Measures of Central Tendency | ICSE Class 10 Maths Notes

This note covers measures of central tendency, raw and arrayed data, arithmetic mean by the direct, short-cut and step-deviation methods, median and mode, grouped frequency distributions, median and modal classes, continuous and discontinuous classes, histograms, less than ogives, quartiles and interquartile range.

What do mean, median and mode tell us about data?

Data are collected observations, such as marks obtained by students. An observation is one recorded value. A measure of central tendency summarises a set of observations by a single typical or representative value.

Definition: The arithmetic mean is the sum of all observation values divided by the number of observations. The median is the middle value of ordered data. The mode is a value occurring most frequently.

Raw data are observations in their original, unarranged form. An array arranges those observations in order of magnitude. Ascending order means smallest to largest; descending order means largest to smallest. Arranging observations helps locate the median.

How do the three measures differ?

The mean takes all observations into account, but extreme values, which are unusually large or small values, affect it. The median is more appropriate in situations where a typical observation is required and extreme values may be present.

The mode is useful when the most frequent value or most popular item is required. Choosing a measure therefore depends on the question being asked. An average mark and the mark occurring most frequently answer different questions about the same observations.

Do not assume that these three measures must have the same value. Nor should a value be called the mode merely because it is numerically large. Its frequency, meaning its number of occurrences, is what matters.

How is the arithmetic mean calculated for raw data?

Let x̄, read as “x bar”, denote the arithmetic mean. Let x denote an observation and n the total number of observations. The symbol Σ, capital sigma, means “sum of”. Thus Σx means the sum of all observation values.

Result: Mean of individual observations

x̄ = Σx ÷ n. Add every observation, including repeated values, and divide by the number of observations. The denominator counts observations; it is not the number of different values appearing in the list.

1. Identify the quantity being measured and retain its unit where one is given.
2. Count all the observations to obtain the denominator n.
3. Add the observation values to obtain the numerator Σx.
4. Divide the total by n and state what the resulting mean represents.

Worked example 1. Five students obtain 40, 50, 55, 78 and 58 marks in an economics test. Find their arithmetic mean.

Answer: The number of observations is n = 5. Their sum is 40 + 50 + 55 + 78 + 58 = 281. Therefore x̄ = 281 ÷ 5 = 56.2 marks.

The result describes the average mark for this set of students. It does not say that a student actually scored 56.2 marks. The calculation uses the sum of the recorded marks and the number of students.

Why does each observation matter?

In the numerator, every recorded value contributes to the total. Leaving out an observation changes the count and, unless its value is zero, changes the total. Replacing repeated values by a single copy also changes the data being averaged.

Ordering the observations is not needed for addition, although a neat list can help prevent omissions. By contrast, ordering is essential when identifying the middle observation for the median. Keep these two procedures distinct when the same question asks for both measures.

How do frequencies simplify the calculation of a mean?

A frequency distribution records values alongside their frequencies. Write f for the frequency of a value x. The product fx means f multiplied by x. It gives that row's contribution to the sum of all observations.

The sum Σf is the total frequency, equal to n. The sum Σfx is obtained by adding the row products. For values listed separately with their frequencies, use x̄ = Σfx ÷ Σf.

What must be multiplied before adding?

Multiply each value by its own frequency. A value occurring several times contributes several copies of itself to the total. Adding the distinct values alone would ignore the information supplied by the frequency column.

Worked example 2. Find the mean marks for the complete distribution in the table above.

Answer: Σf = 30 and Σfx = 1779. Hence x̄ = 1779 ÷ 30 = 59.3 marks. This is the exact mean of the listed marks, because each individual mark and its frequency are known.

A discrete frequency distribution lists separate values, as this table does. A grouped distribution instead combines observations into intervals. Before using a formula, check which form has been supplied, because an interval needs a representative value.

How is the direct method used for grouped data?

A class interval is a range into which observations are grouped. Its lower and upper limits specify the ends of that range. Its class mark is its midpoint, calculated as the average of those limits.

