MathsICSE Class 10 · Histograms and Ogive

Median and Quartiles from an Ogive, with the Interquartile Range (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

On a less than ogive with total frequency $N$, mark $\frac{N}{2}$ on the cumulative frequency axis, draw a line across to the curve and down to the x-axis: that is the median. Do the same at $\frac{N}{4}$ for the lower quartile $Q_1$ and $\frac{3N}{4}$ for $Q_3$. The interquartile range is $Q_3 - Q_1$.

Go across from N/4, N/2 and 3N/4 to the curve, then straight down to read Q1, median and Q3; IQR is the gap Q3 − Q1.

Illustration drawn for QuestionBuddy.

Understand Median and Quartiles from an Ogive

The CISCE 2027 syllabus asks you to find the median, lower quartile and upper quartile (and similar values) from the ogive, and to calculate the interquartile range. Both recent ogive questions asked for these read-offs (2024 Q9(iii), 2025 Q10(iii)).

What the quartiles are. Quartiles split ordered data into four equal parts:

$Q_{1}$ (lower quartile): a quarter of the data lie below it;
$Q_{2}$ = the median: half lie below it;
$Q_{3}$ (upper quartile): three quarters lie below it.

Positions on the ogive (ICSE uses these for grouped data):

$Q_{1}$ at the $\frac{N}{4}$ th observation;
median at the $\frac{N}{2}$ th observation;
$Q_{3}$ at the $\frac{3 N}{4}$ th observation.

Reading. Mark the position on the y-axis (c.f.), draw a horizontal line to the curve, then a perpendicular down to the x-axis. Read the value and write it on the graph. Show every line; the lines carry marks.

Interquartile range. $IQR = Q_{3} - Q_{1}$ . It is the spread of the middle half of the data: half of all observations lie between $Q_{1}$ and $Q_{3}$ . A smaller IQR means the middle values are more bunched, so performance is more consistent. (Half of the IQR is sometimes called the semi-interquartile range.)

Other positions ('etc.'). The same method gives any position: for example, the value exceeded by the top 10% is read at c.f. $0.9 N$ .

Quartile classes from a printed ogive. The median class is the class in which the c.f. first reaches $\frac{N}{2}$ ; the lower quartile class is where it first reaches $\frac{N}{4}$ . If the c.f. reaches the position exactly at a boundary, that observation is the last one of the class ending there. Write classes in full (₹6000–₹8000, not 6–8).

Checking a reading. Inside a class, the curve is close to a straight line, so value $\approx L + \frac{position - c.f. before}{f} \times h$ . Use this only to check that your graph reading is sensible.

Median and Quartiles from an Ogive: formulas to remember

Median position: $\frac{N}{2} th observation$; Read across, then down.
Quartile positions: $Q_{1} : \frac{N}{4} th, Q_{3} : \frac{3 N}{4} th$; $N$ = total frequency.
Interquartile range: $IQR = Q_{3} - Q_{1}$; Spread of the middle 50%.
Straight-line check inside a class: $L + \frac{position - c.f. before}{f} \times h$; For checking a graph reading.

How to solve Median and Quartiles from an Ogive questions

Positions first, then lines on the graph, then values, then the IQR.

1Find $N$ (the last c.f.) and write the positions $\frac{N}{4}$ , $\frac{N}{2}$ and $\frac{3 N}{4}$ .
2Mark each position on the c.f. axis using the given scale.
3Draw a horizontal line from each mark to the curve, then a perpendicular to the x-axis.
4Read and label each value on the graph: $Q_{1}$ , median, $Q_{3}$ .
5Subtract to get the IQR; give the unit.
6If a class is asked (median class, quartile class), name the class in full instead of a value.

Median and Quartiles from an Ogive 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.

Spread of study time

Suppose 40 students record their daily study time: 0–1 hour (2), 1–2 (10), 2–3 (14), 3–4 (10), 4–5 (4).

From the ogive, estimate the median, the quartiles and the IQR.

Try it first, then show the solutionSolution

c.f.: 2, 12, 26, 36, 40. Positions: 10th, 20th, 30th.
$Q_{1} \approx 1 + \frac{8}{10} = 1.8$ h; median $\approx 2 + \frac{8}{14} \approx 2.57$ h; $Q_{3} \approx 3 + \frac{4}{10} = 3.4$ h.
IQR $\approx 3.4 - 1.8 = 1.6$ h.

