Measures of Central Tendency

ICSE Class 10 · Mathematics · Chapter 24

What this chapter covers

What a measure of central tendency does
Mean of raw data
Mean of arrayed data
Median of raw and arrayed data
Mode of raw and arrayed data
Grouped data and class marks
Mean of grouped data: the direct method
Mean of grouped data: the short-cut method
Mean of grouped data: the step-deviation method
Median class and modal class of grouped data
Comparing the three measures

Key points

A measure of central tendency is one value that represents the centre of a set of data; the three measures are the mean, the median and the mode.
Mean of raw data: barx = (Sigma x)/(n); mean of arrayed data: barx = (Sigma fx)/(Sigma f).
Adding k to every observation adds k to the mean; multiplying every observation by k multiplies the mean by k; Sigma(x - barx) = 0.
Median of ordered raw data: the ((n+1)/(2))th value when n is odd; the mean of the ((n)/(2))th and ((n)/(2)+1)th values when n is even.
For a frequency table, cumulative frequencies locate the median position.
The mode is the value with the highest frequency; data can be unimodal, bimodal or have no mode.
For grouped data, the class mark (lower limit + upper limit)/(2) represents every value in the class.
Direct method: barx = (Sigma fx)/(Sigma f).
Short-cut method: barx = A + (Sigma fd)/(Sigma f), with d = x - A.
Step-deviation method: barx = A + (Sigma ft)/(Sigma f) × i, with t = (x - A)/(i).
The median class is the first class whose cumulative frequency is at least (n)/(2); the modal class is the class with the highest frequency.
Convert discontinuous classes to class boundaries before naming the median class; the class marks are unchanged by the conversion.

Questions answered

Define the arithmetic mean of a set of observations.

The arithmetic mean is the sum of all the observations divided by the number of observations, barx = (Sigma x)/(n).

Find the mean of 18, 22, 25, 15, 30 and 22.

Sigma x = 132 and n = 6, so barx = (132)/(6) = 22.

The mean of 6 numbers is 15. If one number is removed, the mean of the remaining numbers is 14. Find the number removed.

The total of 6 numbers is 90 and the total of 5 numbers is 70, so the number removed is 90 - 70 = 20.

The mean of 12 observations is 40. If 5 is added to every observation, what is the new mean? What is the new mean if, instead, every observation is doubled?

Adding 5 to each value gives a mean of 45. Doubling each value gives a mean of 80.

Find the median of 34, 27, 41, 19, 50, 38, 23 and 45.

In order: 19, 23, 27, 34, 38, 41, 45, 50. With n = 8, the median is (34 + 38)/(2) = 36.

Find the mode of 5, 8, 6, 5, 9, 8, 5, 7, 8, 6. Is the data unimodal?

Both 5 and 8 occur three times, more than any other value, so the data is bimodal with modes 5 and 8. It is not unimodal.

Find the mean and the median of the distribution: x = 1, 2, 3, 4, 5 with f = 4, 6, 9, 7, 4.

Sigma f = 30 and Sigma fx = 4 + 12 + 27 + 28 + 20 = 91, so barx = (91)/(30) ≈ 3.03. The cumulative frequencies are 4, 10, 19, 26, 30; the 15th and 16th observations both lie at x = 3, so the median is 3.

Write the class marks and the class boundaries of the classes 21–30 and 31–40.

Class marks 25.5 and 35.5; class boundaries 20.5–30.5 and 30.5–40.5.

Past board exam questions

ICSE 2027 sample paper · 1 mark

Consider the following distribution 30, 34, 35, 36, 37, 38, 39, 40. If the value 35 is removed from the given data, then the median changes by:

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0.5

ICSE 2027 sample paper · 3 marks

The mean of the following distribution is 27 and the sum of all the frequencies is 50. Find the missing frequencies x and y. | Class | 0 − 10 | 10 − 20 | 20 − 30 | 30 − 40 | 40 − 50 | |---|---|---|---|---|---| | Frequency | 8 | x | 12 | 13 | y |

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Class marks: 5, 15, 25, 35, 45. Σ f = 8 + x + 12 + 13 + y = 50 ⇒ x + y = 17 Σ fx = 40 + 15x + 300 + 455 + 45y = 795 + 15x + 45y Mean = (795 + 15x + 45y)/(50) = 27 ⇒ 15x + 45y = 555 ⇒ x + 3y = 37 Subtracting: 2y = 20 ⇒ y = 10, x = 7.

ICSE 2026 board paper · 1 mark

Rakhi's mobile number has the following integers: 1, 6, 9, 8, 9, 1, 7, 8, 9 The mode of the above given data is:

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ICSE 2026 board paper · 3 marks

Find the mean of the following frequency distribution using step-deviation method. Take assumed mean = 28 | Class Interval | 0 – 8 | 8 – 16 | 16 – 24 | 24 – 32 | 32 – 40 | 40 – 48 | |---|---|---|---|---|---|---| | Frequency | 10 | 20 | 14 | 16 | 18 | 22 |

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A = 28, h = 8, u = (x - 28)/(8). | Class | x | f | u | fu | |---|---|---|---|---| | 0 – 8 | 4 | 10 | −3 | −30 | | 8 – 16 | 12 | 20 | −2 | −40 | | 16 – 24 | 20 | 14 | −1 | −14 | | 24 – 32 | 28 | 16 | 0 | 0 | | 32 – 40 | 36 | 18 | 1 | 18 | | 40 – 48 | 44 | 22 | 2 | 44 | Sigma f = 100, Sigma fu = -22 Mean = A + h · (Sigma fu)/(Sigma f) = 28 + 8 × (-22)/(100) = 28 - 1.76 = 26.24.

ICSE 2025 compartment paper · 1 mark

The mean of 1², 2², 3³ and 4⁴ is:

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ICSE 2025 sample paper · 1 mark

Given that the sum of the squares of the first seven natural numbers is 140, then their mean is:

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ICSE 2025 board paper · 1 mark

Assertion(A): The mean of first 9 natural numbers is 4.5. Reason(R): Mean = (Sum of all the observations)/(Total number of observations)

(A) is true, (R) is false.
(A) is false, (R) is true.
Both (A) and (R) are true, and (R) is the correct reason for (A).
Both (A) and (R) are true, and (R) is the incorrect reason for (A).

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(A) is false, (R) is true.

ICSE 2025 sample paper · 4 marks

Using step-deviation method, find mean for the following frequency distribution | Class | 0 – 15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 | |---|---|---|---|---|---|---| | Frequency | 3 | 4 | 7 | 6 | 8 | 2 |

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Assumed mean A = 52.5, class width i = 15, u = (x - 52.5)/(15). | Class | x | u | f | fu | |---|---|---|---|---| | 0 – 15 | 7.5 | −3 | 3 | −9 | | 15 – 30 | 22.5 | −2 | 4 | −8 | | 30 – 45 | 37.5 | −1 | 7 | −7 | | 45 – 60 | 52.5 | 0 | 6 | 0 | | 60 – 75 | 67.5 | 1 | 8 | 8 | | 75 – 90 | 82.5 | 2 | 2 | 4 | | Total | | | 30 | −12 | Mean = A + (Sigma fu)/(Sigma f) × i = 52.5 + (-12)/(30) × 15 = 52.5 - 6 = 46.5.