MathsCBSE Class 10 · Statistics

Median of Grouped Data for Class 10 Statistics

Q: Is the median class the same as the modal class?

Not necessarily. The modal class has the highest frequency, while the median class is chosen from cumulative frequency. They often differ.

Q: What if a cumulative frequency equals exactly n/2?

NCERT's rule picks the class whose cf is greater than and nearest to n/2, which is the next class. If you use the class whose cf equals n/2 instead, the formula still gives the same median (its upper limit), so the median value does not change. Follow the NCERT wording when a question asks you to name the median class.

Q: Do I need a graph to find the median?

No. For the board exam, use the cumulative frequency table and the median formula.

Q: Why is the grouped median only an estimate?

The formula assumes observations are spread evenly inside the median class. Real values may not be, so the answer is an approximation.

Updated 14 October 2026 · Fact-checked

The median of grouped data is the value that splits the data into two equal halves. You find it by making a cumulative frequency column, locating the median class at position n/2 and applying the median formula. Board questions ask you to find the median, convert cumulative tables or find missing frequencies from a given median.

The median class is the first class whose cf passes n/2; the median sits 7 of 15 observations into it: 20 + (7/15) × 10 ≈ 24.7.

Illustration drawn for QuestionBuddy.

Understand Median of Grouped Data

The median is the middle value of data arranged in order. Half the observations lie below it and half lie above it. In ungrouped data you simply count to the middle. In grouped data you cannot see individual values, only classes and their frequencies, so you need a method to estimate where the middle value sits.

First you find where the middle is. Let $n = \sum f_{i}$ . The middle observation is at position $\frac{n}{2}$ . To see which class holds it, you build a cumulative frequency (cf) column. Each cf is the running total of frequencies up to that class. The median class is the first class whose cf is greater than $\frac{n}{2}$ (the cf greater than and nearest to it).

Next you find how far into that class the median lies. The formula is $Median = l + \frac{\frac{n}{2} - c f}{f} \times h$ . Up to the lower limit $l$ you have already counted $c f$ observations (the cf of the class before the median class). You need $\frac{n}{2} - c f$ more. The median class has $f$ observations spread over width $h$ . If you assume they are spread evenly, then moving $\frac{n}{2} - c f$ of them takes you that fraction of the width beyond $l$ . This even-spread assumption is why the answer is an estimate.

The method needs continuous classes. If classes are inclusive, like 1–10, 11–20, close the gaps first by subtracting $0.5$ from each lower limit and adding $0.5$ to each upper limit. Otherwise $l$ will be wrong. Data may also arrive as a 'less than' or 'more than' cumulative table; subtract consecutive values to recover ordinary frequencies.

The median class is chosen by cumulative frequency, not by highest frequency, so it can differ from the modal class. The median often pairs with the mean and mode in one question, and through $3 Median = Mode + 2 Mean$ you can find one average from the other two. A quick check on every answer: the median must lie inside the median class.

Median of Grouped Data: formulas to remember

Median of grouped data: $Median = l + \frac{\frac{n}{2} - c f}{f} \times h$; l = lower limit of median class, cf = cumulative frequency of the class before it, f = frequency of median class, h = class width.
Median position: $\frac{n}{2}, n = \sum f_{i}$; Use n/2 for grouped data, not (n+1)/2.
Median class rule: $first class whose c f is greater than (and nearest to) \frac{n}{2}$; This is the NCERT wording. If a cf equals n/2 exactly, the median is the upper limit of that class, and taking either neighbouring class in the formula gives the same median value.
Frequency from cumulative frequency: $f_{i} = c f_{i} - c f_{i - 1}$; Works for less than tables top-down; for more than tables subtract each value from the one above it.
Inclusive to continuous classes: $adjustment = \frac{gap}{2}$; Subtract from lower limits, add to upper limits. Usually 0.5.
Location check: $l \leq Median \leq l + h$; If your median is outside the median class, you used a wrong cf or class.
Empirical relation: $3 Median = Mode + 2 Mean$; Rearranged: Mode = 3 Median − 2 Mean, or Mean = (3 Median − Mode)/2. It holds only approximately, for moderately skewed data, so it gives an estimate, not the exact value. Use it when a question gives two averages, or asks you to use the relation.

