Statistics

Statistics

Statistics carries 6 marks in every recent board paper (the 2026-27 sample paper, 2025 and both 2026 papers).

The marks usually come as one or two 1-mark MCQs or an Assertion-Reason item (the empirical relation, median class, modal class), plus either a 5-mark question (mean and mode, or missing frequencies from a given median or mean) or a 4-mark case study on a grouped table. A 5-mark statistics question appeared in three of the last four papers.

Where marks are usually lost:

• swapping $f_0$ and $f_2$ in the mode formula;
• in the median formula, using the cumulative frequency of the median class itself instead of the class before it;
• a sign slip in $\sum f_i u_i$, or forgetting to multiply by $h$ in the step-deviation method;
• not writing the table of $x_i$, $f_i x_i$ (or $u_i$) and cumulative frequencies, which carries marks.

Revise in 5 Minutes

Mean of grouped data ($x_i$ = class mark = $\frac{\text{lower limit} + \text{upper limit}}{2}$)

Take $a$ as a class mark near the middle, so the $u_i$ are small numbers like $-2, -1, 0, 1, 2$. All three methods give the same answer.

Mode

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

• Modal class: the class with the highest frequency.
• $f_1$ = frequency of the modal class, $f_0$ = of the class before it, $f_2$ = of the class after it.

Median

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

• Find $\frac{n}{2}$; the median class is the first class whose cumulative frequency is greater than $\frac{n}{2}$.
• $cf$ = cumulative frequency of the class before the median class; $f$ = frequency of the median class.

Empirical relation: 3 × Median = Mode + 2 × Mean.

Missing frequencies: use the total $n$ for one equation and the given mean, median or mode for the other.

Cumulative table given ("less than 20: 12, less than 30: 25")? Subtract neighbours to get frequencies: $25 - 12 = 13$ for 20-30.

Classes like 10-19, 20-29? Make them continuous first (9.5-19.5, 19.5-29.5) before using $l$ and $h$.

Traps

• $f_0$ and $f_2$ swapped in the mode formula.
• Using the median class's own $cf$ in place of the previous class's.
• Forgetting $\times h$ in step-deviation, or a sign slip in $\sum f_i u_i$.
• Picking the median class by the highest frequency (that is the modal class).

How to use this page: try each question on paper first, then read the answer. The marks against each step show what an examiner looks for. The 1-mark MCQs and Assertion-Reason questions are in the quiz at the end, together with questions that test how well you understand the chapter; every quiz answer comes with its explanation.

Short Answer Questions (2 and 3 Marks)

Question 1 (2 marks)

The number of books read by 20 students of a class during the summer vacation is given below. Find the mean number of books read.

Answer.

1. $\sum f_i x_i = 6 + 20 + 48 + 24 + 10 = 108$ and $\sum f_i = 20$ — 1 mark
2. Mean $= \frac{108}{20} = 5.4$ books — 1 mark

Question 2 (2 marks)

The runs scored by 40 batters in a school cricket tournament are given below. Find the mode.

Answer.

1. Make the classes continuous: 9.5-19.5, 19.5-29.5, …; modal class 29.5-39.5 with $l = 29.5$, $f_1 = 14$, $f_0 = 9$, $f_2 = 9$, $h = 10$ — 1 mark
2. Mode $= 29.5 + \frac{14 - 9}{28 - 9 - 9} \times 10 = 29.5 + 5 = 34.5$ runs — 1 mark

Question 3 (2 marks)

Arjun found the mode of the distribution below as follows.

"Modal class is 30-40, $l = 30$, $f_1 = 14$, $f_0 = 10$, $f_2 = 8$, $h = 10$. Mode $= 30 + \frac{14 - 10}{28 - 10 - 8} \times 10 = 34$."

Find his mistake and the correct mode.

Answer.

1. He swapped $f_0$ and $f_2$: $f_0$ is the class before the modal class, so $f_0 = 8$ and $f_2 = 10$ — 1 mark
2. Mode $= 30 + \frac{14 - 8}{28 - 8 - 10} \times 10 = 30 + 6 = 36$ — 1 mark

Question 4 (2 marks)

The lengths (in mm) of 40 leaves of a plant are given below as a "more than or equal to" cumulative frequency table. Write the frequency of each class 0-10, 10-20, …, 40-50, and find the modal class and the median class.

Answer.

