Class mark: $x_{i} = \frac{lower limit + upper limit}{2}$; Represents each class in all mean calculations. It does not change when you convert inclusive classes to continuous ones.
Mean: direct method: $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}}$; Best for small class marks and for finding a missing frequency when the mean is given.
Mean: assumed mean method: $\overset{x}{ˉ} = a + \frac{\sum f _{i} d _{i}}{\sum f _{i}}, d_{i} = x_{i} - a$; Use when class marks are large. Take a as a middle class mark.
Mean: step deviation method: $\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}}, u_{i} = \frac{x _{i} - a}{h}$; Use when classes have equal width h. Never forget to multiply by h.
Mode of grouped data: $Mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$; l, h, f1 belong to the modal class (highest frequency); f0 is the class before, f2 the class after. Uses frequencies only.
Median of grouped data: $Median = l + \frac{\frac{n}{2} - c f}{f} \times h$; Median class is the first class with cumulative frequency at least n/2. cf is the cumulative frequency of the class before the median class.
Median position: $Middle position = \frac{n}{2}, n = \sum f_{i}$; For grouped data use n/2 (not (n+1)/2) to locate the median class.
Empirical relation: $3 Median = Mode + 2 Mean$; Approximate for moderately skewed data; treat as exact in board questions when two averages are given.
Inclusive to continuous classes: $adjustment = \frac{gap between classes}{2}$; Subtract it from every lower limit and add it to every upper limit before finding the median or mode.
Obtaining frequency from cumulative frequency: $f_{i} = c f_{i} - c f_{i - 1}$; Use when the data is given as a less than or more than cumulative frequency table.
Location checks: $Mode \in modal class, Median \in median class$; A quick check: if your answer lies outside its class, you used a wrong value.

How the board asks Statistics

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Find the mean by a stated method	A frequency table is given and you are asked for the mean, sometimes naming the direct, assumed mean or step deviation method. Often set in a context such as wages, marks or rainfall.	Make a full table with class marks and the required column ( $f_{i} x_{i}$ , $f_{i} d_{i}$ or $f_{i} u_{i}$ ). Use the method asked. Write the formula before substituting, and give units in the final answer.
Find a missing frequency when the mean is given	One frequency is written as $p$ or $k$ and the mean is stated. Sometimes the total frequency is also given and two frequencies are missing.	Write $\sum f_{i}$ and $\sum f_{i} x_{i}$ (or $\sum f_{i} u_{i}$ ) in terms of the unknown, equate the mean formula to the given mean and solve the linear equation. Verify by substitution.
Find the mode, or a frequency from the mode	Find the mode of a table, or find a missing frequency of the modal class or its neighbour when the mode is given.	Identify the modal class, write $l, h, f_{1}, f_{0}, f_{2}$ clearly, then substitute. If the mode is given, the modal class is the one containing it; substitute and solve for the unknown.
Find the median from a frequency or cumulative frequency table	A frequency table, or a less than / more than cumulative table, is given and you must find the median.	Convert a cumulative table to frequencies by subtraction, add a cf column, find $\frac{n}{2}$ , pick the first class with cf at least $\frac{n}{2}$ , and apply the median formula.
Two missing frequencies from the median and total	A long-answer question gives the total frequency and the median and hides two frequencies.	Get one equation from the total. Locate the median class from the given median, write its cf in terms of the unknown, and use the median formula for the second equation.
Mean, median and mode of one table	A single distribution where you must find two or all three averages, or find two and use the empirical relation for the third.	Build one table with $x_{i}$ , $f_{i} x_{i}$ (or $u_{i}$ ) and cf columns so every average comes from the same work. Check consistency with $3 Median = Mode + 2 Mean$ .
Empirical relation MCQ or assertion-reason	Two of mean, median and mode are given; find the third, or judge a statement about them.	Write $3 Median = Mode + 2 Mean$ first, substitute, and solve. Do not reorder the coefficients.
Case study with real-life data	A Section E passage describes data such as delivery times or electricity use, with sub-parts asking for the modal class, median class, an average, or an interpretation.	Read which average fits the question: mean for a typical total share, median for the middle value, mode for the most common value. Answer the 1-mark parts quickly and show full working in the 2-mark part.

Tricks to remember Statistics

Short memory aids. Say each one aloud once, then use it on a question the same day.

Mode: own minus before, over two drops
Mode $= l + \frac{drop on the left}{drop on the left + drop on the right} \times h$ , where the drops are $f_{1} - f_{0}$ and $f_{1} - f_{2}$ .
Use it when: Rebuild the mode formula without memorising $2 f_{1} - f_{0} - f_{2}$ blindly, and you will never swap $f_{0}$ and $f_{2}$ .
Median: half minus before, over own
Median $= l + \frac{half - cf before}{own f} \times h$ .
Use it when: Reminds you that cf is of the class before, and f is the median class's own frequency.
Three medians for the middle
The median sits in the middle, so it gets the biggest coefficient: $3 Median = Mode + 2 Mean$ .
Use it when: Stops you from writing 3 Mean or 3 Mode in the empirical relation.
Highest f for mode, cf for median
Mode class: look down the f column for the biggest number. Median class: look down the cf column for the first number at least $\frac{n}{2}$ .
Use it when: Choosing the right class first, before any formula.
Inside-the-class check
Mode must lie inside the modal class and the median inside the median class; the mean must lie between the smallest and largest class limits.
Use it when: A 10-second check at the end of every answer that catches wrong $l$ , wrong cf or a missing $h$ .

Statistics: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Modal class and median class together · 1 mark · MCQBoard level

For the distribution below, what is the sum of the lower limits of the modal class and the median class?
Classes: 0–5, 5–10, 10–15, 15–20, 20–25
Frequencies: 10, 15, 12, 20, 9
(a) 15
(b) 20
(c) 25
(d) 30

Try it first, then show the solutionSolution

The highest frequency is 20, so the modal class is 15–20. Its lower limit is 15.
$n = 66$ , so $\frac{n}{2} = 33$ .
Cumulative frequencies: 10, 25, 37, ... The first value that is at least 33 is 37, so the median class is 10–15. Its lower limit is 10.
The sum is $15 + 10 = 25$ .

Answer: (c) 25

Example 2: Mean of a symmetric distribution · 1 mark · Assertion-reasonBoard level

Assertion (A): The mean of the distribution with classes 0–10, 10–20, 20–30 and frequencies 3, 4, 3 is 15.
Reason (R): If all the classes have equal width and the frequencies are symmetric about the middle class, the mean equals the class mark of the middle class.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

Check A: $\sum f_{i} x_{i} = 3 (5) + 4 (15) + 3 (25) = 150$ and $\sum f_{i} = 10$ , so the mean is $\frac{150}{10} = 15$ . A is true.
Check R: with equal class widths the class marks are equally spaced, so with symmetric frequencies every deviation $d_{i}$ on one side of the middle class mark is cancelled by an equal and opposite one on the other side. So $\sum f_{i} d_{i} = 0$ and the mean equals the middle class mark. R is true.
The frequencies 3, 4, 3 are symmetric about 10–20, whose class mark is 15. So R explains A.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Example 3: Missing frequency from the mode · 3 marks · Short answerBoard level

The mode of the following distribution is 25. Find the missing frequency $x$ , given that the modal class is 20–30.
Classes: 0–10, 10–20, 20–30, 30–40, 40–50
Frequencies: 5, $x$ , 20, 15, 8

Try it first, then show the solutionSolution

The modal class is 20–30, so $l = 20$ , $h = 10$ , $f_{1} = 20$ , $f_{0} = x$ , $f_{2} = 15$ . (1 mark)
Mode $= 20 + \frac{20 - x}{40 - x - 15} \times 10 = 25$ , so $\frac{20 - x}{25 - x} = \frac{1}{2}$ . (1 mark)
Cross-multiply: $40 - 2 x = 25 - x$ , so $x = 15$ . (½ mark)
Check: $20 + \frac{5}{10} \times 10 = 25$ . The value 20 is still the highest frequency, so the modal class is correct. (½ mark)

Answer: $x = 15$

Example 4: Assumed mean with unequal class widths · 3 marks · Short answerBoard level

The time (in minutes) taken by 40 employees of an office to commute is given below. Find the mean commuting time using the assumed mean method.
Time (min): 0–10, 10–30, 30–40, 40–60, 60–70
Number of employees: 6, 10, 12, 8, 4

Try it first, then show the solutionSolution

The class widths are not all equal, so use the class mark of each class. Class marks: $5, 20, 35, 50, 65$ . (½ mark)
Take $a = 35$ . Then $d_{i} = x_{i} - 35 = - 30, - 15, 0, 15, 30$ . (½ mark)
$f_{i} d_{i} = - 180, - 150, 0, 120, 120$ , so $\sum f_{i} d_{i} = - 90$ and $\sum f_{i} = 40$ . (1 mark)
Mean $= 35 + \frac{- 90}{40} = 35 - 2.25 = 32.75$ minutes. (1 mark)
Check by the direct method: $\sum f_{i} x_{i} = 30 + 200 + 420 + 400 + 260 = 1310$ , and $\frac{1310}{40} = 32.75$ .