How to solve Median of Grouped Data questions

This routine handles every median question, from a plain table to two missing frequencies.

1Make the classes continuous if they are inclusive, and convert any cumulative table back to ordinary frequencies.
2Draw a table with columns: class, frequency $f$ , cumulative frequency $c f$ .
3Find $n = \sum f_{i}$ and write $\frac{n}{2}$ .
4Pick the median class: the first class whose $c f$ is greater than $\frac{n}{2}$ .
5Write $l$ , $c f$ (of the previous class), $f$ and $h$ on separate lines.
6Write the formula, then substitute and simplify; give units.
7If frequencies are missing, get one equation from the total and one from the formula, after fixing the median class from the given median.
8Check that the median lies inside the median class.

Median of Grouped Data in real life

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.

Median of Grouped Data

Suppose your class records last month's mobile data use (in GB) to suggest a shared plan size. When the sheet is collected, one count has been smudged, but the class representative remembers that the median use was 5.2 GB.

Data use (GB): 0–2: 6, 2–4: 12, 4–6: $x$ , 6–8: 14, 8–10: 8 students. The median is 5.2 GB. Find $x$ , the number of students in the class, and the number of heavy users (6 GB or more).

Try it first, then show the solutionSolution

Median 5.2 lies in 4–6, so the median class is 4–6.
$n = 40 + x$ , $\frac{n}{2} = 20 + \frac{x}{2}$ ; $l = 4$ , $c f = 6 + 12 = 18$ , $f = x$ , $h = 2$ .
$5.2 = 4 + \frac{20 + \frac{x}{2} - 18}{x} \times 2 = 4 + \frac{4 + x}{x}$ .
$1.2 x = 4 + x$ , so $0.2 x = 4$ and $x = 20$ .
Class size $n = 40 + 20 = 60$ ; check: cf 6, 18, 38, so $\frac{n}{2} = 30$ lies in 4–6, and median $= 4 + \frac{30 - 18}{20} \times 2 = 5.2$ GB.
Heavy users $= 14 + 8 = 22$ .

Answer: $x = 20$ ; there are 60 students in the class; 22 students are heavy users.

Memory hook: A remembered median can rebuild a smudged count.

Median of Grouped Data

Imagine your school trip organiser notes how long 50 students took to reach the meeting point. The register keeps 'at least' totals: how many took 0 minutes or more, 20 minutes or more, and so on. One total is smudged.

Took at least 0 min: 50, at least 20 min: 45, at least 40 min: $k$ , at least 60 min: 15, at least 80 min: 6, at least 100 min: 0. The organiser remembers that the 40–60 minute group had twice as many students as the 20–40 minute group. Find $k$ and the median travel time.

Try it first, then show the solutionSolution

20–40 group $= 45 - k$ ; 40–60 group $= k - 15$ .
$k - 15 = 2 (45 - k)$ , so $3 k = 105$ and $k = 35$ .
Frequencies: 0–20: 5, 20–40: 10, 40–60: 20, 60–80: 9, 80–100: 6.
Less than cf: 5, 15, 35, 44, 50; $\frac{n}{2} = 25$ , so the median class is 40–60.
$l = 40$ , $c f = 15$ , $f = 20$ , $h = 20$ .
Median $= 40 + \frac{25 - 15}{20} \times 20 = 40 + 10 = 50$ minutes.

Answer: $k = 35$ ; the median travel time is 50 minutes.

Memory hook: 'At least' totals: subtract down the list, then find the middle traveller.

Tricks to remember Median of Grouped Data

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

L-C-F-H order
'Look, Count, Find, Hop': l is where you start, cf is what you have counted, f is the crowd in the class, h is the hop width.
Use it when: Write the four values in this order before substituting.
Before, not inside
cf in the formula is the total of the class BEFORE the median class, the people already passed.
Use it when: Stops you using the median class's own cf.
Half means n/2
Grouped data: halve n, no plus one.
Use it when: Locating the median class.
Stay in your class
The median must land between l and l + h.
Use it when: Instant check of every answer.