Class mark = (lower class limit + upper class limit) ÷ 2. For grouped calculations, x now represents a class mark and f represents the frequency of that class. The direct method remains x̄ = Σfx ÷ Σf.

What assumption is made when using class marks?

It is assumed that the frequency of each class interval is centred around its midpoint. The class mark therefore represents the observations in that interval. The resulting grouped mean is an approximation to the mean of the original observations.

Worked example 3. Find the grouped mean marks from the table above using the direct method.

Answer: For the first class, x = (10 + 25) ÷ 2 = 17.5 and fx = 2 × 17.5 = 35.0. Adding all products gives Σfx = 1860.0. Thus x̄ = 1860.0 ÷ 30 = 62 marks.

Here an observation at an upper class limit belongs to the next class. The four students scoring 40 marks therefore belong to 40 to 55, not 25 to 40. This avoids counting a boundary observation twice.

Note: The exact mean of the original marks is 59.3, whereas the grouped mean is 62. The difference comes from the midpoint assumption, not from using a different definition of mean.

How does the short-cut or assumed mean method work?

The assumed mean, denoted by A, is a chosen reference value used to simplify arithmetic. A centrally located class mark can make the differences smaller. It need not equal the final mean.

A deviation, denoted by d, is the difference between a value and A. Here d = x − A, where x is the class mark. The product fd multiplies each deviation by the corresponding class frequency.

Result: Mean from deviations

x̄ = A + Σfd ÷ Σf. The quantity Σfd ÷ Σf is the mean deviation from A. Adding A restores the amount subtracted when the deviations were formed.

Worked example 4. Use A = 47.5 to find the mean marks for the distribution above by the short-cut method.

Answer: The weighted deviation total is Σfd = 435 and Σf = 30. Therefore x̄ = 47.5 + 435 ÷ 30 = 47.5 + 14.5 = 62 marks.

Why does choosing A not change the answer?

Subtracting the same A from every observation shifts the mean by A. Adding A at the end reverses that shift. The choice affects the size of the intermediate numbers, but the calculated mean for the same distribution remains unchanged.

Retain negative signs throughout the calculation. Values below A have negative deviations; values above A have positive deviations. The positive and negative products must be added algebraically, which means adding them with their signs.

How does the step-deviation method reduce the arithmetic?

The step-deviation method divides deviations by a convenient common scale. Let h denote that non-zero scale, and let u = (x − A) ÷ h denote a scaled deviation. For equal class widths, h can be the class width.

The class width is the upper boundary minus the lower boundary of a continuous class. Class boundaries are the dividing values between adjoining classes. The marks intervals used here have width 15, so take h = 15 and A = 47.5.

Result: Mean from scaled deviations

x̄ = A + h × (Σfu ÷ Σf). First find the frequency-weighted mean of u. Multiply it by h to undo the scaling, then add A to undo the subtraction.

Worked example 5. Calculate the mean of the marks distribution above using step-deviation.

Answer: A = 47.5, h = 15, Σfu = 29 and Σf = 30. Hence x̄ = 47.5 + 15 × (29 ÷ 30) = 47.5 + 14.5 = 62 marks.

All three methods give the same mean for the same grouped table. The short-cut and step-deviation methods are simplified forms of the direct method. They do not remove the approximation introduced by replacing observations with class marks.

Step-deviation is convenient when the deviations have a common factor. Unequal class widths do not prevent its use: a suitable fixed divisor can still simplify the deviations. Do not divide different rows by different widths while applying this single-scale formula.

How do you find the median of raw and discrete data?

The median is found from positions after arranging observations in ascending order. With n observations, an odd n gives one middle position, (n + 1) ÷ 2. An even n gives two middle positions, n ÷ 2 and (n ÷ 2) + 1.

For an odd number of observations, read the value at the middle position. For an even number, take the arithmetic mean of the two middle values. A position identifies an observation; it is not itself the median value.

How does cumulative frequency locate the middle values?

