Measures of Central Tendency Class 11: Mean, Median and Mode with Solved Numericals

What Measures of Central Tendency is about

Weightage: This chapter is part of Unit 3, Statistical Tools and Interpretation, which carries 25 of the 80 theory marks in the school annual exam.Source: CBSE Economics (030) syllabus 2026-27, Class XI course structure, Unit 3 Statistical Tools and Interpretation

Key concepts

Arithmetic mean: individual series

NCERT section 2

Direct method: $\bar{X}=\frac{\sum X}{N}$

Assumed mean (short-cut) method: choose any value A (usually a middle value); find deviations d = X − A; then $\bar{X}=A+\frac{\sum d}{N}$.

Both methods give the same answer; the short-cut saves work with large numbers.

Properties: the sum of deviations from the mean is zero, $\sum (X-\bar{X})=0$; the mean is affected by every value, including extreme ones.

Earnings of 8 auto drivers in a day:

Driver	Earnings X (₹)	d = X − 700
1	640	-60
2	720	20
3	580	-120
4	900	200
5	760	60
6	680	-20
7	820	120
8	700	0
Total	5800	200

Direct: 5800 ÷ 8 = ₹725. Assumed mean: 700 + 200 ÷ 8 = ₹725.

Worked example

Find the mean daily earning of the 8 auto drivers (₹640, 720, 580, 900, 760, 680, 820, 700) by both methods and interpret it.

Try it first, then show the solutionSolution

1. Direct: ΣX = 5800; mean = 5800 ÷ 8 = ₹725
2. Assumed mean A = 700: Σd = 200; mean = 700 + 200 ÷ 8 = ₹725

Answer: ₹725. Interpretation: on average a driver earned ₹725 that day; total earnings of ₹5,800 would give each ₹725 if shared equally.

Arithmetic mean: discrete and continuous series (direct, assumed mean, step deviation)

NCERT section 2

Discrete series: $\bar{X}=\frac{\sum fX}{\sum f}$, or with assumed mean, $\bar{X}=A+\frac{\sum fd}{\sum f}$.

Continuous series: use the mid-points (m) of classes as X:

• Direct: $\bar{X}=\frac{\sum fm}{\sum f}$
• Assumed mean: $\bar{X}=A+\frac{\sum fd}{\sum f}$, where d = m − A
• Step deviation: $\bar{X}=A+\frac{\sum f{d}^{\prime }}{\sum f}\times h$, where d' = (m − A) ÷ h and h is the common class width.

The step deviation method is the quickest with equal class widths. In grouped data we assume all values in a class are at its mid-point.

Daily wages of 50 workers:

Wages (₹)	f	Mid-point m	fm	d' = (m − 350) ÷ 100	fd'	c.f.
100-200	5	150	750	-2	-10	5
200-300	12	250	3000	-1	-12	17
300-400	18	350	6300	0	0	35
400-500	10	450	4500	1	10	45
500-600	5	550	2750	2	10	50
Total	50		17300		-2

Worked example

Find the mean daily wage of the 50 workers above by the direct and step deviation methods.

Try it first, then show the solutionSolution

1. Direct: Σfm = 17300; mean = 17300 ÷ 50 = ₹346
2. Step deviation (A = 350, h = 100): Σfd' = -2; mean = 350 + (-2 ÷ 50) × 100 = ₹346

Answer: ₹346. Interpretation: the average worker earns ₹346 a day; the total daily wage bill is ₹17,300.

Weighted arithmetic mean

NCERT section 2

When items are not equally important, each is given a weight (W):

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

Uses: average price of goods consumed in different quantities, index numbers (Chapter 7), grade point averages with credits, average wage when groups have different numbers of workers.

Subject	Marks X	Credits W	WX
Accountancy	82	4	328
Economics	75	4	300
English	88	3	264
Business Studies	70	3	210
Physical Education	95	2	190
Total		16	1292

Weighted mean = 1292 ÷ 16 = 80.75; simple mean = 410 ÷ 5 = 82. The weighted mean is lower because the high-credit subjects carry lower marks.

Worked example

Suppose a family buys rice at ₹50/kg (20 kg a month), wheat at ₹30/kg (30 kg) and pulses at ₹110/kg (5 kg). Find the average price paid per kg.