Answer: The mean commuting time is 32.75 minutes.

Example 5: Mean, median and mode of one table · 5 marks · Long answerHarder board level

The ages (in years) of 50 patients at a health camp are given below. Find the mean, median and mode of the data. Then check whether they satisfy the empirical relation $3 Median = Mode + 2 Mean$ approximately.
Age (years): 10–20, 20–30, 30–40, 40–50, 50–60, 60–70
Number of patients: 4, 8, 10, 12, 10, 6

Try it first, then show the solutionSolution

Mean: class marks are $15, 25, 35, 45, 55, 65$ . Take $a = 45$ and $h = 10$ , so $u_{i} = - 3, - 2, - 1, 0, 1, 2$ . (½ mark)
$f_{i} u_{i} = - 12, - 16, - 10, 0, 10, 12$ , so $\sum f_{i} u_{i} = - 16$ and $\sum f_{i} = 50$ . (½ mark)
Mean $= 45 + 10 \times \frac{- 16}{50} = 45 - 3.2 = 41.8$ years. (1 mark)
Median: cumulative frequencies are 4, 12, 22, 34, 44, 50. Since $\frac{n}{2} = 25$ , the median class is 40–50. (½ mark)
$l = 40$ , $c f = 22$ , $f = 12$ , $h = 10$ . Median $= 40 + \frac{25 - 22}{12} \times 10 = 40 + 2.5 = 42.5$ years. (1 mark)
Mode: the modal class is 40–50, with $f_{1} = 12$ , $f_{0} = 10$ , $f_{2} = 10$ . Mode $= 40 + \frac{2}{24 - 20} \times 10 = 45$ years. (1 mark)
Check: $3 \times 42.5 = 127.5$ and $45 + 2 \times 41.8 = 128.6$ . These are close, so the relation holds approximately. (½ mark)

Answer: Mean = 41.8 years, median = 42.5 years, mode = 45 years. The empirical relation holds approximately (127.5 compared with 128.6).

Example 6: Two missing frequencies from the mean, then the median · 5 marks · Long answerHarder board level

The weekly hours spent by 50 students on a hobby are given below. The mean is 24.6 hours.
Hours: 0–10, 10–20, 20–30, 30–40, 40–50
Number of students: 8, $p$ , 14, $q$ , 6
(i) Find $p$ and $q$ .
(ii) Find the median number of hours.

Try it first, then show the solutionSolution

Total: $8 + p + 14 + q + 6 = 50$ , so $p + q = 22$ . (½ mark)
Class marks are $5, 15, 25, 35, 45$ . Take $a = 25$ and $h = 10$ , so $u_{i} = - 2, - 1, 0, 1, 2$ . (½ mark)
$\sum f_{i} u_{i} = - 16 - p + 0 + q + 12 = q - p - 4$ (½ mark)
Mean: $25 + 10 \times \frac{q - p - 4}{50} = 24.6$ , so $\frac{q - p - 4}{5} = - 0.4$ and $q - p = 2$ . (1 mark)
Solving $p + q = 22$ and $q - p = 2$ gives $q = 12$ and $p = 10$ . (½ mark)
Median: frequencies are 8, 10, 14, 12, 6, so cumulative frequencies are 8, 18, 32, 44, 50. Since $\frac{n}{2} = 25$ , the median class is 20–30. (1 mark)
Median $= 20 + \frac{25 - 18}{14} \times 10 = 20 + 5 = 25$ hours. (1 mark)

Answer: (i) $p = 10$ , $q = 12$ . (ii) The median is 25 hours.

Example 7: Find the error: median with inclusive classes · 3 marks · Short answerBoard level

The heights (in cm) of 50 girls are given below.
Height (cm): 140–144, 145–149, 150–154, 155–159, 160–164
Number of girls: 5, 10, 20, 12, 3
A student wrote: "Median class is 150–154, so $l = 150$ , $h = 5$ , $c f = 15$ , $f = 20$ . Median $= 150 + \frac{25 - 15}{20} \times 5 = 152.5$ cm."
Identify the error and find the correct median.

Try it first, then show the solutionSolution

Error: the classes are inclusive, with a gap of 1 between 144 and 145. The median formula needs continuous classes, so $l$ must be the true lower boundary. The student used 150. (1 mark)
Adjust by $\frac{1}{2} = 0.5$ : the median class becomes 149.5–154.5, so $l = 149.5$ and $h = 5$ . The cumulative frequencies 5, 15, 35 show that $c f = 15$ and $f = 20$ . (1 mark)
Median $= 149.5 + \frac{25 - 15}{20} \times 5 = 149.5 + 2.5 = 152$ cm. (1 mark)

Answer: The student did not convert to class boundaries. The correct median is 152 cm, not 152.5 cm.

Example 8: Median and mode from a 'more than' table · 5 marks · Long answerHOTS / competency

A survey of 50 households recorded their monthly electricity bills (in Rs) as a 'more than or equal to' cumulative table:
Bill ≥ 0: 50 households; ≥ 100: 47; ≥ 200: 35; ≥ 300: 15; ≥ 400: 7; the highest bill is below 500.
(i) Write the frequency distribution.
(ii) Find the median bill and the modal bill.
(iii) Which of the two would you quote as the 'typical' bill, and why?

Try it first, then show the solutionSolution

Each frequency is the difference of consecutive cumulative values: 0–100: $50 - 47 = 3$ ; 100–200: $47 - 35 = 12$ ; 200–300: $35 - 15 = 20$ ; 300–400: $15 - 7 = 8$ ; 400–500: 7. (1 mark)
The 'less than' cumulative frequencies are 3, 15, 35, 43, 50. Since $\frac{n}{2} = 25$ , the median class is 200–300. (½ mark)
Median $= 200 + \frac{25 - 15}{20} \times 100 = 200 + 50$ , so the median bill is Rs 250. (1½ marks)
The modal class is 200–300, with $f_{1} = 20$ , $f_{0} = 12$ , $f_{2} = 8$ . (½ mark)
Mode $= 200 + \frac{20 - 12}{40 - 12 - 8} \times 100 = 200 + 40$ , so the modal bill is Rs 240. (1 mark)
The mode, Rs 240, is the bill paid most often, so it is the 'typical' bill. The median, Rs 250, tells you that half the households pay less than this. (½ mark)

Answer: Frequencies: 3, 12, 20, 8, 7. Median = Rs 250 and mode = Rs 240. The mode (Rs 240) is the most typical bill. The median (Rs 250) splits the households into two equal halves.

