Graphical Representation. Histograms and Less than Ogive | ICSE Class 10 Maths Notes

This note covers grouped data, continuous class boundaries, histograms, graphical estimation of mode, cumulative frequency tables, less-than ogives, graphical readings of median and quartiles, interquartile range, and interpretation of frequencies from graphs.

What information must be understood before drawing a statistical graph?

Data are collected observations, such as students' marks or measured lengths. A variable is the quantity whose values are recorded. A frequency is the number of observations having a particular value or belonging to a particular group.

A grouped frequency distribution arranges observations into intervals and records the frequency of each interval. A class interval is one such group. Its lower and upper limits indicate the values at the beginning and end of that group.

How are values assigned to intervals?

In a continuous grouping such as 0 to 20, followed by 20 to 40, the first class includes 0 but excludes 20. The value 20 belongs to the next class. This convention prevents the same observation from being counted twice.

The class width is the upper boundary minus the lower boundary. Class boundaries are the dividing values between adjacent continuous classes. For the continuous classes below, the printed limits already serve as boundaries, so the common width is 20 marks.

Here, total frequency means the sum of all class frequencies: 6 + 5 + 33 + 14 + 6 = 64. This is the total number of students represented. The intervals tell us where their marks lie, but do not reveal each student's exact mark.

A histogram represents class frequencies through rectangles over continuous intervals. A less-than ogive is a graph showing the accumulated number of observations below successive upper boundaries. These graphs use the same distribution to answer different questions.

How are discontinuous intervals made continuous?

Intervals such as 118 to 126 and 127 to 135 do not have a shared printed limit. When measurements are recorded to the nearest millimetre, abbreviated mm, their continuous boundaries lie halfway between the neighbouring recorded limits.

Find the gap between the upper limit of one class and the lower limit of the next. Half this gap is subtracted from each lower limit and added to each upper limit. The resulting classes meet without changing the recorded frequencies.

How does the correction work for measured leaves?

The following distribution gives lengths of 40 leaves, measured correct to the nearest millimetre. The first two recorded classes have a gap of 127 − 126 = 1 mm, so the boundary correction is 0.5 mm.

Worked example 1. Convert the recorded leaf-length class 136 to 144 mm, followed by 145 to 153 mm, into continuous form and find its width.

Answer: Half the gap is (145 − 144) ÷ 2 = 0.5 mm. The boundaries become 135.5 and 144.5 mm. The class width is 144.5 − 135.5 = 9 mm; its frequency remains 9.

The adjustment changes the interval labels used for the graph, not the number of leaves. The first histogram rectangle therefore starts at 117.5 mm, and its top represents a frequency of 3.

Continuous intervals must be checked before plotting either graph. Do not apply a 0.5 correction automatically to every table. The marks classes 0 to 20 and 20 to 40 already meet, so they need no correction.

How is a histogram constructed and read?

Definition: A histogram is a set of adjacent rectangles whose bases represent continuous class intervals and whose areas are proportional to the corresponding class frequencies.

The horizontal axis, also called the x-axis, shows the variable and its class boundaries. The vertical axis, also called the y-axis, shows frequency when class widths are equal. An axis scale specifies how much numerical change each equal graph-paper distance represents.

What steps produce an equal-width histogram?

1. Check that the classes are continuous and calculate their widths.
2. Label the horizontal axis with the variable and its unit, and mark every class boundary in numerical order.
3. Label the vertical axis with frequency and choose a uniform scale that accommodates the largest frequency.
4. Draw a rectangle over each interval, with height corresponding to its frequency when all widths are equal.
5. Keep neighbouring rectangles adjacent and check each top against the frequency table.

Worked example 2. Construct the histogram for marks classes 0 to 20, 20 to 40, 40 to 60, 60 to 80 and 80 to 100, with frequencies 6, 5, 33, 14 and 6 respectively.

Answer: Mark boundaries 0, 20, 40, 60, 80 and 100 on the horizontal axis. All widths equal 20 marks. Draw adjacent rectangles with heights 6, 5, 33, 14 and 6 on the frequency scale.

The rectangle over 40 to 60 is the tallest in this distribution. It represents 33 students with marks at least 40 and less than 60. Its height does not mean that their marks are 33, or that all 33 students scored the same mark.

