Measures of Central Value: ISC Class 11 Mean, Median and Mode

What Measures of Central Value is about

Weightage: Measures of Central Value is part of the unit Statistics of the ISC Class XI Economics syllabus. CISCE gives no Class XI unit weightage for Economics: Paper I (Theory) carries 80 marks, and the school sets the paper.Source: CISCE ISC Economics syllabus, examination year 2028, Class XI paper pattern

Key concepts

Average: meaning and qualities of a good average

An average (measure of central tendency or central value) is a single value that represents a whole set of data; it lies somewhere in the middle, where values tend to cluster.

Purposes: to give a brief picture of large data; to compare groups; to help decisions and policy; to be the base for further analysis (dispersion, skewness).

Qualities of a good average:

1. Rigidly defined: one clear value, not a matter of judgement.
2. Easy to understand and calculate.
3. Based on all observations.
4. Not unduly affected by extreme values.
5. Capable of further algebraic treatment (for example, combining means of groups).
6. Sampling stability: different samples give similar values.

No average has all these qualities; the choice depends on the data and the purpose.

The three averages in the syllabus: arithmetic mean (the sum divided by the number), median (the middle value), mode (the most frequent value).

Worked example

Which quality of a good average does the mean lack when a dataset contains one very large value? [1 mark]

Try it first, then show the solutionSolution

1. The mean uses every value, so one extreme value pulls it up.

Answer: It is unduly affected by extreme values.

Arithmetic mean: individual and discrete series

Individual series: $\bar{X}=\frac{\sum X}{N}$

Short-cut (assumed mean) method: $\bar{X}=A+\frac{\sum d}{N}$, where $d=X-A$.

Discrete series (values with frequencies): $\bar{X}=\frac{\sum fX}{\sum f}$

Example (individual): marks 45, 52, 38, 61, 54: $\sum X=250$, N = 5, mean = 50.

Example (discrete): number of children per household in 50 households:

Children (X)	0	1	2	3	4
Households (f)	6	14	18	8	4
fX	0	14	36	24	16

$\sum fX=90$, $\sum f=50$, mean = 90 ÷ 50 = 1.8 children.

Properties: (1) the sum of deviations from the mean is zero, $\sum (X-\bar{X})=0$; (2) the sum of squared deviations is least about the mean; (3) combined mean of groups: ${\bar{X}}_{12}=\frac{N_1{\bar{X}}_1+N_2{\bar{X}}_2}{N_1+N_2}$.

Worked example

Suppose Section A has 30 students with mean marks 62 and Section B has 20 students with mean marks 72. Find the combined mean. [2 marks]

Try it first, then show the solutionSolution

1. Totals: 30 × 62 = 1,860; 20 × 72 = 1,440.
2. Combined mean = (1,860 + 1,440) ÷ 50 = 3,300 ÷ 50 = 66.

Answer: 66 marks.

Arithmetic mean: continuous series and weighted mean

Continuous series: use class marks (mid-points) m.

• Direct: $\bar{X}=\frac{\sum fm}{\sum f}$
• Step-deviation: $\bar{X}=A+\frac{\sum f{d}^{\prime }}{\sum f}\times i$, where $d^{\prime }=(m-A)÷i$ and i is the class width.

Example: daily wages (₹) of 50 workers:

Wages	100-200	200-300	300-400	400-500	500-600	600-700
f	5	8	15	12	6	4
m	150	250	350	450	550	650
d' (A = 350, i = 100)	−2	−1	0	1	2	3
fd'	−10	−8	0	12	12	12

$\sum f{d}^{\prime }=18$; mean = 350 + (18 ÷ 50) × 100 = ₹386.

Weighted arithmetic mean: when items differ in importance, each is multiplied by its weight W:

${\bar{X}}_w=\frac{\sum WX}{\sum W}$

Example: a student's marks: English 70 (weight 2), Maths 85 (4), Economics 78 (3), Accounts 64 (3). $\sum WX=140+340+234+192=906$; $\sum W=12$; weighted mean = 75.5, while the simple mean is 74.25. Weights matter: Maths, the heaviest-weighted subject, has the highest marks.

Weighted means are used in index numbers, cost-of-living calculations and combined results.