Example 9: Both missing frequencies from the mean and the median · 5 marks · Long answerHarder board level

The mean of the distribution below is 24.2 and its median is 24. The total frequency is not given. Find $p$ and $q$ , and confirm that your median class is correct.
Classes: 0–10, 10–20, 20–30, 30–40, 40–50
Frequencies: 5, $p$ , 20, $q$ , 5

Try it first, then show the solutionSolution

Let the total be $n = 30 + p + q$ . Take $a = 25$ and $h = 10$ , so $u_{i} = - 2, - 1, 0, 1, 2$ and $\sum f_{i} u_{i} = - 10 - p + 0 + q + 10 = q - p$ . (1 mark)
The median 24 lies in 20–30, so take 20–30 as the median class: $l = 20$ , $f = 20$ , $h = 10$ , $c f = 5 + p$ and $\frac{n}{2} = \frac{30 + p + q}{2}$ . (½ mark)
Median: $24 = 20 + \frac{\frac{30 + p + q}{2} - ( 5 + p )}{20} \times 10$ , so $8 = \frac{30 + p + q}{2} - 5 - p$ . (1 mark)
Multiply by 2: $16 = 30 + p + q - 10 - 2 p = 20 + q - p$ , so $p - q = 4$ . (½ mark)
Mean: $25 + 10 \times \frac{q - p}{n} = 24.2$ . Since $q - p = - 4$ , $\frac{- 40}{n} = - 0.8$ , so $n = 50$ and $p + q = 20$ . (1 mark)
Solving $p + q = 20$ and $p - q = 4$ gives $p = 12$ and $q = 8$ . (½ mark)
Test the median class: frequencies 5, 12, 20, 8, 5 give cumulative frequencies 5, 17, 37, 45, 50. Since $\frac{n}{2} = 25$ and $17 < 25 \leq 37$ , the class 20–30 is indeed the median class. Mean check: $\sum f_{i} x_{i} = 25 + 180 + 500 + 280 + 225 = 1210$ and $\frac{1210}{50} = 24.2$ . (½ mark)

Answer: $p = 12$ and $q = 8$ (total frequency 50); the median class 20–30 is confirmed.

Example 10: Comparing two sections: mean, median and grace marks · 4 marks · Case-basedHOTS / competency

Suppose a teacher compares the marks (out of 50) of two sections in a unit test. Each section has 40 students.
Marks: 0–10, 10–20, 20–30, 30–40, 40–50
Section A: 2, 8, 20, 6, 4
Section B: 6, 10, 10, 8, 6
(i) Write the median class of Section A. (1 mark)
(ii) Which section has more students scoring 30 marks or more, and by how many? (1 mark)
(iii) Find the mean and the median of each section, and say which section did better on average and which has the higher middle score. (2 marks)
OR
(iii) Every student of Section A is given 2 bonus marks, and for this bonus no mark is capped, so a total above 50 is allowed. Without making a new table, find the new mean and median of Section A and explain your reasoning. (2 marks)

Try it first, then show the solutionSolution

(i) Section A: cumulative frequencies are 2, 10, 30, 36, 40. Since $\frac{n}{2} = 20$ , the first cf that is at least 20 is 30, so the median class is 20–30. (1 mark)
(ii) Scoring 30 or more: Section A has $6 + 4 = 10$ students and Section B has $8 + 6 = 14$ . So Section B has 4 more. (1 mark)
(iii) Section A: with $a = 25$ , $h = 10$ , $\sum f_{i} u_{i} = - 4 - 8 + 0 + 6 + 8 = 2$ , so Mean $= 25 + 10 \times \frac{2}{40} = 25.5$ . Median $= 20 + \frac{20 - 10}{20} \times 10 = 25$ . (1 mark)
Section B: $\sum f_{i} u_{i} = - 12 - 10 + 0 + 8 + 12 = - 2$ , so Mean $= 25 - 0.5 = 24.5$ . Cumulative frequencies 6, 16, 26, 34, 40 give median class 20–30 and Median $= 20 + \frac{20 - 16}{10} \times 10 = 24$ . (½ mark)
Inference: Section A is ahead on both the mean (25.5 against 24.5) and the median (25 against 24). Section B has more high scorers but also more low scorers, so its marks are more spread out. (½ mark)
OR (iii): Since no mark is capped, adding 2 marks to every student moves every observation, every class limit and every class mark up by 2. In the mean, each $x_{i}$ becomes $x_{i} + 2$ , so the mean rises by 2 to 27.5. In the median formula only $l$ changes, from 20 to 22, while $c f$ , $f$ and $h$ stay the same, so the median rises by 2 to 27. (2 marks)

Answer: (i) 20–30. (ii) Section B, by 4 students (14 against 10). (iii) Section A: mean 25.5 marks, median 25 marks; Section B: mean 24.5 marks, median 24 marks. Section A did better on average and in the middle. OR (iii): new mean 27.5 marks and new median 27 marks.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Household waste survey in a ward

Suppose that, as part of a ward cleanliness drive, a ward officer surveyed 40 households. She recorded the waste (in kg) that each household produced in a day. The data are given below.
Waste (kg): 0–2, 2–4, 4–6, 6–8, 8–10
Number of households: 6, 10, 14, 7, 3
The officer wants to know the most common amount of waste, the middle value and the average. The ward has 800 households like these, and each collection truck can carry 500 kg of waste per trip. This will help her plan collection trucks and compost bins for the ward.

(i) · 1 mark
Write the modal class of the data.
Answer
Answer: 4–6 kg
The highest frequency is 14, which belongs to the class 4–6. So the modal class is 4–6 kg.
(ii) · 1 mark
A councillor claims that half the households produce less than 4 kg of waste a day. Use the median class to say whether the claim is correct.
Answer
Answer: The claim is not correct; only 16 of the 40 households produce less than 4 kg.
$n = 40$ , so $\frac{n}{2} = 20$ . The cumulative frequencies are 6, 16, 30, ... The first value that is at least 20 is 30, so the median class is 4–6 kg and the median is more than 4 kg. Only 16 households, fewer than half, produce less than 4 kg, so the claim is wrong.
(iii) · 2 marks
Find the mean daily waste per household using the step deviation method. Using this mean, how many truck trips are needed each day to clear the waste of all 800 households?
Answer
Answer: Mean = 4.55 kg; 8 truck trips a day
Class marks are $1, 3, 5, 7, 9$ . Take $a = 5$ and $h = 2$ , so $u_{i} = - 2, - 1, 0, 1, 2$ and $\sum f_{i} u_{i} = - 12 - 10 + 0 + 7 + 6 = - 9$ . (½ mark)
Mean $= 5 + 2 \times \frac{- 9}{40} = 5 - 0.45 = 4.55$ kg (½ mark)
Total daily waste of the ward $= 800 \times 4.55 = 3640$ kg (½ mark)
Trips $= \frac{3640}{500} = 7.28$ . Seven trips would leave waste behind, so 8 trips are needed each day. (½ mark)

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the mode of the following data. Classes: 0–20, 20–40, 40–60, 60–80, 80–100; frequencies: 5, 8, 15, 12, 6. (2 marks)

Step	Marks
Modal class is 40–60, since its frequency 15 is the highest	0.5
Writing $l = 40$ , $h = 20$ , $f_{1} = 15$ , $f_{0} = 8$ , $f_{2} = 12$	0.5
Substitution: Mode $= 40 + \frac{15 - 8}{30 - 8 - 12} \times 20 = 40 + \frac{7}{10} \times 20$	0.5
Mode $= 40 + 14 = 54$	0.5

Find the mean of the following distribution using the step deviation method. Classes: 25–35, 35–45, 45–55, 55–65, 65–75; frequencies: 6, 10, 14, 12, 8. (3 marks)

Step	Marks
Class marks $30, 40, 50, 60, 70$ with $a = 50$ , $h = 10$	0.5
$u_{i} = - 2, - 1, 0, 1, 2$ and $f_{i} u_{i} = - 12, - 10, 0, 12, 16$	0.5
$\sum f_{i} = 50$ and $\sum f_{i} u_{i} = 6$	1
Substitution: $\overset{x}{ˉ} = 50 + 10 \times \frac{6}{50}$	0.5
$\overset{x}{ˉ} = 50 + 1.2 = 51.2$	0.5

The median of the following distribution is 50 and the total frequency is 50. Find $x$ and $y$ . Classes: 0–20, 20–40, 40–60, 60–80, 80–100; frequencies: 6, $x$ , 16, $y$ , 4. (5 marks)

Step	Marks
Cumulative frequency column: $6, 6 + x, 22 + x, 22 + x + y, 26 + x + y$	1
From total: $26 + x + y = 50$ , so $x + y = 24$	1
Median 50 lies in 40–60, so $l = 40$ , $h = 20$ , $f = 16$ , $c f = 6 + x$ , $\frac{n}{2} = 25$	1
Substitution: $50 = 40 + \frac{25 - ( 6 + x )}{16} \times 20$ , giving $\frac{( 19 - x ) \times 20}{16} = 10$	1
$19 - x = 8$ , so $x = 11$	0.5
$y = 24 - 11 = 13$	0.5

Where students lose marks in Statistics

The slips examiners see most often, and what to write instead.