Median of Grouped Data: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Median from a frequency table · 3 marks · Short answerBoard level

The daily study time (in minutes) of 50 students is given. Find the median study time.
Time (min): 0–10, 10–20, 20–30, 30–40, 40–50
Students: 5, 11, 20, 10, 4

Try it first, then show the solutionSolution

cf column: 5, 16, 36, 46, 50. (1 mark)
$n = 50$ , $\frac{n}{2} = 25$ ; first cf greater than $25$ is 36, so median class is 20–30. (½ mark)
$l = 20$ , $c f = 16$ , $f = 20$ , $h = 10$ . (½ mark)
Median $= 20 + \frac{25 - 16}{20} \times 10 = 20 + 4.5 = 24.5$ . (1 mark)

Answer: Median study time = 24.5 minutes.

Example 2: Median with inclusive classes · 3 marks · Short answerBoard level

The scores of 60 players in a quiz are given. Find the median score.
Score: 1–10, 11–20, 21–30, 31–40, 41–50
Players: 6, 10, 20, 16, 8

Try it first, then show the solutionSolution

Gap is 1, so adjust by 0.5: 0.5–10.5, 10.5–20.5, 20.5–30.5, 30.5–40.5, 40.5–50.5. (1 mark)
cf: 6, 16, 36, 52, 60; $\frac{n}{2} = 30$ , median class 20.5–30.5. (½ mark)
$l = 20.5$ , $c f = 16$ , $f = 20$ , $h = 10$ . (½ mark)
Median $= 20.5 + \frac{30 - 16}{20} \times 10 = 20.5 + 7 = 27.5$ . (1 mark)

Answer: Median score = 27.5.

Example 3: Median from a 'more than' table · 3 marks · Short answerBoard level

Marks of 50 students: marks $\geq 0$ : 50, $\geq 20$ : 46, $\geq 40$ : 38, $\geq 60$ : 22, $\geq 80$ : 10, and nobody scored 100 or more. Find the median marks.

Try it first, then show the solutionSolution

Frequencies: 0–20: $50 - 46 = 4$ , 20–40: $46 - 38 = 8$ , 40–60: $38 - 22 = 16$ , 60–80: $22 - 10 = 12$ , 80–100: 10. (1 mark)
Less than cf: 4, 12, 28, 40, 50; $\frac{n}{2} = 25$ , median class 40–60. (½ mark)
$l = 40$ , $c f = 12$ , $f = 16$ , $h = 20$ . (½ mark)
Median $= 40 + \frac{25 - 12}{16} \times 20 = 40 + 16.25 = 56.25$ . (1 mark)

Answer: Median marks = 56.25.

Example 4: Two missing frequencies · 5 marks · Long answerHarder board level

The median of the distribution is 52 and total frequency is 50. Find $p$ and $q$ .
Class: 0–20, 20–40, 40–60, 60–80, 80–100
Frequency: 6, $p$ , 15, $q$ , 5

Try it first, then show the solutionSolution

Total: $6 + p + 15 + q + 5 = 50$ , so $p + q = 24$ . (1 mark)
cf: 6, $6 + p$ , $21 + p$ , $21 + p + q$ , 50. (½ mark)
Median 52 lies in 40–60, so median class is 40–60. (½ mark)
$l = 40$ , $c f = 6 + p$ , $f = 15$ , $h = 20$ , $\frac{n}{2} = 25$ . (½ mark)
$52 = 40 + \frac{25 - 6 - p}{15} \times 20$ . (1 mark)
$12 = \frac{4 ( 19 - p )}{3}$ , so $9 = 19 - p$ , $p = 10$ . (½ mark)
$q = 24 - 10 = 14$ . (½ mark)
Check: cf before class 16, median $= 40 + \frac{9}{15} \times 20 = 52$ . (½ mark)

Answer: $p = 10$ , $q = 14$ .