Cumulative frequency is a running total of frequencies up to and including the current value or class. It tells you how many observations have been counted so far. A discrete table can therefore locate middle observations without writing every repetition separately.

Worked example 6. The table gives marks out of 50 for 100 students. Find the median mark.

Answer: Since n = 100, use the 50th and 51st observations. The cumulative frequency reaches 50 at 28 marks, so the 50th observation is 28. The next observation is 29. Median = (28 + 29) ÷ 2 = 28.5 marks.

The median of 28.5 conveys that about 50% of these students scored below it and another 50% scored above it. Here the two middle values differ, so neither value alone is the median.

Check that the final cumulative frequency equals the total number of observations. Then identify both central positions when n is even. Stopping at the first central value would give an incomplete calculation for this distribution.

How do you locate the median class and modal class?

For observations grouped into intervals, the original individual values are not shown. A median class is the class containing the central part of the distribution. A modal class is the class with the greatest frequency.

To locate the median class, add the frequencies cumulatively and calculate n ÷ 2. Select the class whose cumulative frequency is greater than, and nearest to, n ÷ 2. This uses cumulative frequency, whereas identifying the modal class uses the individual class frequencies.

Worked example 7. Locate the median class and modal class for the 53 students in the table above.

Answer: n ÷ 2 = 53 ÷ 2 = 26.5. The first cumulative frequency greater than 26.5 is 29, belonging to 60 to 70. This is the median class. The highest class frequency is 9, so the modal class is 70 to 80.

Why is a class different from a single value?

The interval 60 to 70 identifies the median class; it does not identify the median as either endpoint or the class mark. Likewise, 9 is the modal-class frequency, not the mode of the marks.

Keep the two frequency columns clearly labelled. The final cumulative frequency is the total number of observations and is therefore the largest running total. It must not be mistaken for evidence that the last class is the modal class.

How is the mode found from raw data and a histogram?

For raw data, count the occurrences of each value. The mode is the value with the greatest frequency. Arranging the data or making a discrete frequency table makes repeated values easier to count.

Worked example 8. A bowler takes 2, 6, 4, 5, 0, 2, 1, 3, 2 and 3 wickets in ten matches. Find the mode.

Answer: The value 2 occurs three times, 3 occurs twice, and each other value occurs once. Therefore the mode is 2 wickets. The frequency of the mode is 3 matches.

More than one value may share the greatest frequency. Data with two modes are called bimodal; data with more than two modes are called multimodal. There may be no mode when no value occurs more frequently than any other.

How is a histogram used?

A histogram represents continuous class intervals by adjacent rectangles. Put class boundaries on the horizontal axis and frequency on the vertical axis when class widths are equal. For unequal widths, rectangle heights use frequency density, meaning class frequency divided by class width.

For equal-width classes with a single modal rectangle and neighbours on both sides, join its upper left corner to the upper right corner of the following rectangle. Join its upper right corner to the upper left corner of the preceding rectangle.

Draw a vertical line from the intersection of these two joining lines to the horizontal axis. The value read there gives the graphical estimate of the mode. Read the horizontal value, not the vertical frequency.

Adjacent rectangles represent daily wages on the horizontal axis and numbers of wage earners on the vertical axis. Two sloping lines cross inside the tallest rectangle. A dotted vertical line meets the wage axis at the marked mode, 80.6.

See Fig. 4.5 in your NCERT textbook

How are continuous and discontinuous grouped classes handled?

Continuous classes meet at common boundaries. With the lower-inclusive, upper-exclusive convention, a value at the shared boundary belongs to the following class. Discontinuous classes have a gap between a class's stated upper limit and the next class's stated lower limit.

For the grouped mean, find each class mark by averaging its stated limits. Inclusive intervals include both stated endpoints. This changes how observations are assigned, but does not change the midpoint calculation used in finding a grouped mean.

How do you convert classes before drawing a graph?