An ordinary bar diagram separates bars, and their widths are not used to represent numerical class widths. In a histogram, the base intervals matter. Leaving decorative gaps between these consecutive intervals would misrepresent the continuous grouping.

Why does rectangle area matter when class widths differ?

Comparing rectangle heights directly with frequencies works for equal-width classes because each base has the same width. If widths differ, using unadjusted frequencies as heights gives wider intervals extra area simply because their bases are longer.

Property: Histogram area represents frequency

Frequency density means class frequency divided by class width. It gives frequency per unit of the variable. Plotting density as height ensures that the product of base width and height represents the class frequency.

Frequency density = class frequency ÷ class width. The area calculation then multiplies the width by this density. Since the width divides and then multiplies the same frequency, the frequency is recovered.

This rule covers both equal and unequal widths. With equal widths, replacing density by frequency multiplies every height by the same factor, so the comparative shape remains unchanged. With unequal widths, the factors would differ, so an adjustment is necessary.

Worked example 3. For the marks class 40 to 60 with frequency 33, calculate its frequency density and check the represented frequency.

Answer: The width is 60 − 40 = 20 marks. Frequency density is 33 ÷ 20 = 1.65 students per mark. Multiplying this height by the width gives 1.65 × 20 = 33 students.

The calculation shows why a histogram must preserve both dimensions. A rectangle has a numerical width and a numerical height, and its area combines them. Read the vertical-axis label before interpreting a graph, because a density height is different from a frequency height.

Equal-width condition: the mode construction in the next section uses equal continuous intervals. Its tallest rectangle therefore identifies the class with the greatest frequency directly. Do not transfer that frequency comparison unchanged to an unequal-width graph.

How can the mode be estimated from a histogram?

The mode is the value occurring most frequently. For grouped data with equal class widths, the modal class is the class with the greatest frequency. The modal class is an interval; the graphical mode is an estimated value within it.

How are the construction lines drawn?

1. Draw the equal-width histogram accurately and identify the tallest rectangle.
2. Join the top-left corner of the tallest rectangle to the top-left corner of the rectangle immediately to its right.
3. Join the top-right corner of the tallest rectangle to the top-right corner of the rectangle immediately to its left.
4. Mark the intersection of these two lines inside the tallest rectangle.
5. Draw a vertical line from the intersection to the horizontal axis. Read the variable value at its foot as the estimated mode.

The construction uses the heights of the modal rectangle and its two neighbours. It therefore gives more information than simply choosing the middle of the modal interval. A taller neighbouring rectangle affects where the two lines intersect.

Adjacent shaded rectangles show daily wages horizontally and numbers of wage earners vertically. Two sloping lines cross inside the tallest rectangle. A dotted vertical line descends from their intersection to the wage axis.

See Fig. 4.5 in your NCERT textbook

Worked example 4. Estimate the mode for marks classes 0 to 20, 20 to 40, 40 to 60, 60 to 80 and 80 to 100 with frequencies 6, 5, 33, 14 and 6.

Answer: The modal class is 40 to 60 because its frequency is 33. Draw the two crossing lines using its neighbouring heights 5 and 14. Their intersection projects to approximately 52 marks on the horizontal axis.

Graphical estimates should reflect the accuracy of the drawing. The answer is approximately 52 marks, not a claim about an exactly observed individual mark. Also distinguish the modal interval 40 to 60, the highest frequency 33, and the estimated mode.

How is a less-than cumulative frequency table prepared?

Cumulative frequency is the accumulated frequency up to a specified boundary. In a less-than table, each entry counts all observations below the stated upper boundary. Include the frequency of the current class as well as those of the preceding classes.

Property: Successive cumulative frequencies do not decrease

Moving to a higher upper boundary includes the earlier observations and may add further observations. Thus, a less-than cumulative total cannot decrease. The final cumulative frequency is the total number of observations in the distribution.

Consider these marks, out of 100, obtained by 53 students. Each row pairs a class with its own frequency and the cumulative total at its upper boundary.

Worked example 5. For the 53-student distribution above, obtain the cumulative frequencies below 20 and below 30 marks.

Answer: Below 20, add the first two frequencies: 5 + 3 = 8. Below 30, include the next frequency: 8 + 4 = 12. Thus 12 students scored less than 30 marks.