Answer: Median ≈ 2.57 h; IQR ≈ 1.6 h

Memory hook: The middle half of the class studies between 1.8 and 3.4 hours.

Choosing the more reliable driver

Imagine a courier company compares two drivers. Driver P's delivery times have $Q_{1} = 29$ and $Q_{3} = 35$ minutes; driver Q's have $Q_{1} = 22$ and $Q_{3} = 37$ minutes.

Which driver is more consistent?

Try it first, then show the solutionSolution

IQR of P $= 35 - 29 = 6$ minutes; IQR of Q $= 37 - 22 = 15$ minutes.
The smaller IQR means P's middle half of deliveries vary less.

Answer: Driver P

Memory hook: Small IQR, steady performer.

Tricks to remember Median and Quartiles from an Ogive

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

Quarter, half, three-quarters
$Q_{1}$ , median, $Q_{3}$ sit at a quarter, a half and three quarters of the way up the c.f. axis.
Use it when: Finding the three positions.
Across, then down
Positions go on the y-axis; answers come from the x-axis.
Use it when: Avoiding reading a c.f. as the answer.
IQR = the middle half
Cut off the bottom quarter and the top quarter; the width of what is left is the IQR.
Use it when: Interpreting spread.

Median and Quartiles from an Ogive: solved examples

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

Example 1: Quartile position · 1 mark · MCQBoard level

For a distribution with $N = 120$ , the upper quartile is read on the ogive at the:
(a) 30th observation
(b) 60th observation
(c) 90th observation
(d) 100th observation

Try it first, then show the solutionSolution

$\frac{3 N}{4} = \frac{3 \times 120}{4} = 90$ .

Answer: (c)

Example 2: Median, quartiles and IQR · 4 marks · Short answerBoard level

The masses (kg) of 60 students are given. Draw an ogive (2 cm = 5 kg, 2 cm = 10 students) and use it to estimate the median, the quartiles and the interquartile range.

Mass	30–35	35–40	40–45	45–50	50–55	55–60
Students	5	9	16	15	10	5

Try it first, then show the solutionSolution

c.f.: 5, 14, 30, 45, 55, 60. Plot (30, 0), (35, 5), …, (60, 60) with a kink; join free-hand.
Median: 30th observation; the curve passes through (45, 30), so median $= 45$ kg.
$Q_{1}$ : 15th; just past 40, about 40.3 kg. $Q_{3}$ : 45th; the curve passes through (50, 45), so $Q_{3} = 50$ kg.
IQR $\approx 50 - 40.3 = 9.7$ kg.

Answer: Median 45 kg; $Q_{1}$ ≈ 40.3 kg; $Q_{3}$ = 50 kg; IQR ≈ 9.7 kg

Example 3: Quartile classes from a printed ogive · 4 marks · Short answerHarder board level

A less than ogive of the monthly electricity use (units) of some homes passes through (0, 0), (100, 8), (200, 20), (300, 44), (400, 62), (500, 72) and (600, 80). Find:
(a) the number of homes;
(b) the median class;
(c) the lower quartile class;
(d) the upper quartile class.

Try it first, then show the solutionSolution

(a) Last point: $N = 80$ homes.
(b) 40th home: c.f. passes 40 between 20 and 44, so the class is 200–300 units.
(c) 20th home: c.f. reaches 20 exactly at 200, so the 20th home is the last one in 100–200 units.
(d) 60th home: c.f. passes 60 between 44 and 62: 300–400 units.

Answer: (a) 80 (b) 200–300 (c) 100–200 (d) 300–400

Example 4: Seven classes, all read-offs · 5 marks · Long answerHarder board level

The daily water use (litres) of 100 homes is shown. From an ogive, estimate the median, $Q_{1}$ , $Q_{3}$ and the IQR to 1 decimal place, and the number of homes using more than 110 litres.

Litres	0–20	20–40	40–60	60–80	80–100	100–120	120–140
Homes	4	10	18	26	22	14	6

Try it first, then show the solutionSolution

c.f.: 4, 14, 32, 58, 80, 94, 100. Positions 25th, 50th, 75th.
Median (50th) in 60–80: $\approx 60 + \frac{18}{26} \times 20 \approx 73.8$ L.
$Q_{1}$ (25th) in 40–60: $\approx 40 + \frac{11}{18} \times 20 \approx 52.2$ L. $Q_{3}$ (75th) in 80–100: $\approx 80 + \frac{17}{22} \times 20 \approx 95.5$ L.
IQR $\approx 95.45 - 52.22 \approx 43.2$ L.
At 110, c.f. $\approx 80 + 7 = 87$ , so more than 110 L: $100 - 87 = 13$ homes.