For measurements recorded to the nearest millimetre, the intervals below become continuous by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit. A millimetre, abbreviated mm, is the length unit used in this example.

Worked example 9. Locate the median and modal classes for the 40 leaf lengths in the table.

Answer: The cumulative frequencies are 3, 8, 17, 29, 34, 38 and 40. Since n ÷ 2 = 20, the median class is 145 to 153. Its frequency 12 is also the highest, making it the modal class. The corresponding continuous interval is 144.5 to 153.5 mm.

Conversion changes class boundaries, not frequencies. The leaves are still the same observations. For a histogram or an ogive, use continuous class boundaries. An ogive is a cumulative frequency curve, explained in the next section.

Note: The adjustment of 0.5 belongs to these whole-millimetre intervals. Do not automatically attach it to every grouped table. Examine the stated class limits and the measurement convention first.

How does a less than ogive give the median and quartiles?

A less than ogive plots the running total below each upper class boundary. The horizontal axis shows the measured quantity, and the vertical axis shows cumulative frequency. Label both axes and choose scales that allow the points to be read clearly.

1. Arrange the continuous classes in increasing order and calculate their cumulative frequencies.
2. Start by plotting the lowest class boundary at cumulative frequency zero. Then plot each upper class boundary against its corresponding less than cumulative frequency.
3. Draw the cumulative frequency curve through the plotted points.
4. From n ÷ 2 on the vertical axis, move horizontally to the curve and vertically down to read the median.

The quartiles divide an ordered distribution into four parts. The lower quartile, Q₁, marks the first quarter; the median is the second quartile, Q₂; and the upper quartile, Q₃, marks the third quarter.

On a less than ogive, read Q₁ at cumulative frequency n ÷ 4 and Q₃ at 3n ÷ 4. The interquartile range is the difference between these readings: interquartile range = Q₃ − Q₁. It describes the width occupied by the central half of the distribution.

What should you check when reading the curve?

The cumulative frequency scale need not match the horizontal scale. A cumulative frequency is a count, whereas the answer read on the horizontal axis is a value of the measured quantity. Preserve its unit when reporting a median, quartile or interquartile range.

The graphs place marks in mathematics horizontally and cumulative frequency vertically. The less than curve rises, the more than curve falls, and the combined graph marks their intersection with a dotted vertical guide towards the marks axis.

See Fig. 4.8 in your NCERT textbook

A less than ogive is never decreasing because the running total cannot fall as successive class frequencies are added. Read graphical answers to the precision supported by the scale. Median class identification and a graphical median reading are different tasks.

Glossary

• Central tendency — A way of summarising observations using a single typical or representative value.
• Raw data — Recorded observations in their original form before arranging them in order.
• Array — A set of observations arranged in ascending or descending order of magnitude.
• Arithmetic mean — The sum of all observation values divided by the total number of observations.
• Median — The central value obtained after arranging observations, averaging two middle values when necessary.
• Mode — A value occurring with the greatest frequency in a given set of observations.
• Frequency — The number of times a value occurs, or the number of observations within a class.
• Class mark — The midpoint of a class interval, found by averaging its lower and upper limits.
• Assumed mean — A chosen reference value subtracted from observations or class marks to simplify mean calculations.
• Deviation — The signed difference obtained by subtracting a reference value from an observation or class mark.
• Cumulative frequency — A running total including the frequency of the current value or class and all preceding ones.
• Median class — The class identified using half the total frequency and the cumulative frequency distribution.
• Modal class — A class interval having the greatest frequency in a grouped frequency distribution.
• Less than ogive — A cumulative frequency curve plotting less than cumulative frequencies against upper class boundaries.
• Interquartile range — The upper quartile minus the lower quartile, describing the width of the central half of a distribution.