The entry 12 does not describe just the class 20 to 30. That class contains 4 students; the accumulated total contains the students in all three classes below 30. Mixing these two columns would produce an incorrect graph.

Keep the same order while adding frequencies and labelling boundaries. A running total paired with the wrong boundary changes its meaning even if the addition is correct. Finally, compare the last entry with the sum of the original frequency column.

How can class frequencies be recovered from cumulative frequencies?

Property: A difference of cumulative totals gives an interval frequency

A later cumulative total contains all the observations in the earlier total and the observations added between the two boundaries. Subtracting the earlier total therefore leaves the number in the intervening interval.

Interval frequency = later cumulative frequency − earlier cumulative frequency. For a less-than table, the interval includes its lower boundary and excludes its upper boundary. Use the stated boundaries carefully when translating the subtraction into words.

The heights of 51 girls are recorded below. Here cm means centimetres, the unit of height. The first class is described as below 140 because its lower boundary is not supplied.

Worked example 6. There are 11 girls below 145 cm and 29 below 150 cm. Find the number with heights at least 145 cm and less than 150 cm.

Answer: Subtract the earlier accumulated count from the later count: 29 − 11 = 18 girls. The 11 girls below 145 cm have been removed from the larger group.

Check the recovered frequencies by adding them: 4 + 7 + 18 + 11 + 6 + 5 = 51. This restores the final cumulative total. It is a useful check that no interval has been omitted or counted twice.

Do not invent a lower limit for the first height class. The points at supplied upper thresholds can be plotted on an ogive, but an additional zero-frequency starting point needs a known lower boundary.

How is a less-than ogive drawn accurately?

An ogive is a cumulative frequency curve. For the less-than form, plot the upper class boundaries on the horizontal axis and the corresponding cumulative frequencies on the vertical axis. The two axes can use different numerical scales.

An ordered pair gives the horizontal coordinate first and the vertical coordinate second. Thus the point (60, 44) means 44 observations below a variable value of 60. It does not mean that the class ending at 60 has frequency 44.

What is the plotting sequence?

1. Make the classes continuous and prepare the less-than cumulative frequency column.
2. Label the horizontal axis with the variable and unit, and the vertical axis with cumulative frequency.
3. Choose uniform scales covering the boundary values and the total frequency.
4. Plot each upper boundary with its matching cumulative frequency. Include the first lower boundary with zero cumulative frequency when that boundary is known.
5. Join the plotted points in boundary order to form the less-than ogive, and check that it does not decrease.

Worked example 7. Draw a less-than ogive for marks classes 0 to 20, 20 to 40, 40 to 60, 60 to 80 and 80 to 100 with frequencies 6, 5, 33, 14 and 6.

Answer: The cumulative frequencies are 6, 11, 44, 58 and 64. Plot (0, 0), (20, 6), (40, 11), (60, 44), (80, 58) and (100, 64), then join them in order.

The left-hand graph labels marks in Mathematics horizontally and frequency vertically. Its joined points rise from the origin, passing the cumulative values 6, 11, 44, 58 and 64 at the successive upper limits.

See Fig. 4.8(a) in your NCERT textbook

Class midpoints, the averages of the two class boundaries, are not the horizontal plotting values for this graph. Using them shifts cumulative totals away from the boundaries to which they belong. Check the first coordinate of every plotted point against the table.

How are the median and quartiles read from an ogive?

The median is the central value of ordered data. Quartiles divide an ordered distribution into four equal parts by number of observations. The lower quartile, written Q₁, marks the first quarter; the upper quartile, written Q₃, marks the first three quarters.

Let N denote the total frequency. For a grouped less-than ogive, use the cumulative-frequency levels N ÷ 4 for Q₁, N ÷ 2 for the median, and 3N ÷ 4 for Q₃. Here 3N means three multiplied by N.

How is a cumulative-frequency level turned into a value?

1. Calculate the required fraction of the total frequency.
2. Locate that cumulative-frequency level on the vertical axis.
3. Move horizontally to meet the less-than ogive.
4. From the intersection, move vertically down to the horizontal axis.
5. Read and report the variable value with its unit, using appropriate graphical approximation.