Mistake: Using the cumulative frequency of the median class instead of the class before it
The cf of the median class is the number you just used to find the class, so it is the one in front of your eyes.
Wrong working
Median $= 20 + \frac{20 - 26}{16} \times 10 = 16.25$
Right working
Median $= 20 + \frac{20 - 10}{16} \times 10 = 26.25$
Do this: Circle the cf of the class just above the median class. If $\frac{n}{2} - c f$ is negative, you used the wrong cf.
Usually costs: Usually 1 to 2 marks, as substitution and answer are both wrong.
Mistake: Swapping $f_{0}$ and $f_{2}$ in the mode formula
Students remember the letters but not which neighbour goes in the numerator.
Wrong working
Mode $= 30 + \frac{20 - 14}{40 - 14 - 10} \times 10 = 33.75$
Right working
Mode $= 30 + \frac{20 - 10}{40 - 10 - 14} \times 10 = 36.25$
Do this: Write $f_{0}$ = class before and $f_{2}$ = class after in words beside the table before substituting.
Usually costs: Usually 1 mark for substitution and the answer.
Mistake: Not converting inclusive classes before finding the median or mode
Classes like 21–30 look fine, so students take $l = 21$ directly.
Wrong working
Modal class 21–30, $l = 21$ , Mode $= 21 + \frac{6}{10} \times 10 = 27$
Right working
Modal class 20.5–30.5, $l = 20.5$ , Mode $= 20.5 + \frac{6}{10} \times 10 = 26.5$
Do this: Whenever there is a gap between one class's upper limit and the next class's lower limit, subtract and add half the gap first.
Usually costs: Usually 1 mark, sometimes more if the class widths also come out wrong.
Mistake: Forgetting to multiply by $h$ in the step deviation method
The $u_{i}$ column hides the width, so $h$ slips out of the final formula.
Wrong working
$\overset{x}{ˉ} = 50 + \frac{2}{50} = 50.04$
Right working
$\overset{x}{ˉ} = 50 + 20 \times \frac{2}{50} = 50.8$
Do this: Write the full formula $\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}}$ before substituting. Check that the mean looks sensible for the data.
Usually costs: Usually 1 mark.
Mistake: Dividing by the number of classes instead of the total frequency
Habit from finding the mean of a list of numbers.
Wrong working
$\overset{x}{ˉ} = \frac{2540}{5} = 508$
Right working
$\overset{x}{ˉ} = \frac{2540}{50} = 50.8$
Do this: Always write $\sum f_{i}$ separately in the table and divide by it. A mean of 508 marks out of 100 is an instant warning.
Usually costs: Usually 1 to 1½ marks.
Mistake: Taking the class with the highest frequency as the median class
Confusing the rule for the modal class with the rule for the median class.
Wrong working
Median class 30–40 (highest $f = 13$ ): Median $= 30 + \frac{25 - 31}{13} \times 10$
Right working
cf: 8, 19, 31, ... First cf $\geq 25$ is 31, so median class is 20–30: Median $= 20 + \frac{25 - 19}{12} \times 10 = 25$
Do this: Always build a cf column and locate $\frac{n}{2}$ in it. The median class comes only from cf.
Usually costs: Often 2 or more marks, since everything after the class choice is wrong.
Mistake: Misremembering the empirical relation
Students mix up which average gets the coefficient 3.
Wrong working
Mode $= 3 Mean - 2 Median$ : $36 = 90 - 2 Median$ , Median $= 27$
Right working
$3 Median = Mode + 2 Mean = 36 + 60 = 96$ , Median $= 32$
Do this: Remember the median lies between mean and mode. Here $32$ lies between $30$ and $36$ ; $27$ does not, so it must be wrong.
Usually costs: The full 1 mark in an MCQ, or 1 mark in a longer question.

NCERT exercise guide: Statistics

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 13.1

What it tests: Finding the mean of grouped data by the direct method, the assumed mean method and the step deviation method. It also has questions where the mean is given and you find a missing frequency, and word problems built on real data.

How to approach it: First write the class mark $x_{i}$ of every class. Use the direct method when the class marks are small. Use the assumed mean method when they are large, and the step deviation method when all the deviations share a common factor $h$ . For a missing frequency, keep it as a letter, build $\sum f_{i}$ and $\sum f_{i} u_{i}$ in terms of it, set the mean formula equal to the given mean and solve.

Like the missing-frequency questions in Exercise 13.1

The daily time (in minutes) that students of a class spend on homework is given below. The mean time is 46 minutes. Find the missing frequency $f$ .
Time (min): 20–30, 30–40, 40–50, 50–60, 60–70
Number of students: 4, 6, $f$ , 8, 5

Show the solutionSolution

Class marks: $x_{i} = 25, 35, 45, 55, 65$ . Take assumed mean $a = 45$ and class width $h = 10$ .
$u_{i} = \frac{x _{i} - 45}{10} = - 2, - 1, 0, 1, 2$
$f_{i} u_{i} = - 8, - 6, 0, 8, 10$ , so $\sum f_{i} u_{i} = 4$
$\sum f_{i} = 4 + 6 + f + 8 + 5 = 23 + f$
Mean: $45 + 10 \times \frac{4}{23 + f} = 46$
So $\frac{40}{23 + f} = 1$ , which gives $23 + f = 40$ and $f = 17$ .
Check by the direct method: $\sum f_{i} = 40$ and $\sum f_{i} x_{i} = 100 + 210 + 765 + 440 + 325 = 1840$ . Then $\frac{1840}{40} = 46$ , which is correct.

Answer: The missing frequency is $f = 17$ .

Exercise 13.2

What it tests: Finding the mode of grouped data with the formula, sometimes alongside the mean of the same data, and saying what the mode and mean tell you about the situation.

How to approach it: Find the modal class, which is the class with the highest frequency. Write $l$ , $h$ , $f_{1}$ (frequency of the modal class), $f_{0}$ (the class before it) and $f_{2}$ (the class after it) before you substitute. Always state the modal class in your answer. When asked to compare, remember that the mode is the most typical value, while the mean is the average.

Like the questions in Exercise 13.2 that ask for the mode and mean together

The monthly electricity consumption of 60 consumers in a locality is given below. Find the mode and the mean of the data, and interpret both.
Units consumed: 60–80, 80–100, 100–120, 120–140, 140–160
Number of consumers: 8, 12, 18, 14, 8

Show the solutionSolution

The highest frequency is 18, so the modal class is 100–120.
$l = 100$ , $h = 20$ , $f_{1} = 18$ , $f_{0} = 12$ , $f_{2} = 14$
Mode $= 100 + \frac{18 - 12}{36 - 12 - 14} \times 20 = 100 + \frac{6}{10} \times 20 = 112$ units
For the mean, class marks are $70, 90, 110, 130, 150$ . Take $a = 110$ and $h = 20$ , so $u_{i} = - 2, - 1, 0, 1, 2$ .
$f_{i} u_{i} = - 16, - 12, 0, 14, 16$ , so $\sum f_{i} u_{i} = 2$ and $\sum f_{i} = 60$
Mean $= 110 + 20 \times \frac{2}{60} = 110 + \frac{2}{3} \approx 110.67$ units
Interpretation: the most common consumption is about 112 units. The average consumption is a little lower, about 110.67 units.

Answer: Mode = 112 units and mean ≈ 110.67 units. Most consumers use about 112 units a month, and the average is slightly lower.

Exercise 13.3

What it tests: Finding the median of grouped data, including data in inclusive classes and data given as a cumulative frequency table. It also has questions where the median is given and you find missing frequencies, and questions that ask for the mean, median and mode of one table.

How to approach it: Make a cumulative frequency column and find $\frac{n}{2}$ . The median class is the first class whose cumulative frequency is at least $\frac{n}{2}$ . If the classes are inclusive, such as 11–20 and 21–30, convert them to continuous boundaries first. Use $c f$ of the class before the median class, not of the median class itself.