1. Frequencies: $40 - 35 = 5$, $35 - 28 = 7$, $28 - 15 = 13$, $15 - 7 = 8$, and 7 for 40-50 — 1 mark
2. Modal class 20-30 (frequency 13); $\frac{n}{2} = 20$, 'less than' cumulative frequencies 5, 12, 25, …, so the median class is 20-30 — 1 mark

Question 5 (2 marks)

The mean of 20 observations was found to be 15. Later it was found that one observation, 25, had been wrongly copied as 45. Find the correct mean.

Answer.

1. Wrong total $= 20 \times 15 = 300$; correct total $= 300 - 45 + 25 = 280$ — 1 mark
2. Correct mean $= \frac{280}{20} = 14$ — 1 mark

Question 6 (3 marks)

The distance (in km) covered in a day by 50 food-delivery riders in Bengaluru is given below. Find the mean distance.

Answer.

1. Class marks 10, 30, 50, 70, 90; take $a = 50$, $h = 20$, $u_i = \frac{x_i - 50}{20}$: $-2, -1, 0, 1, 2$ — 1 mark
2. $\sum f_i u_i = -14 - 12 + 0 + 10 + 12 = -4$, $\sum f_i = 50$ — 1 mark
3. Mean $= 50 + \frac{-4}{50} \times 20 = 50 - 1.6 = 48.4$ km — 1 mark

Question 7 (3 marks)

The ages (in years) of 70 people who came for a free eye check-up camp in Nashik are given below. Find the mode of the ages.

Answer.

1. Modal class 30-40: $l = 30$, $f_1 = 20$, $f_0 = 16$, $f_2 = 14$, $h = 10$ — 1 mark
2. Mode $= 30 + \frac{20 - 16}{40 - 16 - 14} \times 10$ — 1 mark
3. $= 30 + \frac{4}{10} \times 10 = 34$ years — 1 mark

Question 8 (3 marks)

The number of hours that 40 students spent on sports in a week is given below. Find the median.

Answer.

1. Cumulative frequencies: 5, 14, 26, 34, 40; $\frac{n}{2} = 20$ — 1 mark
2. Median class 10-15: $l = 10$, $cf = 14$, $f = 12$, $h = 5$ — 1 mark
3. Median $= 10 + \frac{20 - 14}{12} \times 5 = 10 + 2.5 = 12.5$ hours — 1 mark

Question 9 (3 marks)

The waiting time (in minutes) of customers at a bank counter is given below. If the mean waiting time is 26 minutes, find the missing frequency $x$.

Answer.

1. Class marks 5, 15, 25, 35, 45; $\sum f_i = 32 + x$ — 1 mark
2. $\sum f_i x_i = 30 + 120 + 25x + 420 + 270 = 840 + 25x$ — 1 mark
3. $\frac{840 + 25x}{32 + x} = 26$ gives $840 + 25x = 832 + 26x$, so $x = 8$ — 1 mark

Question 10 (3 marks)

The scores (out of 100) of teams in an inter-school quiz are given below. The mode of the scores is 52. Find the missing frequency $x$.

Answer.

1. Mode 52 lies in 40-60, so this is the modal class: $l = 40$, $f_1 = x$, $f_0 = 16$, $f_2 = 20$, $h = 20$ — 1 mark
2. $40 + \frac{x - 16}{2x - 36} \times 20 = 52$, so $\frac{x - 16}{2x - 36} \times 20 = 12$ — 1 mark
3. $20x - 320 = 24x - 432$, so $4x = 112$ and $x = 28$ — 1 mark

Long Answer and Case-Based Questions

Question 11 (5 marks)

The ages (in years) of the people who joined a heritage walk in Hampi are given below. The mean age is 39 years. Find the missing frequency $x$, and then find the mode of the ages.

Answer.

Model answer:

Missing frequency. The class marks are 10, 20, 30, 40, 50 and 60.

Mean $= \frac{2180 + 20x}{52 + x} = 39$, so $2180 + 20x = 2028 + 39x$, which gives $19x = 152$ and $x = 8$. (So 60 people joined the walk.)

Mode. With $x = 8$, the highest frequency is 16, so the modal class is 35-45, with $l = 35$, $f_1 = 16$, $f_0 = 10$, $f_2 = 12$, $h = 10$.

Mode $= 35 + \frac{16 - 10}{32 - 10 - 12} \times 10 = 35 + \frac{6}{10} \times 10 = 41$ years.