Worked example 8. In the 53-student marks distribution, the ogive passes through (60, 22) and (70, 29). Estimate the median from this part of the graph.

Answer: The median level is 53 ÷ 2 = 26.5. It lies between cumulative frequencies 22 and 29. Reading horizontally to the ogive and then vertically down gives approximately 66.4 marks, within the class 60 to 70.

About half the students scored below this median and the other half above it. The number 26.5 is a position on the cumulative-frequency axis; it is not the median mark. A fractional position is usable even though the total number of students is a whole number.

Worked example 9. A 64-student marks ogive passes through (0, 0), (20, 6), (40, 11), (60, 44), (80, 58) and (100, 64). Estimate its quartiles.

Answer: Read Q₁ at 64 ÷ 4 = 16 and Q₃ at 3 × 64 ÷ 4 = 48 on the cumulative-frequency axis. Joining successive points by straight segments gives Q₁ approximately 43.0 marks and Q₃ approximately 65.7 marks.

Reading between plotted points is interpolation: estimating an intermediate value from neighbouring known points. The graph does not recover the original individual observations. Report the quartile readings as estimates, and keep the construction lines visible so the reading can be checked.

How are interquartile range and other counts interpreted?

The interquartile range is the upper quartile minus the lower quartile. It measures the interval occupied by the central half of the distribution. Both quartiles must be variable values read from the horizontal axis, not cumulative-frequency positions.

Interquartile range = Q₃ − Q₁. The result has the same unit as the variable. For marks it is measured in marks, and for heights recorded in centimetres it is measured in centimetres.

Worked example 10. An ogive gives estimated quartiles Q₁ = 43.0 marks and Q₃ = 65.7 marks. Find the interquartile range.

Answer: Subtract the lower quartile from the upper quartile: 65.7 − 43.0 = 22.7 marks. This is an approximate spread because the quartiles were graph readings.

How can the ogive answer counting questions?

For a specified mark, move vertically from the horizontal axis to the ogive and then horizontally to the cumulative-frequency axis. This reverses the direction used when finding a median or quartile. At a plotted boundary, the count comes directly from the cumulative table.

For the 64-student distribution, the cumulative frequency below 60 marks is 44. Therefore, the number scoring at least 60 is 64 − 44 = 20. The subtraction removes everybody below 60 from the whole group.

The number scoring at least 40 but less than 80 is 58 − 11 = 47. The later cumulative total includes all students below 80; removing those below 40 leaves the required interval. State the endpoints carefully when giving the answer.

Interpretation check: a cumulative frequency is a count, a median or quartile is a variable value, and an interquartile range is a difference between variable values. Reading the correct axis and attaching the correct unit keeps these answers distinct.

Glossary

• Frequency — The number of observations with a specified value or within a specified class interval.
• Class interval — A group of values between specified limits in a grouped frequency distribution.
• Class boundary — A dividing value at which one continuous class ends and the next begins.
• Class width — The difference between the upper and lower boundaries of a class interval.
• Histogram — Adjacent rectangles over continuous intervals, with areas proportional to the corresponding class frequencies.
• Frequency density — Class frequency divided by class width, used to set histogram heights when widths differ.
• Modal class — For equal-width grouped data, the interval having the greatest class frequency.
• Mode — The most frequently occurring value, estimated within a modal interval for grouped data.
• Cumulative frequency — An accumulated count of observations up to a specified boundary in a distribution.
• Less-than ogive — A graph of less-than cumulative frequencies plotted against the corresponding upper class boundaries.
• Median — The central value, read at half the total cumulative frequency on a grouped ogive.
• Lower quartile — The value marking the first quarter of an ordered distribution by number of observations.
• Upper quartile — The value marking the first three quarters of an ordered distribution by number of observations.
• Interquartile range — The upper quartile minus the lower quartile, describing the spread of the central half.
• Interpolation — Estimation of an intermediate value using known values on either side of it.