Like the inclusive-class median questions in Exercise 13.3

The marks (out of 50) of 50 students in a class test are given below. Find the median marks.
Marks: 1–10, 11–20, 21–30, 31–40, 41–50
Number of students: 5, 10, 20, 12, 3

Show the solutionSolution

The classes have a gap of 1, so subtract 0.5 from each lower limit and add 0.5 to each upper limit: 0.5–10.5, 10.5–20.5, 20.5–30.5, 30.5–40.5, 40.5–50.5.
Cumulative frequencies: 5, 15, 35, 47, 50
$n = 50$ , so $\frac{n}{2} = 25$ .
The first $c f$ that is at least 25 is 35, so the median class is 20.5–30.5.
$l = 20.5$ , $c f = 15$ , $f = 20$ , $h = 10$
Median $= 20.5 + \frac{25 - 15}{20} \times 10 = 20.5 + 5 = 25.5$

Answer: The median is 25.5 marks.

Statistics: previous year questions

Standard: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.8 marks per paper; 50 questions in all, counting OR alternatives. In 4 of 4 sample papers.
Basic: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6.1 marks per paper; 56 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Statistics

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026 Basic, set 430/1/1, Q16 · 1 mark · MCQ

The class mark of the median class of the following data is :

Class Interval	10 – 25	25 – 40	40 – 55	55 – 70	70 – 85	85 – 100
Frequency	2	3	7	6	6	6

(A) $40$
(B) $55$
(C) $47 \cdot 5$
(D) $62 \cdot 5$

Try it first, then show our solutionSolution

$N = 30$ , so $\frac{N}{2} = 15$ . The cumulative frequencies are 2, 5, 12, 18, …
½ mark
The first cumulative frequency above 15 is 18, so the median class is 55–70. Its class mark is $\frac{55 + 70}{2} = 62.5$ .
½ mark

Answer: 62.5, option (D)

Where the marks are

Finds $N /2 = 15$ and the cumulative frequencies (½ mark)
Median class 55–70, class mark 62.5 (½ mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/1/1, Q35 · 5 marks · Long answer

The following data gives the information on the observed lifetime (in hours) of 200 electrical components :

Lifetime (in hours)	Number of electrical components
0 – 20	10
20 – 40	35
40 – 60	50
60 – 80	60
80 – 100	30
100 – 120	15

Find the mean lifetime (in hours) of the electrical components.

Try it first, then show our solutionSolution

Use the step-deviation method. Take the assumed mean $a = 50$ and class size $h = 20$ , so $u_{i} = \frac{x _{i} - 50}{20}$ .
Lifetime $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
0 – 20 10 10 –2 –20
20 – 40 30 35 –1 –35
40 – 60 50 50 0 0
60 – 80 70 60 1 60
80 – 100 90 30 2 60
100 – 120 110 15 3 45
Total $\sum f_{i} = 200$ $\sum f_{i} u_{i} = 110$
2 marks
Apply the formula:
$\overset{x}{ˉ} = a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 50 + \frac{110}{200} \times 20$
2 marks
$\overset{x}{ˉ} = 50 + 11 = 61$
So the mean lifetime of the electrical components is 61 hours.
1 mark

Lifetime	$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
0 – 20	10	10	–2	–20
20 – 40	30	35	–1	–35
40 – 60	50	50	0	0
60 – 80	70	60	1	60
80 – 100	90	30	2	60
100 – 120	110	15	3	45
Total		$\sum f_{i} = 200$		$\sum f_{i} u_{i} = 110$

Answer: Mean lifetime = 61 hours

Where the marks are

Correct table with class marks $x_{i}$ , $u_{i}$ and $f_{i} u_{i}$ , giving $\sum f_{i} = 200$ and $\sum f_{i} u_{i} = 110$ (2 marks)
Substitutes correctly into $\overset{x}{ˉ} = a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h$ (2 marks)
Obtains mean = 61 hours (1 mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/4/1, Q37 · 4 marks · Case-basedCompetency-based

The Air Quality Index (AQI) is a scale from 0 to 500 that indicates air quality, with higher numbers signifying more pollution and greater health concerns. Mansi collected the daily data of AQI of her city for a month and presented it as given below :

AQI Range	1 – 100	101 – 200	201 – 300	301 – 400	401 – 500
Number of Days	3	9	12	4	2

Based on the above, answer the following questions :

(i)
Convert the data to continuous frequency distribution.
1 mark
(ii)
What is the quality of air in most of the days of the month ?
1 mark
(iii)
(a) Using table formed in part (i), find mode of the data.
2 marks
OR
(iii)
(b) Using table formed in part (i), find median of the data.
2 marks

Figure in the paper: Table titled 'Central Pollution Control Board's Air Quality Standards': AQI 0-50 Good; 51-100 Satisfactory; 101-200 Moderate; 201-300 Poor; 301-400 Very Poor; 401-500 Severe. See it in the official paper.

Try it first, then show our solutionSolution

(i) The gap between consecutive classes is $101 - 100 = 1$ . Subtract $0.5$ from each lower limit and add $0.5$ to each upper limit.
AQI (class) Number of days
0.5 – 100.5 3
100.5 – 200.5 9
200.5 – 300.5 12
300.5 – 400.5 4
400.5 – 500.5 2
1 mark
(ii) The highest number of days (12) falls in the range 201–300. By the CPCB scale this range is 'Poor', so the air quality was Poor on most days.
1 mark
(iii)(a) The modal class is $200.5 - 300.5$ , since its frequency 12 is the highest.
Here $l = 200.5$ , $f_{1} = 12$ , $f_{0} = 9$ , $f_{2} = 4$ and $h = 100$ .
Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h = 200.5 + (\frac{12 - 9}{24 - 9 - 4}) \times 100 = 200.5 + \frac{300}{11}$
1½ marks
$= 200.5 + 27.27 = 227.77$ (approx.)
½ mark
(iii)(b) Cumulative frequency table:
Class f cf
0.5 – 100.5 3 3
100.5 – 200.5 9 12
200.5 – 300.5 12 24
300.5 – 400.5 4 28
400.5 – 500.5 2 30
$N = 30$ , so $\frac{N}{2} = 15$ . The cf first exceeds 15 in the class $200.5 - 300.5$ , so this is the median class.
½ mark
Here $l = 200.5$ , $c f = 12$ , $f = 12$ and $h = 100$ .
Median $= l + (\frac{\frac{N}{2} - c f}{f}) \times h = 200.5 + (\frac{15 - 12}{12}) \times 100$
1 mark
$= 200.5 + 25 = 225.5$
½ mark

AQI (class)	Number of days
0.5 – 100.5	3
100.5 – 200.5	9
200.5 – 300.5	12
300.5 – 400.5	4
400.5 – 500.5	2

Class	f	cf
0.5 – 100.5	3	3
100.5 – 200.5	9	12
200.5 – 300.5	12	24
300.5 – 400.5	4	28
400.5 – 500.5	2	30

Answer: (i) Classes 0.5–100.5, 100.5–200.5, 200.5–300.5, 300.5–400.5, 400.5–500.5 with frequencies 3, 9, 12, 4, 2; (ii) Poor; (iii)(a) Mode ≈ 227.77; (iii)(b) Median = 225.5

Where the marks are

Writes the correct continuous class intervals with their frequencies (1 mark)
States that the air quality was Poor on most days (1 mark)
Identifies the modal class and substitutes correctly into the mode formula (1½ marks)
Gets mode ≈ 227.77 (½ mark)
Makes the correct cumulative frequency table and median class (½ mark)
Substitutes correctly into the median formula (1 mark)
Gets median = 225.5 (½ mark)

See it in the full 2026 board paper, set 430/4/1

Every Statistics question, by year

Standard (63 questions, sample papers included)

For 2027

Sample paper 2026-27: Q12 (1 mark, solved), Q16 (1 mark, solved), Q38 (4 marks, solved)

2026

Set 30/1/1: Q17 (1 mark, solved), Q35 (5 marks, solved)

Set 30/2/1: Q17 (1 mark, solved), Q35 (5 marks, solved)

Set 30/3/1: Q16 (1 mark, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 30/4/1: Q18 (1 mark, solved), Q35(a) (5 marks, solved), Q35(b) (5 marks, solved)

Set 30/5/1: Q6 (1 mark, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Sample paper 2025-26: Q17 (1 mark, solved), Q35(A) (5 marks, solved), Q35(B) (5 marks, solved)