Worked example

Suppose a family buys rice at ₹50/kg (20 kg), dal at ₹120/kg (5 kg) and oil at ₹160/kg (4 kg). Find the weighted average price per kg. [2 marks]

Try it first, then show the solutionSolution

1. ΣWX = 50 × 20 + 120 × 5 + 160 × 4 = 1,000 + 600 + 640 = 2,240; ΣW = 29.
2. Weighted mean = 2,240 ÷ 29 = ₹77.24 per kg (the simple mean of prices would be ₹110).

Answer: ₹77.24 per kg.

Median

The median is the middle value of a series arranged in order; it divides the data into two equal halves. It is a positional average.

Individual series: arrange in order; median = size of the $\frac{N+1}{2}$th item. If N is even, take the mean of the two middle items.

Example: 12, 7, 15, 9, 20, 11, 18, 14 → ordered 7, 9, 11, 12, 14, 15, 18, 20; N = 8; median = (12 + 14) ÷ 2 = 13.

Discrete series: find cumulative frequencies; median = value at the $\frac{N+1}{2}$th item. Children example: N = 50, (50 + 1) ÷ 2 = 25.5th item; cumulative frequencies 6, 20, 38, ... so the 25.5th item is 2 children.

Continuous series: median class contains the $\frac{N}{2}$th item.

$\text{Median}=L_1+\frac{\frac{N}{2}-c.f.}{f}\times i$

where $L_1$ = lower limit of the median class, c.f. = cumulative frequency of the class before it, f = frequency of the median class, i = class width.

Wages example: N ÷ 2 = 25; cumulative frequencies 5, 13, 28, ...; median class 300-400 (c.f. before = 13, f = 15): median = 300 + (25 − 13) ÷ 15 × 100 = ₹380.

Merits: not affected by extreme values; can be found for open-ended classes; can be located graphically (ogives). Demerits: not based on all values; not suitable for algebraic treatment.

Worked example

Find the median of 23, 41, 17, 35, 29, 50, 38. [1 mark]

Try it first, then show the solutionSolution

1. Ordered: 17, 23, 29, 35, 38, 41, 50; N = 7; (7 + 1) ÷ 2 = 4th item = 35.

Answer: 35.

Mode

The mode is the value that occurs most often: the most typical value (the most common shoe size, the usual wage).

Individual and discrete series: by inspection, the value with the highest frequency. Children example: highest frequency 18 → mode = 2 children. (If frequencies are irregular or two values are close, the grouping method is used.)

Continuous series: identify the modal class (highest frequency), then:

$\text{Mode}=L_1+\frac{f_1-f_0}{2f_1-f_0-f_2}\times i$

where $f_1$ = frequency of the modal class, $f_0$ = of the class before it, $f_2$ = of the class after it.

Wages example: modal class 300-400 ($f_1$ = 15, $f_0$ = 8, $f_2$ = 12): mode = 300 + (15 − 8) ÷ (30 − 8 − 12) × 100 = 300 + 7 ÷ 10 × 100 = ₹370.

Graphical location: in a histogram, join the top corners of the tallest bar to the adjacent bars' corners crosswise; drop a perpendicular from the intersection.

Merits: easy; unaffected by extremes; useful for qualitative data (most popular colour). Demerits: may not exist or there may be several modes (bimodal); not based on all values; unsuitable for algebra.

Conditions: classes must be equal and exclusive before applying the formula.

Worked example

Suppose shoe sizes sold in a day are: 6, 7, 8, 7, 9, 7, 8, 6, 7, 10. Find the mode and say why a shopkeeper prefers it to the mean. [1 mark]

Try it first, then show the solutionSolution

1. 7 occurs 4 times, more than any other size: mode = 7.

Answer: Mode = size 7; the shopkeeper stocks the most demanded size, and a mean size (7.5) may not even exist.

Relationship between mean, median and mode; skewness

Symmetrical distribution: the curve is bell-shaped and balanced: Mean = Median = Mode.

Positively skewed (long tail to the right; a few very high values): Mean > Median > Mode. Example: incomes, where a few very rich people pull the mean up. Our wage data: mean ₹386 > median ₹380 > mode ₹370: positively skewed.

Negatively skewed (long tail to the left; a few very low values): Mean < Median < Mode. Example: marks in an easy test, where most score high and a few score very low.

Empirical relationship (Karl Pearson) for moderately skewed distributions:

Mode = 3 Median − 2 Mean

The median always lies between the mean and the mode, and is about one-third of the way from the mean towards the mode.

Which average to use? Mean for symmetrical data and further calculations; median for skewed data such as incomes and for open-ended classes; mode for the most typical value or qualitative data.