Example 5: Case study: bus shelter decision · 4 marks · Case-basedHOTS / competency

A town transport cell records the waiting times of passengers at a bus stop. One entry in the register is smudged and shown as $x$ . The median waiting time is known to be 13 minutes.
Wait (min): 0–5, 5–10, 10–15, 15–20, 20–25
Passengers: 8, 14, $x$ , 16, 12
(i) Write the median class, with a reason. (1 mark)
(ii) How many passengers waited less than 10 minutes? (1 mark)
(iii) Find $x$ . The cell builds a shelter if more than 60% of passengers wait 10 minutes or more. Should it build one? (2 marks)

Try it first, then show the solutionSolution

(i) The median 13 lies between 10 and 15, so the median class is 10–15. (1 mark)
(ii) Less than 10 minutes $= 8 + 14 = 22$ passengers. (1 mark)
(iii) $n = 50 + x$ , $\frac{n}{2} = 25 + \frac{x}{2}$ ; $l = 10$ , $c f = 22$ , $f = x$ , $h = 5$ . (½ mark)
$13 = 10 + \frac{25 + \frac{x}{2} - 22}{x} \times 5$ , so $3 = \frac{5 ( 6 + x )}{2 x}$ . (½ mark)
$6 x = 30 + 5 x$ , so $x = 30$ and $n = 80$ . Check: cf 8, 22, 52; $\frac{n}{2} = 40$ lies in 10–15. (½ mark)
Waited 10 minutes or more $= 30 + 16 + 12 = 58$ , and $\frac{58}{80} \times 100 = 72.5%$ , which is more than 60%, so yes, the shelter should be built. (½ mark)

Answer: (i) 10–15 (ii) 22 passengers (iii) $x = 30$ ; 72.5% of passengers wait 10 minutes or more, so the shelter should be built.

Example 6: Mode from the empirical relation · 3 marks · Short answerHarder board level

The daily distance (in km) cycled by 50 riders is given. The mean distance is 25.2 km. Find the median, and then use the empirical relation to estimate the mode.
Distance (km): 0–10, 10–20, 20–30, 30–40, 40–50
Riders: 5, 10, 20, 9, 6

Try it first, then show the solutionSolution

cf: 5, 15, 35, 44, 50; $\frac{n}{2} = 25$ ; first cf greater than $25$ is 35, so the median class is 20–30. (1 mark)
$l = 20$ , $c f = 15$ , $f = 20$ , $h = 10$ : Median $= 20 + \frac{25 - 15}{20} \times 10 = 25$ km. (1 mark)
$3 Median = Mode + 2 Mean$ gives Mode $= 3 (25) - 2 (25.2) = 75 - 50.4 = 24.6$ km. (1 mark)

Answer: Median = 25 km; Mode is about 24.6 km.

Example 7: MCQ: picking the median class from cumulative frequencies · 1 mark · MCQBoard level

For classes 0–10, 10–20, 20–30, 30–40, 40–50, the cumulative frequencies are 6, 14, 24, 31, 40. The median class is
(a) 10–20
(b) 20–30
(c) 30–40
(d) 40–50

Try it first, then show the solutionSolution

$n = 40$ , so $\frac{n}{2} = 20$ . The cumulative frequencies before and after 20 are 14 and 24, and the first cf greater than 20 is 24, which belongs to 20–30. (1 mark)

Answer: (b) 20–30

Example 8: Assertion-reason: median class versus modal class · 1 mark · Assertion-reasonBoard level

Assertion (A): For classes 0–10, 10–20, 20–30, 30–40 with frequencies 3, 9, 15, 13, the median class is 20–30.
Reason (R): The median class is always the class with the highest frequency.
(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

cf: 3, 12, 27, 40; $\frac{n}{2} = 20$ ; first cf greater than $20$ is 27, so the median class is 20–30. A is true. (½ mark)
The highest frequency fixes the modal class, not the median class; the median class comes from cumulative frequency. R is false. (½ mark)

Answer: (c) A is true but R is false.