2025

Set 30/1/1: Q12 (1 mark, solved), Q35 (5 marks, solved)

Set 30/2/1: Q35 (5 marks, solved)

Set 30/3/1: Q15 (1 mark, solved), Q38 (4 marks, solved)

Set 30/4/1: Q12 (1 mark, solved), Q13 (1 mark, solved), Q35 (5 marks, solved)

Set 30/5/1: Q16 (1 mark, solved), Q17 (1 mark, solved), Q35 (5 marks, solved)

Set 30/6/1: Q15 (1 mark, solved), Q16 (1 mark, solved), Q35 (5 marks, solved)

Sample paper 2024-25: Q9 (1 mark, solved), Q16 (1 mark, solved), Q35(A) (5 marks, solved), Q35(B) (5 marks, solved)

2024

Set 30/1/1: Q9 (1 mark, solved), Q14 (1 mark, solved), Q37 (4 marks, solved)

Set 30/2/1: Q12 (1 mark, solved), Q18 (1 mark, solved), Q30 (3 marks, solved)

Set 30/3/1: Q16 (1 mark, solved), Q37 (4 marks, solved)

Set 30/4/1: Q7 (1 mark, solved), Q35 (5 marks, solved)

Set 30/5/1: Q6 (1 mark, solved), Q38 (4 marks, solved)

Sample paper 2023-24: Q18 (1 mark, solved), Q31 (3 marks, solved), Q35 (5 marks, solved)

2023

Set 30/1/1: Q11 (1 mark, solved), Q15 (1 mark, solved), Q37 (4 marks, solved)

Set 30/2/1: Q34 (5 marks, solved)

Set 30/4/1: Q6 (1 mark, solved), Q35 (5 marks, solved)

Set 30/5/1: Q9 (1 mark, solved), Q10 (1 mark, solved), Q18 (1 mark, solved), Q31 (3 marks, solved)

Set 30/6/1: Q35 (5 marks, solved)

Basic (70 questions, sample papers included)

For 2027

Sample paper 2026-27: Q10 (1 mark, solved), Q12 (1 mark, solved), Q30(A) (3 marks, solved), Q30(B) (3 marks, solved)

2026

Set 430/1/1: Q16 (1 mark, solved), Q17 (1 mark, solved), Q35 (5 marks, solved)

Set 430/2/1: Q16 (1 mark, solved), Q17 (1 mark, solved), Q35(a) (5 marks, solved), Q35(b) (5 marks, solved)

Set 430/3/1: Q16 (1 mark, solved), Q17 (1 mark, solved), Q35 (5 marks, solved)

Set 430/4/1: Q8 (1 mark, solved), Q37 (4 marks, solved)

Set 430/5/1: Q20 (1 mark, solved), Q35 (5 marks, solved)

Sample paper 2025-26: Q13 (1 mark, solved), Q14 (1 mark, solved), Q27(A) (3 marks, solved), Q27(B) (3 marks, solved)

2025

Set 430/1/1: Q16 (1 mark, solved), Q17 (1 mark, solved), Q35 (5 marks, solved)

Set 430/2/1: Q16 (1 mark, solved), Q17 (1 mark, solved), Q35(a) (5 marks, solved), Q35(b) (5 marks, solved)

Set 430/3/1: Q16 (1 mark, solved), Q17 (1 mark, solved), Q35 (5 marks, solved)

Set 430/4/1: Q19 (1 mark, solved), Q33 (5 marks, solved)

Set 430/5/1: Q15 (1 mark, solved), Q35(a) (5 marks, solved), Q35(b) (5 marks, solved)

Set 430/6/1: Q15 (1 mark, solved), Q35 (5 marks, solved)

Sample paper 2024-25: Q17 (1 mark, solved), Q25 (2 marks, solved), Q29 (3 marks, solved)

2024

Set 430/1/1: Q11 (1 mark, solved), Q13 (1 mark, solved), Q37 (4 marks, solved)

Set 430/2/1: Q11 (1 mark, solved), Q37 (4 marks, solved)

Set 430/3/1: Q13 (1 mark, solved), Q18 (1 mark, solved), Q32 (5 marks, solved)

Set 430/4/1: Q4 (1 mark, solved), Q15 (1 mark, solved), Q36 (4 marks, solved)

Set 430/5/1: Q4 (1 mark, solved), Q35 (5 marks, solved)

Sample paper 2023-24: Q16 (1 mark, solved), Q18 (1 mark, solved), Q35 (5 marks, solved)

2023

Set 430/1/1: Q34 (5 marks, solved)

Set 430/2/1: Q17 (1 mark, solved), Q33 (5 marks, solved)

Set 430/4/1: Q13 (1 mark, solved), Q34 (5 marks, solved)

Set 430/5/1: Q15 (1 mark, solved), Q18 (1 mark, solved), Q28 (3 marks, solved)

Set 430/6/1: Q1 (1 mark, solved), Q11 (1 mark, solved), Q18 (1 mark, solved), Q27 (3 marks, solved)

