Graphical Representation (Histograms and Ogives)

ICSE Class 10 · Mathematics · Chapter 23

What this chapter covers

Why data is drawn as a graph
Grouped data and class intervals
The histogram
Histograms from discontinuous classes and class marks
Finding the mode from a histogram
Cumulative frequency
The less than ogive
Reading the median from an ogive
Quartiles from an ogive
The inter-quartile range
Other readings from an ogive
Common mistakes and how to avoid them

Key points

Graphical representation shows the pattern in data and lets you estimate values; graph readings are approximate.
Classes must be continuous before any graph is drawn: convert inclusive classes using the adjustment factor, half the gap between consecutive classes.
When only class marks are given, the width is the difference between consecutive class marks; subtract and add half the width to get the classes.
A histogram has touching rectangles on the class intervals, with heights equal to the frequencies.
A kink on an axis shows that the scale does not start from zero.
The mode from a histogram: in the tallest rectangle, join its top right corner to the top right corner of the rectangle on its left, and its top left corner to the top left corner of the rectangle on its right; drop a perpendicular from the crossing point.
The cumulative frequency of a class counts all observations less than its upper boundary; the last one equals n.
A less than ogive plots cumulative frequency against upper boundary, starts at (lower boundary of the first class, 0) and is a smooth freehand curve.
From an ogive: median at (n)/(2), lower quartile Q₁ at (n)/(4), upper quartile Q₃ at (3n)/(4); go across to the curve, then down to the x-axis.
Inter-quartile range = Q₃ - Q₁, the spread of the middle half of the data.
"More than a value" is n minus the ogive reading at that value; "between two values" is the difference of the two readings.

Questions answered

Define a histogram. How is it different from a bar graph?

A histogram is a graph of a grouped frequency distribution in which each class is shown by a rectangle on the horizontal axis, with width equal to the class width and height proportional to its frequency. The rectangles touch each other because the classes are continuous, while the bars of a bar graph stand for separate categories and have gaps between them.

Convert the classes 5–9, 10–14, 15–19 into continuous classes.

The gap is 1, so the adjustment factor is 0.5. The classes become 4.5–9.5, 9.5–14.5 and 14.5–19.5.

The class marks of a distribution are 15, 25, 35, 45. Write the class intervals.

The width is 25 - 15 = 10, and half of it is 5. The classes are 10–20, 20–30, 30–40 and 40–50.

In a histogram, the rectangles over the classes 40–50, 50–60 and 60–70 have heights 8, 15 and 8, and 50–60 is the tallest. Without drawing, where will the mode lie?

The two neighbours are equally tall, so the crossing lines are symmetrical and the mode lies at the centre of the modal class, at 55.

Why is the mode estimated from a histogram nearer the lower boundary of the modal class when the class before it has a larger frequency than the class after it?

A larger frequency just before the modal class means the data crowds towards its left side. The crossing lines of the construction tilt towards that side, so the perpendicular meets the x-axis nearer the lower boundary.

Define the cumulative frequency of a class, and write the cumulative frequencies for the frequencies 6, 10, 14, 7, 3.

The cumulative frequency of a class is the sum of its frequency and the frequencies of all the classes before it. The cumulative frequencies are 6, 16, 30, 37 and 40.

For the classes 0–10, 10–20, 20–30 with frequencies 4, 9, 7, write the points you would plot for a less than ogive.

(0, 0), (10, 4), (20, 13) and (30, 20).

Why does a less than ogive begin with the point (lower boundary of the first class, 0)?

No observation is less than the lower boundary of the first class, so the cumulative frequency there is 0. Without this point the curve would not start from the x-axis and readings in the first class would be lost.

Past board exam questions

ICSE 2027 sample paper · 5 marks

The distribution given below represents marks obtained by 200 students in a Mathematics Examination with maximum marks 80. | Marks | No. of students | |---|---| | 0 − 10 | 8 | | 10 − 20 | 12 | | 20 − 30 | 16 | | 30 − 40 | 24 | | 40 − 50 | 29 | | 50 − 60 | 38 | | 60 − 70 | 42 | | 70 − 80 | 31 | Using graph paper draw an ogive for the given distribution. Along one axis take 2 cm equal to 10 marks and along the other axis take 1cm equal to 10 students. Use your graph to find the following: (a) Median (b) Number of students who scored above 75 marks (c) If the pass percentage is 35%, find the number of students who scored 35% and below.