Where students lose marks in Median of Grouped Data

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

Mistake: Using (n+1)/2
Ungrouped habit carried over.
Wrong working
$n = 50$ , position $= 25.5$ , then $\frac{n + 1}{2}$ put in formula.
Right working
Use $\frac{n}{2} = 25$ throughout.
Do this: Grouped data always uses n/2.
Usually costs: ½ to 1 mark
Mistake: Taking the highest-frequency class as median class
Confusing median with mode.
Wrong working
Frequencies 14, 5, 8, 7, 6 on 0–10 to 40–50 ( $n = 40$ ): the student picks 0–10 because 14 is highest, so median $= 0 + \frac{20 - 0}{14} \times 10 \approx 14.3$ , which is not even inside 0–10.
Right working
cf: 14, 19, 27, 34, 40; first cf greater than $20$ is 27, so the median class is 20–30. Median $= 20 + \frac{20 - 19}{8} \times 10 = 21.25$ .
Do this: Always draw the cf column first.
Usually costs: Up to all method marks
Mistake: Using cf of the median class itself
Misreading which cf the formula needs.
Wrong working
$20 + \frac{25 - 40}{20} \times 10 = 12.5$
Right working
$20 + \frac{25 - 20}{20} \times 10 = 22.5$
Do this: cf belongs to the class before.
Usually costs: 1 to 2 marks
Mistake: Not making inclusive classes continuous
Gaps look harmless.
Wrong working
Class 21–30: $l = 21$ , median $= 21 + 7 = 28$ .
Right working
Class 20.5–30.5: $l = 20.5$ , median $= 27.5$ .
Do this: Check for gaps before writing l.
Usually costs: 1 mark

Exam tips

Show the cf column clearly; it usually carries a mark.
Write l, cf, f and h on separate lines before substituting.
In missing-frequency questions, state the median class with a reason (the given median lies in it).
For 'less than' or 'more than' tables, first write the ordinary frequency table.
Finish with units and a one-line check that the median lies in its class.