Common errors and misconceptions

• Misconception: A histogram should have gaps like an ordinary bar diagram. Correct: Rectangles over consecutive continuous intervals are adjacent; their widths represent actual class widths.
• Misconception: Every class interval needs a 0.5 boundary correction. Correct: Check the recorded limits and measurement precision. Already continuous classes need no adjustment.
• Misconception: Histogram height directly represents frequency even when widths differ. Correct: Use frequency density for unequal widths so rectangle areas remain proportional to frequencies.
• Misconception: The modal class, its frequency and the mode are the same answer. Correct: They are an interval, a count and a variable value respectively.
• Misconception: An ogive uses class midpoints and ordinary frequencies. Correct: A less-than ogive pairs upper class boundaries with accumulated frequencies.
• Misconception: Half the total frequency is the median value. Correct: It is the vertical-axis level used to locate the median on the horizontal axis.
• Misconception: Interquartile range is found by subtracting the quartile frequency levels. Correct: Subtract the lower quartile value from the upper quartile value after reading both from the graph.
• Misconception: Graphical estimates give exact original observations. Correct: Values read between grouped boundaries are estimates and should be reported with appropriate precision.

Exam-style questions with model answers

Q1. Explain two differences between a histogram and an ordinary bar diagram. [2 marks]

1. Histogram rectangles over consecutive continuous intervals are adjacent, whereas an ordinary bar diagram leaves spaces between its bars.
2. Histogram widths represent class widths and areas represent frequencies; ordinary bar widths do not have this numerical role.

Q2. Leaf lengths are measured to the nearest millimetre. Classes 118 to 126 and 127 to 135 contain 3 and 5 leaves respectively. State their continuous boundaries and frequencies. [2 marks]

1. The gap between 126 and 127 is 1 mm, so half the gap is 0.5 mm. The continuous classes are 117.5 to 126.5 and 126.5 to 135.5 mm.
2. The frequencies remain 3 and 5 respectively because adjusting the boundaries does not change the recorded numbers of leaves.

Q3. Marks classes 0 to 20, 20 to 40, 40 to 60, 60 to 80 and 80 to 100 have frequencies 6, 5, 33, 14 and 6. Describe the histogram and identify the modal class. [3 marks]

1. The classes are continuous, each with width 20 marks. Put marks on the horizontal axis and frequency on the vertical axis, using uniform scales on each axis.
2. Draw adjacent rectangles over the five intervals with heights representing 6, 5, 33, 14 and 6 students respectively.
3. The modal class is 40 to 60 marks, because its frequency 33 is the largest. This identifies an interval, not an exact mode.

Q4. Of 51 girls, 4 have heights below 140 cm, 11 below 145 cm, 29 below 150 cm, 40 below 155 cm, 46 below 160 cm and 51 below 165 cm. Recover all six class frequencies and check the total. [4 marks]

1. The first group, below 140 cm, contains 4 girls. The frequency for heights from 140 cm to below 145 cm is 11 − 4 = 7.
2. The next frequencies are 29 − 11 = 18 for 145 to 150 cm and 40 − 29 = 11 for 150 to 155 cm.
3. The final frequencies are 46 − 40 = 6 for 155 to 160 cm and 51 − 46 = 5 for 160 to 165 cm.
4. Adding the recovered frequencies gives 4 + 7 + 18 + 11 + 6 + 5 = 51, agreeing with the given total.

Q5. Marks classes 0 to 20, 20 to 40, 40 to 60, 60 to 80 and 80 to 100 have frequencies 6, 5, 33, 14 and 6. Give a complete construction of the less-than ogive and state the cumulative-frequency level for its median. [5 marks]

1. Add the class frequencies successively to obtain cumulative frequencies 6, 11, 44, 58 and 64. The total number of students is therefore 64.
2. Label the horizontal axis with marks and the vertical axis with cumulative frequency. Choose uniform scales that include marks from 0 to 100 and cumulative frequency from 0 to 64.
3. Plot (0, 0), (20, 6), (40, 11), (60, 44), (80, 58) and (100, 64), pairing each upper boundary with its cumulative count.
4. Join these points in order to obtain the less-than ogive. Check that the graph does not decrease and finishes at cumulative frequency 64.
5. The median level is 64 ÷ 2 = 32. Read horizontally from this vertical-axis level to the ogive, then vertically down to the marks axis.