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Cumulative frequencies (upper limits 10, 20, …, 80): 8, 20, 36, 60, 89, 127, 169, 200. Plot (10, 8), (20, 20), (30, 36), (40, 60), (50, 89), (60, 127), (70, 169), (80, 200) and (0, 0), and join with a smooth curve. (a) n = 200, median at the 100th student: read from the graph, median ≈ 53 marks (53 ± 1). (b) At 75 marks the ogive reads about 186 students, so above 75 marks ≈ 200 − 186 = 14 students (linear interpolation in 70–80: 169 + 31 × 0.5 = 184.5, i.e. about 15 students). (c) 35% of 80 = 28 marks. At 28 marks the ogive reads ≈ 33 students (33 ± 2).

ICSE 2027 sample paper · 4 marks

The table given below shows a record of the weight in kilogram of 100 students of a school. | Weight(kg) | No. of Students | |---|---| | 30 − 35 | 5 | | 35 − 40 | 7 | | 40 − 45 | 14 | | 45 − 50 | 21 | | 50 − 55 | 25 | | 55 − 60 | 15 | | 60 − 65 | 7 | | 65 − 70 | 6 | Draw a histogram and find the modal weight. [Take 2 cm = 5 kg along one axis and 2 cm = 5 students along the other axis]

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Draw rectangles of width 5 kg and heights 5, 7, 14, 21, 25, 15, 7, 6 on the class intervals 30–35, …, 65–70 (a kink on the x-axis before 30). Modal class 50–55 (highest bar). Join the top-left corner of the tallest bar to the top-left corner of the next bar (55–60), and the top-right corner of the tallest bar to the top-right corner of the previous bar (45–50); from the point where these lines cross, drop a perpendicular to the x-axis. Modal weight ≈ 51.5 kg (by formula: 50 + (25 - 21)/(2(25) - 21 - 15) × 5 = 51.43 kg).

ICSE 2026 sample paper · 3 marks

Study the histogram described below and answer the following: The histogram shows the number of batsmen (y-axis) against the number of runs (x-axis); scale: x - axis, 2 cm = 100 runs; y - axis, 2 cm = 02 batsman. Its bars are: | Number of runs | 300 – 400 | 400 – 500 | 500 – 600 | 600 – 700 | |---|---|---|---|---| | Number of batsmen | 7 | 9 | 3 | 2 | (a) Number of batsmen who scored 500 to 700 runs (b) Modal class interval (c) The value of mode

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(a) 3 + 2 = 5 batsmen. (b) The tallest bar (9 batsmen) is over 400 – 500, so the modal class is 400 – 500. (c) Mode = L + (f₁ - f₀)/(2f₁ - f₀ - f₂) × h = 400 + (9 - 7)/(18 - 7 - 3) × 100 = 400 + 25 = 425 runs. (Graphically, the crossing lines on the tallest bar meet above about 425; the board's draft marking scheme reads it as 430 runs.)

ICSE 2026 board paper · 3 marks

The histogram drawn on the graph represents the number of students of different heights (in cm). The heights and the number of students shown by the histogram are given in the table. | Height (cm) | 100 – 110 | 110 – 120 | 120 – 130 | 130 – 140 | 140 – 150 | 150 – 160 | 160 – 170 | |---|---|---|---|---|---|---|---| | Number of Students | 6 | 2 | 9 | 14 | 12 | 9 | 4 | Using the histogram, answer the following: (a) the number of students whose height is 150 cm and above. (b) the modal height. (c) the total number of students.

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(a) Students of height 150 cm and above = 9 + 4 = 13. (b) The modal class is 130 – 140 (highest bar, 14). Joining the top corners of the modal bar to the adjacent corners of the neighbouring bars, the lines cross above about 137 cm. (Check: Mode = 130 + (14 - 9)/(2(14) - 9 - 12) × 10 = 130 + (50)/(7) ≈ 137.1 cm.) Modal height ≈ 137 cm. (c) Total = 6 + 2 + 9 + 14 + 12 + 9 + 4 = 56 students.

ICSE 2026 board paper · 5 marks

Using a graph paper, draw an ogive for the following distribution which shows a record of weight in kilograms of 100 students. | Weight (in kg) | Number of students | |---|---| | 35 – 40 | 4 | | 40 – 45 | 6 | | 45 – 50 | 10 | | 50 – 55 | 24 | | 55 – 60 | 26 | | 60 – 65 | 17 | | 65 – 70 | 8 | | 70 – 75 | 5 | Use your ogive to estimate the following: (a) the median weight of the students. (b) percentage of students whose weight is 60 kg or more. (c) the weight above which 20% of the students lie.