Practice questions on Median of Grouped Data

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

Q1 · 2 marks · Very short answerBoard level
Find the median of the distribution.
Class: 10–20, 20–30, 30–40, 40–50, 50–60
Frequency: 4, 6, 10, 12, 8
Hint
A cf equal to n/2 picks that class.
Answer
cf: 4, 10, 20, 32, 40; $\frac{n}{2} = 20$ , median class 30–40. Median $= 30 + \frac{20 - 10}{10} \times 10 = 40$ .
Q2 · 3 marks · Short answerBoard level
The number of club members aged at least a given age is shown. Find the median age.
Aged 21 years or more: 50, 26 or more: 44, 31 or more: 34, 36 or more: 16, 41 or more: 6, 46 or more: 0.
(Ages are in whole years, so the classes are 21–25, 26–30, 31–35, 36–40, 41–45.)
Hint
Subtract consecutive 'or more' counts, then make the classes continuous.
Answer
Frequencies: 6, 10, 18, 10, 6. Continuous classes 20.5–25.5, 25.5–30.5, 30.5–35.5, 35.5–40.5, 40.5–45.5; cf: 6, 16, 34, 44, 50; $\frac{n}{2} = 25$ , median class 30.5–35.5. Median $= 30.5 + \frac{25 - 16}{18} \times 5 = 30.5 + 2.5 = 33$ years.
Q3 · 3 marks · Short answerHarder board level
The median of the distribution is 27. Find $k$ .
Class: 0–10, 10–20, 20–30, 30–40, 40–50
Frequency: 2, 8, $k$ , 12, 6
Hint
n = 28 + k; median class is 20–30 with cf = 10.
Answer
$27 = 20 + \frac{\frac{28 + k}{2} - 10}{k} \times 10$ gives $1.4 k = 8 + k$ , so $k = 20$ . Check: $n = 48$ , median $= 20 + \frac{14}{20} \times 10 = 27$ .
Q4 · 5 marks · Long answerHarder board level
Total frequency is 100 and median is 26. Find $a$ and $b$ .
Class: 0–10, 10–20, 20–30, 30–40, 40–50
Frequency: 10, $a$ , 25, $b$ , 15
Hint
a + b = 50; median class 20–30 with cf = 10 + a.
Answer
$26 = 20 + \frac{50 - 10 - a}{25} \times 10$ gives $15 = 40 - a$ , $a = 25$ , $b = 25$ . Check: median $= 20 + \frac{15}{25} \times 10 = 26$ .
Q5 · 3 marks · Short answerHOTS / competency
For classes 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 6, 14, 20, 6, 4, a student wrote: median class 20–30, $c f = 40$ , so median $= 20 + \frac{25 - 40}{20} \times 10 = 12.5$ . Find the error and the correct median.
Hint
Does 12.5 lie in 20–30?
Answer
The student used the median class's own cf (40) instead of the previous cf (20); 12.5 lies outside the class. Correct: $20 + \frac{25 - 20}{20} \times 10 = 22.5$ .
Q6 · 5 marks · Long answerHarder board level
Find the mean (by the step deviation method) and the median of the distribution.
Class: 0–10, 10–20, 20–30, 30–40, 40–50
Frequency: 6, 9, 20, 10, 5
Hint
Take a = 25 and h = 10 for the mean; build the cf column for the median.
Answer
Mean: $u_{i} = - 2, - 1, 0, 1, 2$ ; $\sum f_{i} u_{i} = - 12 - 9 + 0 + 10 + 10 = - 1$ ; mean $= 25 + \frac{- 1}{50} \times 10 = 24.8$ . Median: cf 6, 15, 35, 45, 50; $\frac{n}{2} = 25$ , median class 20–30; median $= 20 + \frac{25 - 15}{20} \times 10 = 25$ .
Q7 · 3 marks · Short answerHOTS / competency
Suppose a shop records the number of units sold per day over 50 days.
Units: 0–20, 20–40, 40–60, 60–80, 80–100
Days: 7, 13, 20, 6, 4
The mean is 44.8 units. The manager claims the most typical (modal) daily sale is above 45 units. Find the median and use the empirical relation to check the claim.
Hint
Find the median first, then use Mode = 3 Median − 2 Mean.
Answer
cf: 7, 20, 40, 46, 50; $\frac{n}{2} = 25$ , median class 40–60. Median $= 40 + \frac{25 - 20}{20} \times 20 = 45$ units. Mode $= 3 (45) - 2 (44.8) = 135 - 89.6 = 45.4$ units, which is above 45, so the claim is supported.
Q8 · 5 marks · Long answerHarder board level
The mean of the distribution is 25.4 and the total frequency is 50. Find $p$ and $q$ , then find the median. Show which class contains the $\frac{n}{2}$ th observation.
Class: 0–10, 10–20, 20–30, 30–40, 40–50
Frequency: 4, $p$ , 20, $q$ , 6
Hint
Use the total and the mean to get two equations; the median class is fixed only after p and q are known.
Answer
$p + q = 20$ . $\sum f_{i} x_{i} = 20 + 15 p + 500 + 35 q + 270 = 790 + 15 p + 35 q = 50 \times 25.4 = 1270$ , so $15 p + 35 q = 480$ . Putting $q = 20 - p$ : $- 20 p = - 220$ , so $p = 11$ , $q = 9$ . cf: 4, 15, 35, 44, 50; $\frac{n}{2} = 25$ is more than 15 and at most 35, so the median class is 20–30. Median $= 20 + \frac{25 - 15}{20} \times 10 = 25$ .
Q9 · 5 marks · Long answerHarder board level
The marks (whole numbers) of 60 students are grouped below. The median mark is 28.5.
Marks: 1–10, 11–20, 21–30, 31–40, 41–50
Students: 5, $x$ , 20, $y$ , 7