Worked example

Suppose a distribution has mean 40 and median 38. Estimate the mode and state the nature of the distribution. [2 marks]

Try it first, then show the solutionSolution

1. Mode = 3 × 38 − 2 × 40 = 114 − 80 = 34.
2. Mean 40 > median 38 > mode 34: positively skewed.

Answer: Mode ≈ 34; positively skewed.

Measures of Central Value in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Topics in this chapter

• Arithmetic Mean: Individual, Discrete, Continuous and Weighted Series
• Median, Mode and the Relationship among Mean, Median and Mode

Measures of Central Value: key facts

Mean (individual)

$\bar{X}=\frac{\sum X}{N}$

Mean (discrete and continuous)

$\bar{X}=\frac{\sum fX}{\sum f}$ or, with step deviation, $\bar{X}=A+\frac{\sum f{d}^{\prime }}{\sum f}\times i$

In a continuous series, X is the class mark.

Weighted mean

${\bar{X}}_w=\frac{\sum WX}{\sum W}$

Combined mean

${\bar{X}}_{12}=\frac{N_1{\bar{X}}_1+N_2{\bar{X}}_2}{N_1+N_2}$

Median (continuous)

$\text{Median}=L_1+\frac{\frac{N}{2}-c.f.}{f}\times i$

Individual or discrete: the (N + 1) ÷ 2 th item.

Mode (continuous)

$\text{Mode}=L_1+\frac{f_1-f_0}{2f_1-f_0-f_2}\times i$

Empirical relationship

Mode = 3 Median − 2 Mean

For moderately skewed distributions.

Measures of Central Value: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Measures of Central Value

The question patterns to expect, and the method that works for each.

Tricks to remember Measures of Central Value

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

• Mean chases the tail

  In skewed data, the mean moves towards the long tail; the mode stays at the peak; the median sits between.

  Use it when: Skewness and order of averages.

• Mode = 3Md − 2M̄

  'Three medians minus two means'.

  Use it when: Empirical relationship.

• Median formula: L + (half − before) ÷ f × i

  Start at the lower limit, see how far into the class the middle item is.

  Use it when: Median of a continuous series.

• Mode formula: difference over two differences

  (f1 − f0) ÷ [(f1 − f0) + (f1 − f2)].

  Use it when: Mode of a continuous series.

Measures of Central Value: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section C style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 2 + 1 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Measures of Central Value

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

• ✗Mistake: Using (N + 1) ÷ 2 in a continuous series

  For grouped continuous data, the median class is located by N ÷ 2.

  ✓Do this: Individual and discrete: (N + 1) ÷ 2; continuous: N ÷ 2.

  Usually costs: 1 mark.

• ✗Mistake: Applying the mode formula to unequal or inclusive classes

  The formula assumes equal, exclusive classes.

  ✓Do this: Convert to exclusive classes and make widths equal first.

  Usually costs: 1-2 marks.

• ✗Mistake: Taking the simple mean when weights differ

  Items of different importance or size must be weighted.

  ✓Do this: Ask: do the items represent different numbers of people or quantities?

  Usually costs: 1-2 marks.

• ✗Mistake: Writing the formula without substituting values

  Step marks are given for showing L1, c.f., f and i (or f0, f1, f2) clearly.

  ✓Do this: List the values of every symbol before substituting.

  Usually costs: 1 mark.

Practice questions on Measures of Central Value

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   For a symmetrical distribution:
   (a) mean > median > mode
   (b) mean < median < mode
   (c) mean = median = mode
   (d) mode = 2 median − 3 mean

   Hint

   Balanced curve.

   Answer

   (c) mean = median = mode.

2. Q2 · 1 mark · Very short answerExam level

   Find the mean of 18, 22, 25, 30, 35.

   Hint

   Sum ÷ 5.

   Answer

   130 ÷ 5 = 26.

3. Q3 · 2 marks · Short answerExam level

   Find the median and mode of: 5, 8, 6, 9, 8, 7, 8, 10, 6, 4.

   Hint

   Order first.

   Answer

   Ordered: 4, 5, 6, 6, 7, 8, 8, 8, 9, 10; median = (7 + 8) ÷ 2 = 7.5; mode = 8 (occurs 3 times).

4. Q4 · 2 marks · Short answerHarder exam level

   Suppose the mean wage of 60 workers in a factory is ₹500. The mean of the 20 women is ₹470. Find the mean wage of the men.

   Hint

   Combined mean in reverse.