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Cumulative frequencies (less than upper boundary): | Weight less than (kg) | 35 | 40 | 45 | 50 | 55 | 60 | 65 | 70 | 75 | |---|---|---|---|---|---|---|---|---|---| | c.f. | 0 | 4 | 10 | 20 | 44 | 70 | 87 | 95 | 100 | Plot (35, 0), (40, 4), (45, 10), (50, 20), (55, 44), (60, 70), (65, 87), (70, 95), (75, 100) (e.g. 2 cm = 5 kg on the x-axis, 2 cm = 10 students on the y-axis) and join with a smooth freehand curve. (a) Median: (N)/(2) = 50th student. From c.f. 50 on the ogive, the weight is about 56.2 kg (between 55 and 60: 55 + (6)/(26) × 5 ≈ 56.2). Median ≈ 56 kg. (b) Students with weight less than 60 kg = 70, so 60 kg or more = 100 − 70 = 30, i.e. 30%. (c) …

ICSE 2025 compartment paper · 4 marks

The table given below shows a record of the weight in kilogram of 200 students of a school. | Weight (kg) | Number of students | |---|---| | 40 – 45 | 8 | | 45 – 50 | 19 | | 50 – 55 | 24 | | 55 – 60 | 45 | | 60 – 65 | 51 | | 65 – 70 | 31 | | 70 – 75 | 22 | Draw a histogram and find the modal weight. [Take 2 cm = 5 kg along one axis and 2 cm = 5 students along the other axis]

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Draw the histogram: weight (kg) on the x-axis (2 cm = 5 kg, with a kink before 40) and number of students on the y-axis (2 cm = 5 students); bars of heights 8, 19, 24, 45, 51, 31, 22 on the classes 40–45, …, 70–75. The highest bar is 60–65 (51). Join the top-left corner of this bar to the top-left corner of the next bar (65–70, height 31), and its top-right corner to the top-right corner of the previous bar (55–60, height 45). From the intersection of these lines drop a perpendicular to the x-axis. By calculation the foot is at 60 + (51 - 45)/(2 × 51 - 45 - 31) × 5 = 60 + (6)/(26) × 5 ≈ 61.15. Modal weight ≈ 61.2 kg (about 61 kg from the graph).

ICSE 2025 sample paper · 5 marks

The marks of 200 students in a test were recorded as follows: | Marks % | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 | 70 - 80 | 80 - 90 | 90 - 100 | |---|---|---|---|---|---|---|---|---|---|---| | No. of students | 5 | 7 | 11 | 20 | 40 | 52 | 36 | 15 | 9 | 5 | Using graph sheet draw ogive for the given data and use it to find the, (a) median, (b) number of students who obtained more than 65% marks (c) number of students who did not pass, if the pass percentage was 35.

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Cumulative frequencies (less than the upper limit): | Marks less than | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 | |---|---|---|---|---|---|---|---|---|---|---| | c.f. | 5 | 12 | 23 | 43 | 83 | 135 | 171 | 186 | 195 | 200 | Plot (0, 0), (10, 5), (20, 12), (30, 23), (40, 43), (50, 83), (60, 135), (70, 171), (80, 186), (90, 195), (100, 200) with marks on the x-axis (2 cm = 10% marks) and c.f. on the y-axis (2 cm = 20 students), and join them by a smooth curve (ogive). (a) N/2 = 100. From the c.f. = 100 on the y-axis, the ogive gives median ≈ 53 marks (by interpolation, 50 + (100 - 83)/(52) × 10 ≈ 53.3). (b) At 65 marks the ogive reads about 154 students (interpolation: 135 + …

ICSE 2025 board paper · 3 marks

Use a graph paper for this question. (Take 2 cm = 10 Marks along one axis and 2 cm = 10 students along another axis). Draw a Histogram for the following distribution which gives the marks obtained by 164 students in a particular class and hence find the Mode. | Marks | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 – 80 | |---|---|---|---|---|---| | Number of Students | 10 | 26 | 40 | 54 | 34 |

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Draw rectangles on the class intervals 30–40, 40–50, 50–60, 60–70, 70–80 with heights 10, 26, 40, 54, 34 (marks on the x-axis, students on the y-axis). The highest bar is 60–70 (modal class). Join the top-left corner of this bar to the top-left corner of the next bar (70–80), and the top-right corner of this bar to the top-right corner of the previous bar (50–60). From the point where these lines cross, drop a perpendicular to the x-axis. Mode ≈ 64 marks. (By formula: 60 + (54 - 40)/(2(54) - 40 - 34) × 10 = 60 + (140)/(34) ≈ 64.1.)