(i) Convert the classes to continuous form and name the median class, giving a reason. (ii) Find $x$ and $y$ . (iii) Find the mean by the assumed mean method. (iv) Is the mean above or below the median?
Hint
The median 28.5 tells you which continuous class is the median class; then use $x + y$ from the total and the median formula for $x$ .
Answer
(i) Classes 0.5–10.5, 10.5–20.5, 20.5–30.5, 30.5–40.5, 40.5–50.5. The median 28.5 lies in 20.5–30.5, so that is the median class. (1)
(ii) $5 + x + 20 + y + 7 = 60$ , so $x + y = 28$ . cf before the median class $= 5 + x$ , $\frac{n}{2} = 30$ , $f = 20$ , $h = 10$ : $28.5 = 20.5 + \frac{30 - ( 5 + x )}{20} \times 10$ , so $8 = \frac{25 - x}{2}$ , $x = 9$ and $y = 19$ . Check: cf 5, 14, 34, 53, 60, and 14 < 30 < 34. (2)
(iii) Class marks 5.5, 15.5, 25.5, 35.5, 45.5; take $a = 25.5$ , $d_{i}$ : $- 20, - 10, 0, 10, 20$ ; $f_{i} d_{i}$ : $- 100, - 90, 0, 190, 140$ ; $\sum f_{i} d_{i} = 140$ . Mean $= 25.5 + \frac{140}{60} \approx 27.83$ . (1½)
(iv) The mean (about 27.83) is a little below the median (28.5). (½)
Q10 · 4 marks · Case-basedHOTS / competency
Case study: In a puzzle contest, the organisers record how many of the 50 students finished in less than 10, 20, 30, 40 and 50 minutes. One entry is smudged.
Finished in less than 10 min: 4; less than 20 min: 12; less than 30 min: $k$ ; less than 40 min: 44; less than 50 min: 50.
The median time is 26.5 minutes.
(i) Write the frequencies of the five classes in terms of $k$ where needed. (1 mark)
(ii) Find $k$ . (2 marks)
(iii) Anu took 26 minutes. She says she was faster than at least half of the students. Is she right? What percentage of students finished in less than 30 minutes? (1 mark)
Hint
Each frequency is the difference of two 'less than' totals. The median 26.5 lies in 20–30, so the cf before that class is 12.
Answer
(i) 0–10: 4, 10–20: $12 - 4 = 8$ , 20–30: $k - 12$ , 30–40: $44 - k$ , 40–50: $50 - 44 = 6$ . (1)
(ii) $\frac{n}{2} = 25$ ; the median class is 20–30 with cf before it 12 and frequency $k - 12$ : $26.5 = 20 + \frac{25 - 12}{k - 12} \times 10$ , so $6.5 (k - 12) = 130$ , $k - 12 = 20$ , $k = 32$ . Check: frequencies 4, 8, 20, 12, 6 add to 50. (2)
(iii) Yes: 26 minutes is less than the median 26.5 minutes, so about half the students took longer than her. $\frac{32}{50} \times 100 = 64%$ finished in less than 30 minutes. (1)
Q11 · 1 mark · MCQBoard level
For classes 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 7, 9, 12, 14, 8, the value of cf used in the median formula is
(a) 7
(b) 16
(c) 28
(d) 12
Hint
Find n/2 first, then take the cf of the class before the median class.
Answer
(b) 16. cf: 7, 16, 28, 42, 50; $\frac{n}{2} = 25$ , median class 20–30, and the cf of the class before it is 16.
Q12 · 1 mark · MCQBoard level
If the mean of a distribution is 24 and its median is 26, its mode by the empirical relation is
(a) 28
(b) 30
(c) 22
(d) 25
Hint
Mode = 3 Median − 2 Mean.
Answer
(b) 30, since Mode $= 3 (26) - 2 (24) = 78 - 48 = 30$ .

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.Median position for grouped data?Show answer
$\frac{n}{2}$
2.How is the median class chosen?Show answer
The class whose cumulative frequency is greater than (and nearest to) n/2.
3.Which cf goes in the formula?Show answer
The cf of the class before the median class.
4.Median formula?Show answer
$l + \frac{\frac{n}{2} - c f}{f} \times h$
5.Adjustment for classes 1–10, 11–20?Show answer
Subtract 0.5 from lower limits, add 0.5 to upper limits.

Median of Grouped Data: frequently asked questions

Is the median class the same as the modal class?

Not necessarily. The modal class has the highest frequency, while the median class is chosen from cumulative frequency. They often differ.

What if a cumulative frequency equals exactly n/2?

NCERT's rule picks the class whose cf is greater than and nearest to n/2, which is the next class. If you use the class whose cf equals n/2 instead, the formula still gives the same median (its upper limit), so the median value does not change. Follow the NCERT wording when a question asks you to name the median class.

Do I need a graph to find the median?

No. For the board exam, use the cumulative frequency table and the median formula.

Why is the grouped median only an estimate?

The formula assumes observations are spread evenly inside the median class. Real values may not be, so the answer is an approximation.