Q6. A marks ogive for 64 students joins the points (0, 0), (20, 6), (40, 11), (60, 44), (80, 58) and (100, 64) by straight segments. The first coordinate is marks and the second is cumulative frequency. Estimate both quartiles and the interquartile range. [5 marks]

1. The total frequency is 64, so the lower-quartile level is 64 ÷ 4 = 16 on the cumulative-frequency axis.
2. Level 16 lies between cumulative frequencies 11 and 44. Read horizontally to the segment from (40, 11) to (60, 44), then vertically down: the lower quartile is approximately 43.0 marks.
3. The upper-quartile level is 3 × 64 ÷ 4 = 48, which lies between cumulative frequencies 44 and 58.
4. Read horizontally to the segment from (60, 44) to (80, 58), then vertically down: the upper quartile is approximately 65.7 marks.
5. The interquartile range is approximately 65.7 − 43.0 = 22.7 marks. It describes the spread between the two quartile values.

Q7. Among 64 students, a less-than marks table gives 11 below 40 marks, 44 below 60 marks and 58 below 80 marks. Find the numbers scoring at least 60, and at least 40 but less than 80. Explain both subtractions. [3 marks]

1. The number scoring at least 60 marks is 64 − 44 = 20 students. Removing everybody below 60 from the total leaves those scoring 60 or above.
2. The number scoring at least 40 but less than 80 is 58 − 11 = 47 students.
3. The second subtraction removes the students below 40 from all those below 80. Its remaining interval includes 40 and excludes 80, matching the requested endpoints.

Q8. Marks classes 0 to 20, 20 to 40, 40 to 60, 60 to 80 and 80 to 100 have frequencies 6, 5, 33, 14 and 6. Explain how to estimate their mode graphically, and give the approximate result. [5 marks]

1. Draw the histogram with equal class widths of 20 marks and heights representing 6, 5, 33, 14 and 6 students respectively.
2. Identify the modal rectangle over 40 to 60 marks. Its frequency 33 is greater than the frequencies 5 and 14 of its immediate neighbours.
3. Join the modal rectangle's top-left corner to the next rectangle's top-left corner, and its top-right corner to the preceding rectangle's top-right corner.
4. Mark where these construction lines intersect inside the modal rectangle. Draw a vertical line from this point to the marks axis.
5. Read approximately 52 marks at the foot of the vertical line. This is the estimated mode; 40 to 60 is the modal class and 33 is its frequency.

Key takeaways

• Check class boundaries before drawing a graph; converting recorded intervals into continuous classes leaves their frequencies unchanged.
• A histogram represents class frequencies through rectangle areas, with bases determined by the continuous class intervals.
• Equal-width histogram heights can represent frequencies directly; unequal-width histograms need heights adjusted through frequency density.
• The graphical mode is read beneath the intersection of construction lines inside the modal rectangle.
• A less-than cumulative frequency includes every class frequency up to and including the class ending at that boundary.
• Plot upper boundaries against cumulative frequencies for a less-than ogive, and check that the graph does not decrease.
• Use one quarter, one half and three quarters of total frequency to locate the lower quartile, median and upper quartile.
• Subtract lower quartile from upper quartile to find interquartile range, retaining the variable's unit and the graph's approximate precision.

Test yourself

What does the width of a histogram rectangle represent?

It represents the width of the continuous class interval, found by subtracting its lower boundary from its upper boundary.

Why are rectangle heights adjusted when class widths differ?

The adjustment keeps rectangle areas proportional to frequencies. Frequency density, obtained by dividing frequency by class width, supplies the required height.

What distinguishes a modal class from a mode?

The modal class is an interval. The mode is a value, estimated within that interval for grouped data.

Which values form the coordinates of a less-than ogive point?

The horizontal coordinate is an upper class boundary; the vertical coordinate is the cumulative frequency below that boundary.

Why can a less-than ogive not decrease?

Increasing the boundary retains all observations already counted and may add more. The cumulative total therefore cannot become smaller.

For a total frequency of 64, which three levels locate the quartiles and median?

Use cumulative-frequency levels 16 for the lower quartile, 32 for the median and 48 for the upper quartile.

If the quartile readings are 43.0 and 65.7 marks, what is the interquartile range?

The interquartile range is approximately 65.7 − 43.0 = 22.7 marks, the difference between the upper and lower quartiles.

If 44 of 64 students scored below 60 marks, how many scored at least 60?

Subtract the number below 60 from the total: 64 − 44 = 20 students scored at least 60 marks.