Common errors and misconceptions

• Misconception: Divide the sum of distinct values by the number of rows to find any mean. Correct: When frequencies are supplied, multiply each value by its frequency and divide the product total by total frequency.
• Misconception: A class limit represents every observation in a grouped mean calculation. Correct: Use the class mark, found by averaging the two limits, under the midpoint assumption.
• Misconception: The assumed mean is the answer. Correct: Add the mean deviation to the assumed mean; in step-deviation, first multiply the mean scaled deviation by the chosen scale.
• Misconception: Deviations can be added without their negative signs. Correct: Preserve the signs of the deviations and their frequency products throughout the calculation.
• Misconception: The middle entry in the original list is the median. Correct: Arrange the observations first. For an even count, average the values at both middle positions.
• Misconception: The largest frequency is the mode. Correct: The mode is the corresponding observation value. For grouped data, the class with the largest frequency is the modal class.
• Misconception: Median class and modal class are found from the same column. Correct: Use cumulative frequency to locate the median class, and individual class frequency to identify the modal class.
• Misconception: A grouped mean must equal the exact mean of the original observations. Correct: Grouping replaces individual values by class marks, so the midpoint assumption can produce a different, approximate mean.

Exam-style questions with model answers

Q1. Find the arithmetic mean of the test marks 40, 50, 55, 78 and 58. Explain what the result represents. [3 marks]

1. There are five observations, one mark for each student. Their sum is 40 + 50 + 55 + 78 + 58 = 281 marks.
2. The arithmetic mean equals the total divided by the number of observations. Thus the mean is 281 ÷ 5 = 56.2 marks.
3. The result is the average mark for these five students. It does not require any individual student to have scored exactly 56.2 marks.

Q2. A bowler takes 2, 6, 4, 5, 0, 2, 1, 3, 2 and 3 wickets in ten matches. Find the mode and distinguish it from its frequency. [2 marks]

1. The value 2 occurs three times, more often than any other value, so the mode is 2 wickets.
2. Its frequency is 3 matches. This counts occurrences; it is not the modal number of wickets.

Q3. Marks 20, 25, 28, 29, 33, 38, 42 and 43 have respective frequencies 6, 20, 24, 28, 15, 4, 2 and 1. Find the median mark. [4 marks]

1. The total frequency is 100. Since this is even, the median is the arithmetic mean of the 50th and 51st observations in ascending order.
2. The cumulative frequencies are 6, 26, 50, 78, 93, 97, 99 and 100.
3. The cumulative frequency reaches 50 at 28 marks, so the 50th observation is 28. The 51st observation belongs to 29 marks.
4. Therefore the median mark is (28 + 29) ÷ 2 = 28.5.

Q4. The marks classes 10 to 25, 25 to 40, 40 to 55, 55 to 70, 70 to 85 and 85 to 100 have respective frequencies 2, 3, 7, 6, 6 and 6. Use the direct method to find the grouped mean and state its underlying assumption. [5 marks]

1. The class marks are the averages of the class limits: 17.5, 32.5, 47.5, 62.5, 77.5 and 92.5. These represent the six intervals in the calculation.
2. Multiplying each class mark by its frequency gives 35.0, 97.5, 332.5, 375.0, 465.0 and 555.0.
3. The sum of the products is 1860.0, while the total frequency is 2 + 3 + 7 + 6 + 6 + 6 = 30.
4. The direct method divides the product total by total frequency. Therefore the grouped mean is 1860.0 ÷ 30 = 62 marks.
5. The calculation assumes each class's frequency is centred around its midpoint. Consequently, this is an approximate mean for the original individual observations.

Q5. Class marks 17.5, 32.5, 47.5, 62.5, 77.5 and 92.5 have respective frequencies 2, 3, 7, 6, 6 and 6. Find the mean by the short-cut method, taking the assumed mean as 47.5. [4 marks]

1. Subtract the assumed mean 47.5 from each class mark. The resulting deviations are −30, −15, 0, 15, 30 and 45.
2. Multiply these deviations by their respective frequencies to obtain −60, −45, 0, 90, 180 and 270.
3. The sum of the frequency-deviation products is 435, and the total frequency is 30. Thus the mean deviation is 435 ÷ 30 = 14.5.
4. Add this mean deviation to the assumed mean: 47.5 + 14.5 = 62. Hence the mean is 62.