   Answer

   Total = 60 × 500 = 30,000; women = 20 × 470 = 9,400; men total = 20,600 for 40 men: mean = ₹515.

5. Q5 · 2 marks · Short answerHarder exam level

   Find the median of the discrete series: X: 10, 20, 30, 40, 50; f: 4, 7, 12, 9, 3.

   Hint

   (N + 1) ÷ 2 th item.

   Answer

   N = 35; (35 + 1) ÷ 2 = 18th item; cumulative frequencies 4, 11, 23, ... → 18th item lies at X = 30. Median = 30.

6. Q6 · 6 marks · Long answerHOTS / competency

   Calculate the mean, median and mode of the marks of 40 students: 0-10: 3, 10-20: 7, 20-30: 12, 30-40: 10, 40-50: 8.

   Hint

   Use the formulas for each.

   Answer

   Mean: Σfm = 15 + 105 + 300 + 350 + 360 = 1,130; mean = 1,130 ÷ 40 = 28.25. Median: N ÷ 2 = 20; c.f. 3, 10, 22 → class 20-30; median = 20 + (20 − 10) ÷ 12 × 10 = 28.33. Mode: modal class 20-30 (f1 = 12, f0 = 7, f2 = 10); mode = 20 + 5 ÷ (24 − 7 − 10) × 10 = 20 + 5 ÷ 7 × 10 = 27.14. The three averages are close (27.14 to 28.33) and follow no clear skewed order, so the distribution is nearly symmetrical.

7. Q7 · 2 marks · Short answerHarder exam level

   Suppose the mode of a distribution is 42 and the mean is 48. Estimate the median.

   Hint

   Rearrange the empirical formula.

   Answer

   Mode = 3 Median − 2 Mean → 42 = 3 Median − 96 → Median = 138 ÷ 3 = 46.

8. Q8 · 2 marks · Short answerHarder exam level

   Suppose a price index uses a weighted mean of price relatives: food 120 (weight 50), fuel 140 (weight 10), clothing 110 (weight 15), rent 130 (weight 25). Calculate the weighted mean.

   Hint

   ΣWX ÷ ΣW.

   Answer

   ΣWX = 6,000 + 1,400 + 1,650 + 3,250 = 12,300; ΣW = 100; weighted mean = 123.

9. Q9 · 4 marks · Short answerHOTS / competency

   Suppose the incomes of five households are ₹20,000, ₹22,000, ₹25,000, ₹28,000 and ₹5,00,000. Calculate the mean and median, and explain which better represents a typical household and why.

   Hint

   Extreme value.

   Answer

   Mean = 5,95,000 ÷ 5 = ₹1,19,000; median = ₹25,000. The median is better: the single extreme income of ₹5 lakh inflates the mean so that four of the five households earn far less than it.

10. Q10 · 2 marks · Short answerHOTS / competency

    Suppose the mean of 10 numbers is 30. If each number is increased by 5 and then multiplied by 2, what is the new mean? Explain the property used.

    Hint

    Mean changes like each value.

    Answer

    Adding 5 raises the mean to 35; multiplying by 2 gives 70. The mean is affected by changes of origin and scale in the same way as each observation.

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. 1.Formula of weighted mean?Show answer

   ΣWX ÷ ΣW.

2. 2.Median position in a continuous series?Show answer

   N ÷ 2 th item.

3. 3.Mode formula (continuous)?Show answer

   L1 + (f1 − f0) ÷ (2f1 − f0 − f2) × i.

4. 4.Order of averages in positive skew?Show answer

   Mean > Median > Mode.

5. 5.Empirical relationship?Show answer

   Mode = 3 Median − 2 Mean.

6. 6.Sum of deviations from the mean?Show answer

   Zero.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Mean = ΣX ÷ N; ΣfX ÷ Σf; A + (Σfd' ÷ Σf) × i.
• ☐Weighted mean = ΣWX ÷ ΣW; combined mean = (N1X̄1 + N2X̄2) ÷ (N1 + N2).
• ☐Median: (N + 1) ÷ 2 th item; continuous: L1 + (N/2 − c.f.) ÷ f × i.
• ☐Mode: highest frequency; continuous: L1 + (f1 − f0) ÷ (2f1 − f0 − f2) × i.
• ☐Symmetrical: all equal; positive skew: mean > median > mode; negative: reverse.
• ☐Mode ≈ 3 Median − 2 Mean.

Measures of Central Value: frequently asked questions