Answer: Median ≈ 73.8 L; $Q_{1}$ ≈ 52.2 L; $Q_{3}$ ≈ 95.5 L; IQR ≈ 43.2 L; about 13 homes

Example 5: Comparing two groups · 3 marks · Short answerHOTS / competency

From their ogives, class A's test scores have median 52, $Q_{1} = 44$ , $Q_{3} = 61$ ; class B's have median 55, $Q_{1} = 38$ , $Q_{3} = 70$ . Which class did better on average, and which was more consistent? Give reasons.

Try it first, then show the solutionSolution

Median: B (55) is higher than A (52), so B did slightly better on average.
IQR of A $= 61 - 44 = 17$ ; IQR of B $= 70 - 38 = 32$ .
A's middle half is much narrower, so A was more consistent.

Answer: B higher on average; A more consistent

Example 6: Case study: a missing frequency from the median · 5 marks · Case-basedHOTS / competency

The table below has one unknown frequency $p$ . On its ogive, the median is read as 36.

Class	10–20	20–30	30–40	40–50	50–60
Frequency	6	10	$p$	12	7

(a) Taking the median class as 30–40, find $p$ using the straight-line check.
(b) Find $Q_{1}$ and $Q_{3}$ and the IQR, to 2 decimal places.

Try it first, then show the solutionSolution

(a) $N = 35 + p$ ; c.f. before 30–40 is 16. $36 = 30 + \frac{\frac{35 + p}{2} - 16}{p} \times 10$ , so $\frac{3 + p}{2 p} = 0.6$ , $3 + p = 1.2 p$ and $p = 15$ . Check: $N = 50$ , 25th in 30–40 ✓
(b) c.f.: 6, 16, 31, 43, 50. $Q_{1}$ (12.5th) in 20–30: $20 + \frac{6.5}{10} \times 10 = 26.5$ .
$Q_{3}$ (37.5th) in 40–50: $40 + \frac{6.5}{12} \times 10 \approx 45.42$ . IQR $\approx 45.42 - 26.5 = 18.92$ .

Answer: (a) $p = 15$ (b) $Q_{1}$ = 26.5, $Q_{3}$ ≈ 45.42, IQR ≈ 18.92

Where students lose marks in Median and Quartiles from an Ogive

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

Mistake: No read-off lines on the graph
The lines show the examiner where each value came from.
Wrong working
Writing 'median = 45' with no lines drawn
Right working
Horizontal from c.f. 30 to the curve, perpendicular to the x-axis, '45' written at the foot
Do this: Draw and label a pair of lines for every value you read.
Usually costs: Reported in the CISCE examiners' comments on 2024 Q9(iii).
Mistake: Giving a value when a class is asked
The median class is an interval; the median is a number.
Wrong working
Median class: ₹7000
Right working
Median class: ₹6000–₹8000
Do this: Check whether the question says 'class'.
Usually costs: Reported in the CISCE examiners' comments on 2025 Q10(iii).
Mistake: Shortened class names and misread totals
The scale may be in thousands; the total is the last c.f., not the top of the axis.
Wrong working
Total 80 (top of the axis); class ₹6–₹8
Right working
Total 75 (last point); class ₹6000–₹8000
Do this: Read the last plotted point and convert the scale back to full values.
Usually costs: Reported in the CISCE examiners' comments on 2025 Q10(iii).
Mistake: Using $\frac{N + 1}{2}$ on an ogive
For grouped data read from an ogive, ICSE uses $\frac{N}{2}$ , $\frac{N}{4}$ , $\frac{3 N}{4}$ .
Wrong working
Median at the 30.5th observation for $N = 60$
Right working
Median at the 30th observation
Do this: Keep $\frac{N + 1}{2}$ for raw data only.
Usually costs: A slightly wrong reading and the accuracy mark.

Exam tips

Asked in 2025 Q10(iii) (4 marks): a printed ogive of factory salaries; total workers, median class, lower quartile class and the number between ₹6000 and ₹10000 were asked.
Asked in 2024 Q9(iii) (4 marks): ogive of the ages of 100 policy holders; median about 42 years.
Examiners accept a small range for graph readings, so a neat curve and clear lines matter more than extra decimals.
State the IQR with its unit and, if asked to interpret, say what a small or large IQR means.