Q6. Marks classes 10 to 25, 25 to 40, 40 to 55, 55 to 70, 70 to 85 and 85 to 100 have frequencies 2, 3, 7, 6, 6 and 6 respectively. Use step-deviation with assumed mean 47.5 and scale 15 to calculate the mean. [5 marks]

1. The class marks are 17.5, 32.5, 47.5, 62.5, 77.5 and 92.5. The assumed mean is 47.5 and the common scale is 15.
2. Subtract 47.5 from each class mark and divide each result by 15. The scaled deviations are −2, −1, 0, 1, 2 and 3.
3. Multiply each scaled deviation by its corresponding frequency. The products are −4, −3, 0, 6, 12 and 18, with total 29.
4. Total frequency is 30. Multiply the mean scaled deviation by the scale to obtain the correction: 15 × (29 ÷ 30) = 14.5.
5. Add the correction to the assumed mean. The grouped mean is 47.5 + 14.5 = 62 marks.

Q7. Leaf lengths, recorded to the nearest millimetre, have classes 118 to 126, 127 to 135, 136 to 144, 145 to 153, 154 to 162, 163 to 171 and 172 to 180. Their frequencies are 3, 5, 9, 12, 5, 4 and 2 respectively. Identify the median and modal classes and give their continuous boundaries. [4 marks]

1. The total frequency is 40, giving half the total as 20. The cumulative frequencies are 3, 8, 17, 29, 34, 38 and 40.
2. The first cumulative frequency greater than 20 is 29. Therefore the median class is 145 to 153 millimetres.
3. The greatest individual class frequency is 12, belonging to the same interval. Thus the modal class is also 145 to 153 millimetres.
4. With lengths recorded to the nearest millimetre, the continuous boundaries for this class are 144.5 and 153.5 millimetres.

Q8. The exact mean of a set of students' marks is 59.3. After the same marks are grouped into intervals and represented by class midpoints, the calculated mean is 62. Explain why these answers can differ. [2 marks]

1. The exact mean uses the individual marks with their actual frequencies, preserving the values of all observations.
2. The grouped calculation replaces observations by their class midpoints. This assumption gives an approximate mean, so 62 can differ from 59.3.

Key takeaways

• Mean uses all observation values, median uses central position after ordering, and mode identifies the value occurring most frequently.
• For a frequency distribution, multiply each value by its frequency before adding, then divide by total frequency.
• A grouped mean uses class midpoints and assumes each class frequency is centred around its midpoint.
• Direct, short-cut and step-deviation methods give the same mean for the same grouped distribution when correctly applied.
• For an even number of observations, the median is the average of the two middle values in the ordered data.
• Use cumulative frequencies to locate the median class; use individual class frequencies to identify the modal class.
• Continuous class boundaries are needed for histograms and ogives; converting boundaries leaves the original frequencies unchanged.
• Read the median and quartiles from the less than ogive, then subtract the lower quartile from the upper quartile for interquartile range.

Test yourself

What does the symbol Σ mean in a mean formula?

It means summation: add all the indicated values or products.

How is the class mark of an interval calculated?

Add its lower and upper limits, then divide their sum by two.

Why must a class mark be multiplied by its frequency?

The frequency records how many observations the class mark represents in the grouped calculation.

Does choosing a different assumed mean change the final mean?

No. Correctly restoring the subtracted reference value gives the same mean for the same distribution.

For an even observation count, how do you obtain the median?

Arrange the values, locate both middle observations, and calculate their arithmetic mean.

How does a modal class differ from its frequency?

The modal class is an interval; its frequency counts the observations belonging to that interval.

What goes on the vertical axis of a less than ogive?

Cumulative frequency goes on the vertical axis; upper class boundaries are plotted horizontally.

What is the interquartile range?

It is the upper quartile minus the lower quartile, measuring the width of the central half.