Marking scheme:

1. Class marks 10, 20, 30, 40, 50, 60; $\sum f_i = 52 + x$ — 1 mark
2. $\sum f_i x_i = 40 + 20x + 300 + 640 + 600 + 600 = 2180 + 20x$ — 1 mark
3. $\frac{2180 + 20x}{52 + x} = 39$ gives $2180 + 20x = 2028 + 39x$, so $19x = 152$ and $x = 8$ — 1 mark
4. Modal class 35-45: $l = 35$, $f_1 = 16$, $f_0 = 10$, $f_2 = 12$, $h = 10$ — 1 mark
5. Mode $= 35 + \frac{16 - 10}{32 - 10 - 12} \times 10 = 35 + 6 = 41$ years — 1 mark

Question 12 (5 marks)

The scores (out of 70) of 70 players in an archery selection trial are given below. The median score is 44. Find the missing frequencies $p$ and $q$.

Answer.

Model answer:

Total frequency: $4 + 9 + p + 20 + q + 7 = 70$, so $40 + p + q = 70$ and $p + q = 30$.

The median, 44, lies in the class 40-50, so 40-50 is the median class, with $l = 40$, $h = 10$, $f = 20$, and $cf = 4 + 9 + p = 13 + p$ for the classes before it. Also $\frac{n}{2} = 35$.

Median $= l + \frac{\frac{n}{2} - cf}{f} \times h$, so $44 = 40 + \frac{35 - 13 - p}{20} \times 10 = 40 + \frac{22 - p}{2}$.

Then $\frac{22 - p}{2} = 4$, so $22 - p = 8$ and $p = 14$. From $p + q = 30$, $q = 16$.

Check: the cumulative frequencies are 4, 13, 27, 47, …, and 35 lies between 27 and 47, so 40-50 is indeed the median class.

Marking scheme:

1. $4 + 9 + p + 20 + q + 7 = 70$, so $p + q = 30$ — 1 mark
2. Median 44 lies in 40-50: $l = 40$, $f = 20$, $cf = 13 + p$, $h = 10$, $\frac{n}{2} = 35$ — 1 mark
3. $44 = 40 + \frac{35 - (13 + p)}{20} \times 10$ — 1 mark
4. $4 = \frac{22 - p}{2}$, so $22 - p = 8$ and $p = 14$ — 1 mark
5. $q = 30 - 14 = 16$ — 1 mark

Question 13 (5 marks)

The heights (in cm) of 80 saplings in a nursery in Dehradun are given below. Find the mean height using the step-deviation method, and also find the median height.

Answer.

Model answer:

Mean. Take $a = 37.5$ and $h = 5$, so $u_i = \frac{x_i - 37.5}{5}$.

Mean $= 37.5 + \frac{28}{80} \times 5 = 37.5 + 1.75 = 39.25$ cm.

Median. $\frac{n}{2} = 40$. The first cumulative frequency greater than 40 is 46, so the median class is 35-40, with $l = 35$, $cf = 26$, $f = 20$, $h = 5$.

Median $= 35 + \frac{40 - 26}{20} \times 5 = 35 + 3.5 = 38.5$ cm.

Marking scheme:

1. $a = 37.5$, $h = 5$, $u_i = -2, -1, 0, 1, 2, 3$; $\sum f_i u_i = -8 - 22 + 0 + 17 + 20 + 21 = 28$ — 1 mark
2. Mean $= 37.5 + \frac{28}{80} \times 5 = 37.5 + 1.75 = 39.25$ cm — 1 mark
3. Cumulative frequencies: 4, 26, 46, 63, 73, 80; $\frac{n}{2} = 40$ — 1 mark
4. Median class 35-40: $l = 35$, $cf = 26$, $f = 20$, $h = 5$ — 1 mark
5. Median $= 35 + \frac{40 - 26}{20} \times 5 = 35 + 3.5 = 38.5$ cm — 1 mark

Question 14 (4 marks)

A fitness centre in Guwahati recorded the number of push-ups each of its 60 members could do in one minute.

(i) Write the modal class. (1 mark)

Answer.

1. 10-20, the class with the highest frequency (18) — 1 mark

(ii) How many members did 30 or more push-ups in a minute? (1 mark)

Answer.

1. $12 + 9 = 21$ members — 1 mark

(iii) Find the mean number of push-ups. (2 marks)

Answer.

1. $\sum f_i x_i = 30 + 270 + 375 + 420 + 405 = 1500$ — 1 mark
2. Mean $= \frac{1500}{60} = 25$ — 1 mark

OR

(iii) Find the mode of the data. (2 marks)

Answer.

1. Modal class 10-20: $l = 10$, $f_1 = 18$, $f_0 = 6$, $f_2 = 15$, $h = 10$ — 1 mark
2. Mode $= 10 + \frac{18 - 6}{36 - 6 - 15} \times 10 = 10 + 8 = 18$ — 1 mark