Practice questions on Median and Quartiles from an Ogive

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 $Q_{1} = 31.5$ and $Q_{3} = 58$ , the interquartile range is:
(a) 13.25
(b) 26.5
(c) 44.75
(d) 89.5
Hint
$Q_{3} - Q_{1}$ .
Answer
(b)
Q2 · 1 mark · Short answerBoard level
For $N = 64$ , at which observations are $Q_{1}$ , the median and $Q_{3}$ read on the ogive?
Hint
$\frac{N}{4}$ , $\frac{N}{2}$ , $\frac{3 N}{4}$ .
Answer
16th, 32nd and 48th.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Half of all the observations lie between the lower and upper quartiles.
Reason (R): The lower and upper quartiles are read at the $\frac{N}{4}$ th and $\frac{3 N}{4}$ th observations.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
$\frac{3 N}{4} - \frac{N}{4}$ .
Answer
(c)
Q4 · 4 marks · Short answerHarder board level
The heights (cm) of 40 seedlings are shown. From an ogive, estimate the median (to 2 decimal places), $Q_{1}$ , $Q_{3}$ and the IQR.
Height 0–4 4–8 8–12 12–16 16–20
Seedlings 3 8 14 10 5
Hint
Positions 10th, 20th, 30th.
Answer
c.f. 3, 11, 25, 35, 40. Median $\approx 8 + \frac{9}{14} \times 4 \approx 10.57$ cm; $Q_{1} \approx 4 + \frac{7}{8} \times 4 = 7.5$ cm; $Q_{3} \approx 12 + \frac{5}{10} \times 4 = 14$ cm; IQR $\approx 6.5$ cm.
Q5 · 3 marks · Short answerHarder board level
An ogive of the minutes by which 60 trains were late passes through (0, 0), (5, 6), (10, 18), (15, 38), (20, 52) and (25, 60). State the median class and the upper quartile class, and find how many trains were at least 10 but less than 20 minutes late.
Hint
Median at 30th, $Q_{3}$ at 45th.
Answer
Median class 10–15 minutes; upper quartile class 15–20 minutes; $52 - 18 = 34$ trains.
Q6 · 4 marks · Short answerHarder board level
From an ogive of the data below, estimate the median and the upper quartile.
Class 1–10 11–20 21–30 31–40 41–50
Frequency 6 12 20 8 4
Hint
Plot at the upper boundaries 10.5, 20.5, ….
Answer
c.f. 6, 18, 38, 46, 50. Median (25th) $\approx 20.5 + \frac{7}{20} \times 10 = 24$ ; $Q_{3}$ (37.5th) $\approx 20.5 + \frac{19.5}{20} \times 10 = 30.25$ .
Q7 · 2 marks · Short answerHOTS / competency
Using the seedling ogive above ( $N = 40$ ), find the height exceeded by only the tallest 10% of the seedlings.
Hint
Read at c.f. $0.9 \times 40 = 36$ .
Answer
36th observation in 16–20: $16 + \frac{1}{5} \times 4 = 16.8$ cm.

Height	0–4	4–8	8–12	12–16	16–20
Seedlings	3	8	14	10	5

Class	1–10	11–20	21–30	31–40	41–50
Frequency	6	12	20	8	4

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.Position of $Q_{1}$ ?Show answer
$\frac{N}{4}$ th
2.Position of the median on an ogive?Show answer
$\frac{N}{2}$ th
3.IQR?Show answer
$Q_{3} - Q_{1}$
4.What fraction of the data lies between $Q_{1}$ and $Q_{3}$ ?Show answer
Half.
5.Which axis do you read the answer from?Show answer
The x-axis (values); positions are marked on the y-axis.

Median and Quartiles from an Ogive: frequently asked questions

How do you find the median from an ogive?

Find N, mark N/2 on the cumulative frequency axis, draw a horizontal line to the ogive and then a perpendicular to the x-axis. The x-axis reading is the median.

How do you find the interquartile range from an ogive?

Read the lower quartile at N/4 and the upper quartile at 3N/4 in the same way as the median, then subtract: IQR = Q₃ − Q₁.

What does the interquartile range tell you?

It is the spread of the middle half of the data. A smaller interquartile range means the middle values are closer together, so the data are more consistent.