Practice questions on Statistics

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If the median and mode of a distribution are 22 and 20 respectively, its mean is:
(a) 21
(b) 23
(c) 24
(d) 26
Hint
Use the empirical relation $3 Median = Mode + 2 Mean$ .
Answer
(b) 23. From $3 (22) = 20 + 2 Mean$ , you get $2 Mean = 46$ , so Mean = 23.
Q2 · 5 marks · Long answerHarder board level
The marks (out of 50) of 40 students in a test are given below. The mean mark is 26.
Marks: 1–10, 11–20, 21–30, 31–40, 41–50
Number of students: 4, $p$ , 12, $q$ , 6
(i) Find $p$ and $q$ .
(ii) Find the median mark.
Hint
Class marks such as 5.5 and 15.5 do not change when you convert the classes, so use them for the mean. For the median, first convert to boundaries such as 20.5–30.5.
Answer
(i) Total: $22 + p + q = 40$ , so $p + q = 18$ .
Class marks are $5.5, 15.5, 25.5, 35.5, 45.5$ . With $a = 25.5$ and $h = 10$ , $u_{i} = - 2, - 1, 0, 1, 2$ and $\sum f_{i} u_{i} = - 8 - p + 0 + q + 12 = q - p + 4$ .
$25.5 + 10 \times \frac{q - p + 4}{40} = 26$ gives $q - p + 4 = 2$ , so $q - p = - 2$ .
Solving, $p = 10$ and $q = 8$ . Check: $\sum f_{i} x_{i} = 22 + 155 + 306 + 284 + 273 = 1040$ and $\frac{1040}{40} = 26$ .
(ii) Boundaries: 0.5–10.5, 10.5–20.5, 20.5–30.5, 30.5–40.5, 40.5–50.5. Cumulative frequencies are 4, 14, 26, 34, 40. Since $\frac{n}{2} = 20$ , the median class is 20.5–30.5.
Median $= 20.5 + \frac{20 - 14}{12} \times 10 = 20.5 + 5 = 25.5$ marks
Q3 · 5 marks · Long answerHOTS / competency
Suppose the daily earnings (in Rs) of 50 auto drivers in a town are given below.
Earnings (Rs): 200–300, 300–400, 400–500, 500–600, 600–700
Number of drivers: 4, 8, 20, 12, 6
(i) Find the mean earning by the step deviation method.
(ii) Find the modal earning.
(iii) Use the empirical relation to estimate the median earning, check it with the median formula, and decide whether a driver who earns Rs 470 a day is in the top half of the drivers.
Hint
Take $a = 450$ and $h = 100$ for the mean. The modal class is the one with frequency 20. Then use $3 Median = Mode + 2 Mean$ .
Answer
(i) $u_{i} = - 2, - 1, 0, 1, 2$ and $f_{i} u_{i} = - 8, - 8, 0, 12, 12$ , so $\sum f_{i} u_{i} = 8$ . Mean $= 450 + 100 \times \frac{8}{50} = 450 + 16 =$ Rs 466.
(ii) Modal class 400–500 with $f_{1} = 20$ , $f_{0} = 8$ , $f_{2} = 12$ . Mode $= 400 + \frac{20 - 8}{40 - 8 - 12} \times 100 = 400 + 60 =$ Rs 460.
(iii) $3 Median = 460 + 2 \times 466 = 1392$ , so Median ≈ Rs 464.
Check: cumulative frequencies are 4, 12, 32, 44, 50, so the median class is 400–500 and Median $= 400 + \frac{25 - 12}{20} \times 100 =$ Rs 465, very close to Rs 464.
Rs 470 is more than the median by either method, so this driver is in the top half.
Q4 · 3 marks · Short answerBoard level
Find the median of the following data.
Marks: 10–19, 20–29, 30–39, 40–49, 50–59
Number of students: 5, 10, 20, 12, 3
Hint
Convert the classes to boundaries such as 29.5–39.5 before you use the formula.
Answer
Cumulative frequencies are 5, 15, 35, 47, 50. Since $\frac{n}{2} = 25$ , the median class is 29.5–39.5.
Median $= 29.5 + \frac{25 - 15}{20} \times 10 = 34.5$ marks
Q5 · 3 marks · Short answerBoard level
The weights (in g) of parcels at a courier counter are given below. The mean weight is 146 g. Find the missing frequency $f$ .
Weight (g): 100–120, 120–140, 140–160, 160–180, 180–200
Number of parcels: 10, $f$ , 15, 10, 5
Hint
Use step deviation with $a = 150$ and $h = 20$ , and keep $f$ as a letter.
Answer
$u_{i} = - 2, - 1, 0, 1, 2$ , so $\sum f_{i} u_{i} = - 20 - f + 0 + 10 + 10 = - f$ , and $\sum f_{i} = 40 + f$ .
$150 - \frac{20 f}{40 + f} = 146$ gives $20 f = 160 + 4 f$ , so $f = 10$ .
Check: $\frac{7300}{50} = 146$
Q6 · 3 marks · Short answerBoard level
The median of the following distribution is 28.5 and the total frequency is 60. Find $x$ and $y$ .
Classes: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60
Frequencies: 5, $x$ , 20, 15, $y$ , 5
Hint
Since 28.5 lies in 20–30, that is the median class, with $c f = 5 + x$ .
Answer
From the total, $45 + x + y = 60$ , so $x + y = 15$ .
$28.5 = 20 + \frac{30 - ( 5 + x )}{20} \times 10$ gives $\frac{25 - x}{2} = 8.5$ , so $x = 8$ .
Then $y = 7$ .
Q7 · 5 marks · Long answerHarder board level
The marks (out of 50) of 50 students are given below.
Marks: 0–10, 10–20, 20–30, 30–40, 40–50
Number of students: 5, 7, 12, 18, 8
(i) Find the mean using the step deviation method.
(ii) Find the mode.
(iii) Use the empirical relation to estimate the median.
Hint
For the mean, take $a = 25$ and $h = 10$ . For the mode, the modal class is 30–40.
Answer
(i) $\sum f_{i} u_{i} = - 10 - 7 + 0 + 18 + 16 = 17$ , so Mean $= 25 + \frac{170}{50} = 28.4$ .
(ii) Mode $= 30 + \frac{6}{36 - 12 - 8} \times 10 = 33.75$
(iii) $3 Median = 33.75 + 56.8 = 90.55$ , so Median ≈ 30.18 marks.
Q8 · 3 marks · Short answerBoard level
The scores of 40 students in a quiz are given as a cumulative frequency table: less than 20: 4, less than 40: 10, less than 60: 24, less than 80: 30, less than 100: 40. Find the modal score.
Hint
First get the class frequencies by subtracting consecutive cumulative frequencies.
Answer
The frequencies for 0–20, 20–40, 40–60, 60–80, 80–100 are 4, 6, 14, 6, 10. The modal class is 40–60.
Mode $= 40 + \frac{14 - 6}{28 - 6 - 6} \times 20 = 40 + 10 = 50$
Q9 · 1 mark · Assertion-reasonBoard level
Assertion (A): The mode of the distribution with classes 0–10, 10–20, 20–30 and frequencies 8, 12, 10 is not 15.
Reason (R): The mode of grouped data equals the class mark of the modal class only when $f_{0} = f_{2}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Work out the mode, then see when the fraction in the mode formula equals one half.
Answer
(a) Mode $= 10 + \frac{4}{24 - 8 - 10} \times 10 \approx 16.67$ , which is not 15, so A is true.
The fraction $\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}$ equals $\frac{1}{2}$ only when $f_{0} = f_{2}$ , so R is true. Here $8 \neq = 10$ , so R explains A.
Q10 · 4 marks · Case-basedHOTS / competency
Suppose a school canteen records the amount (in Rs) spent by some students in a week.
Amount (Rs): 0–20, 20–40, 40–60, 60–80, 80–100
Number of students: 8, 10, 20, 14, 8
(i) Write the modal class and the median class. (1 mark)
(ii) Find the median amount. (1 mark)
(iii) The manager later finds that the count for the class 60–80 was copied wrongly, so the total number of students is not known either; the other four counts are correct. If the true mean amount is Rs 54, find the correct count for 60–80 and the true number of students, and say whether the modal class changes. (2 marks)
Hint
For (iii), call the count $k$ , so the total is $46 + k$ . Use step deviation with $a = 50$ and $h = 20$ , and set the mean equal to 54.
Answer
(i) The highest frequency is 20, so the modal class is 40–60. From the table, $n = 60$ ; cumulative frequencies are 8, 18, 38, 52, 60 and $\frac{n}{2} = 30$ , so the median class is also 40–60.
(ii) Median $= 40 + \frac{30 - 18}{20} \times 20 = 40 + 12$ , which is Rs 52.
(iii) Let the correct count be $k$ . Then $\sum f_{i} = 46 + k$ and, with $u_{i} = - 2, - 1, 0, 1, 2$ , $\sum f_{i} u_{i} = - 16 - 10 + 0 + k + 16 = k - 10$ .
$50 + 20 \times \frac{k - 10}{46 + k} = 54$ gives $20 k - 200 = 184 + 4 k$ , so $16 k = 384$ and $k = 24$ .
The true number of students is $46 + 24 = 70$ . Check: $\sum f_{i} x_{i} = 80 + 300 + 1000 + 1680 + 720 = 3780$ , and $\frac{3780}{70} = 54$ .
Now 60–80 has 24 students, more than the 20 in 40–60, so the modal class changes to 60–80.
Q11 · 5 marks · Long answerHarder board level
The marks (out of 100) of a group of students are given below. The mean mark is 50.8.
Marks: 0–20, 20–40, 40–60, 60–80, 80–100
Number of students: 4, 11, 20, $k$ , 6
(i) Find $k$ .
(ii) Find the median mark.
(iii) Use the empirical relation to estimate the mode, and compare it with the mode given by the formula.
Hint
Total is $41 + k$ . With $a = 50$ and $h = 20$ , $\sum f_{i} u_{i} = k - 7$ . Find $k$ before building the cumulative frequency column.
Answer
(i) $\sum f_{i} = 41 + k$ and $\sum f_{i} u_{i} = - 8 - 11 + 0 + k + 12 = k - 7$ .
$50 + 20 \times \frac{k - 7}{41 + k} = 50.8$ gives $20 (k - 7) = 0.8 (41 + k)$ . Multiply by 5: $100 k - 700 = 164 + 4 k$ , so $96 k = 864$ and $k = 9$ .
Check: $\sum f_{i} = 50$ and $\sum f_{i} x_{i} = 40 + 330 + 1000 + 630 + 540 = 2540$ , and $\frac{2540}{50} = 50.8$ .
(ii) Cumulative frequencies are 4, 15, 35, 44, 50. Since $\frac{n}{2} = 25$ , the median class is 40–60.
Median $= 40 + \frac{25 - 15}{20} \times 20 = 50$ marks
(iii) Mode $= 3 Median - 2 Mean = 150 - 101.6 = 48.4$ marks.
By the formula, the modal class is 40–60 with $f_{1} = 20$ , $f_{0} = 11$ , $f_{2} = 9$ : Mode $= 40 + \frac{9}{40 - 11 - 9} \times 20 = 49$ marks. The two values are close, as the relation is only approximate.
Q12 · 3 marks · Short answerHarder board level
The mode of the following distribution is 24. Find $x$ , and check that your modal class is correct.
Classes: 0–10, 10–20, 20–30, 30–40, 40–50
Frequencies: 4, $x + 2$ , $2 x$ , $x$ , 5
Hint
You cannot see the highest frequency, so use the given mode: 24 lies in 20–30. Here $f_{0}$ , $f_{1}$ and $f_{2}$ all contain $x$ .
Answer
The mode 24 lies in 20–30, so this is the modal class: $l = 20$ , $h = 10$ , $f_{1} = 2 x$ , $f_{0} = x + 2$ , $f_{2} = x$ .
$24 = 20 + \frac{2 x - ( x + 2 )}{4 x - ( x + 2 ) - x} \times 10$ , so $4 = \frac{10 ( x - 2 )}{2 x - 2}$ .
$8 x - 8 = 10 x - 20$ , so $2 x = 12$ and $x = 6$ .
Frequencies are 4, 8, 12, 6, 5. The highest is 12, in 20–30, so the modal class is correct. Check: $20 + \frac{12 - 8}{24 - 8 - 6} \times 10 = 20 + 4 = 24$ .
Q13 · 5 marks · Long answerHarder board level
The total frequency of the distribution below is 50 and its mode is 55.
Classes: 0–20, 20–40, 40–60, 60–80, 80–100
Frequencies: 6, $x$ , 16, $y$ , 4
(i) Find $x$ and $y$ .
(ii) Find the mean and the median.
Hint
The total gives $x + y = 24$ . The mode 55 lies in 40–60, so $f_{0} = x$ and $f_{2} = y$ , and the denominator $32 - x - y$ becomes a number.
Answer
(i) Total: $26 + x + y = 50$ , so $x + y = 24$ .
The mode 55 lies in 40–60, so $l = 40$ , $h = 20$ , $f_{1} = 16$ , $f_{0} = x$ , $f_{2} = y$ .
$55 = 40 + \frac{16 - x}{32 - x - y} \times 20 = 40 + \frac{16 - x}{8} \times 20$ , so $15 = 2.5 (16 - x)$ , $16 - x = 6$ and $x = 10$ . Then $y = 14$ .
Check: 16 is greater than 10 and 14, so 40–60 is the modal class.
(ii) With $a = 50$ , $h = 20$ : $\sum f_{i} u_{i} = - 12 - 10 + 0 + 14 + 8 = 0$ , so Mean $= 50$ . Check: $\frac{60 + 300 + 800 + 980 + 360}{50} = \frac{2500}{50} = 50$ .
Cumulative frequencies are 6, 16, 32, 46, 50. Since $\frac{n}{2} = 25$ , the median class is 40–60.
Median $= 40 + \frac{25 - 16}{16} \times 20 = 40 + 11.25 = 51.25$
Q14 · 3 marks · Short answerHOTS / competency
The time (in minutes) taken by 40 students to finish a lab task is given below.
Time (min): 10–20, 20–30, 30–40, 40–50, 50–60
Number of students: 5, 7, 8, 12, 8
(i) Find the median time.
(ii) Riya says the median class should be 40–50, because the cumulative frequency of 30–40 only equals $\frac{n}{2}$ and does not exceed it. Show that her choice gives the same median, and explain why.
Hint
The cumulative frequency of 30–40 is exactly 20. Work out the median with each class and look at where the 20th observation ends.
Answer
(i) Cumulative frequencies are 5, 12, 20, 32, 40 and $\frac{n}{2} = 20$ . The first cf that is at least 20 is 20, so the median class is 30–40.
Median $= 30 + \frac{20 - 12}{8} \times 10 = 40$ minutes
(ii) With 40–50: $l = 40$ , $c f = 20$ , $f = 12$ , so Median $= 40 + \frac{20 - 20}{12} \times 10 = 40$ minutes, the same.
Exactly 20 observations lie below 40, so the middle position is reached right at the boundary 40, which the two classes share. Both choices land on that boundary.
Q15 · 3 marks · Short answerHarder board level
The marks (out of 50) of 50 students are given below.
Marks: 0–10, 10–20, 20–30, 30–40, 40–50
Number of students: 5, 10, 15, 12, 8
(i) Find the mean mark.
(ii) Some new students join, and all of them score in the class 30–40. The mean of the new table is 28. How many students joined?
Hint
Let $k$ students join. They add $k$ to the frequency of 30–40, where $u = 1$ when $a = 25$ and $h = 10$ .
Answer
(i) With $a = 25$ , $h = 10$ : $\sum f_{i} u_{i} = - 10 - 10 + 0 + 12 + 16 = 8$ , so Mean $= 25 + 10 \times \frac{8}{50} = 26.6$ marks.
(ii) Let $k$ students join. Then $\sum f_{i} u_{i} = 8 + k$ and $\sum f_{i} = 50 + k$ .
$25 + 10 \times \frac{8 + k}{50 + k} = 28$ gives $80 + 10 k = 150 + 3 k$ , so $7 k = 70$ and $k = 10$ .
Check: $\sum f_{i} x_{i} = 1330 + 10 \times 35 = 1680$ and $\frac{1680}{60} = 28$ . So 10 students joined.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the class mark of the class 35–45?Show answer
$\frac{35 + 45}{2} = 40$
2.In the assumed mean method, what is $d_{i}$ ?Show answer
$d_{i} = x_{i} - a$ , the class mark minus the assumed mean.
3.In the step deviation method, what is $u_{i}$ and what must you multiply by at the end?Show answer
$u_{i} = \frac{x _{i} - a}{h}$ ; multiply $\frac{\sum f _{i} u _{i}}{\sum f _{i}}$ by $h$ .
4.How do you identify the modal class?Show answer
It is the class with the highest frequency.
5.How do you identify the median class?Show answer
It is the first class whose cumulative frequency is at least $\frac{n}{2}$ .
6.In the median formula, which class does cf belong to?Show answer
The class just before the median class.
7.State the empirical relation.Show answer
$3 Median = Mode + 2 Mean$
8.If mean = 20 and median = 22, what is the mode?Show answer
Mode $= 3 (22) - 2 (20) = 66 - 40 = 26$

Best order to study it

Revise class marks, frequency tables and how to convert inclusive classes into continuous ones.
Learn the mean by the direct method, then the assumed mean and step deviation methods, checking that all three give the same answer on one table.
Practise missing-frequency questions using the mean, as they test your table and your algebra together.
Learn the mode formula, then cumulative frequency tables and the median formula, including converting less than and more than tables.
Finish with two-unknown median problems, the empirical relation and case-study style questions that need all three averages from one table.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write all three mean formulas (direct, assumed mean, step deviation) without looking?
Can you explain why multiplying by $h$ is needed in step deviation?
Can you convert 1–10, 11–20, 21–30 into continuous classes?
Can you write the mode formula and say what $l, h, f_{1}, f_{0}, f_{2}$ mean?
Can you write the median formula and say whose cf is used?
Can you turn a less than cumulative table into an ordinary frequency table?
Can you find the median class when $\frac{n}{2}$ equals a cumulative frequency exactly?
Can you find two missing frequencies from the total and the median?
Can you find the third average from two using $3 Median = Mode + 2 Mean$ ?
Can you check that your mode and median lie inside their own classes?

Statistics: frequently asked questions

Which method should I use to find the mean if the question does not say?

Any method gets full marks if it is correct and shown properly. Use the direct method for small class marks and the step deviation method when class marks are large and widths are equal. If the question names a method, you must use that one.

Do I need to make the classes continuous to find the mean?

No. The class marks stay the same after conversion, so the mean is unaffected. You must convert to continuous classes before finding the median or mode, because the lower limit is used in those formulas.

Are ogives (cumulative frequency graphs) in the 2026-27 syllabus?

No. Graphical representation of cumulative frequency distribution, including ogives, is not in the 2026-27 syllabus.

Is the relation 3 Median = Mode + 2 Mean always exact?

No, it is an approximate relation that works well for moderately skewed data. In board questions you can use it as exact whenever two averages are given and the third is asked.

What if two classes have the same highest frequency?

That is a bimodal situation, and the syllabus says such cases are to be avoided. Board questions will give a single class with the highest frequency, so you will always have one clear modal class.

Based on: CBSE Mathematics syllabus 2026-27; NCERT Mathematics, Textbook for Class X (Reprint 2026-27)