Try it first, then show the solutionSolution

1. ΣWX = 50 × 20 + 30 × 30 + 110 × 5 = 1,000 + 900 + 550 = 2,450
2. ΣW = 55 kg
3. Weighted mean = 2,450 ÷ 55 ≈ ₹44.55 per kg

Answer: ≈ ₹44.55 per kg (the simple mean of prices, ₹63.33, would overstate it because little of the costly pulses is bought).

Median and its calculation

NCERT section 3

The median is the value of the middle item when data are arranged in order: half the items are below it and half above. It is a positional average and is not affected by extreme values.

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

Discrete series: find cumulative frequencies; Median = value of X whose c.f. first equals or exceeds the $\frac{N+1}{2}$th item.

Continuous series: median class = class whose c.f. first equals or exceeds N/2; then

$Median=L+\frac{\frac{N}{2}-c.f.}{f}\times h$

with L the starting value of the median class, f its frequency and h its width, and c.f. the running total of frequencies up to the class just before it.

For the 50 workers: N/2 = 25; the median class is 300-400 (c.f. reaches 35); L = 300, c.f. = 17, f = 18: Median = 300 + (25 − 17) ÷ 18 × 100 = ₹344.44.

The median can also be found from the intersection of the two ogives.

Worked example

Find the median earning of the 8 auto drivers.

Try it first, then show the solutionSolution

1. Ordered: 580, 640, 680, 700, 720, 760, 820, 900
2. N = 8 (even): middle items are the 4th and 5th
3. Median = (700 + 720) ÷ 2 = ₹710

Answer: ₹710. Interpretation: half the drivers earned ₹710 or less that day.

Quartiles and percentiles

NCERT section 3

Quartiles divide ordered data into four equal parts:

• Q1 (lower quartile): 25% of items below it. Q1 = size of $\frac{N+1}{4}$th item.
• Q2 = median: 50% below.
• Q3 (upper quartile): 75% below. Q3 = size of $\frac{3(N+1)}{4}$th item.

Q1 and Q3 enclose the central 50% of the data. If the position is not a whole number, interpolate: for the 2.75th item, take the 2nd item + 0.75 × (3rd − 2nd).

Percentiles divide data into 100 equal parts (P1 to P99); P50 is the median, P25 = Q1, P75 = Q3. A score at the 90th percentile means 90% of candidates scored less.

For the 11 values 18, 25, 12, 40, 33, 21, 29, 15, 36, 27, 44: ordered 12, 15, 18, 21, 25, 27, 29, 33, 36, 40, 44; Q1 = 3rd item = 18; Q2 = 6th = 27; Q3 = 9th = 36.

Worked example

Find Q1 for the 10 values 8, 10, 13, 15, 17, 20, 24, 26, 30, 35 (already ordered).

Try it first, then show the solutionSolution

1. Position = (10 + 1) ÷ 4 = 2.75th item
2. Q1 = 2nd item + 0.75 × (3rd − 2nd) = 10 + 0.75 × (13 − 10)
3. = 12.25

Answer: Q1 = 12.25: a quarter of the values are at or below 12.25.

Mode

NCERT section 5

The mode (Mo) is the value that occurs most frequently, the point of maximum concentration. A shoe maker wants the modal size; a shirt maker the modal style.

Individual and discrete series: by inspection, the value with the highest frequency. A series may have no mode, one mode (unimodal), two (bimodal) or more.

Continuous series: the modal class is the class with the highest frequency; then

$Mo=L+\frac{D_1}{D_1+D_2}\times h$

where L = lower limit of the modal class, D1 = f(modal) − f(previous class), D2 = f(modal) − f(next class), h = class width. Classes must be continuous and of equal width around the modal class.

For the 50 workers: modal class 300-400 (f = 18); D1 = 18 − 12 = 6; D2 = 18 − 10 = 8; Mo = 300 + 6 ÷ (6 + 8) × 100 = ₹342.86.

The mode can also be located from a histogram: join the top corners of the tallest rectangle to the adjacent corners of its neighbours; drop a perpendicular from the crossing point.

Worked example

Families classified by number of children: 0: 6; 1: 14; 2: 18; 3: 7; 4: 3; 5: 2. Find the mode, the median and the mean.

Try it first, then show the solutionSolution

1. Children (X)	Families (f)	fX	c.f.
   0	6	0	6
   1	14	14	20
   2	18	36	38
   3	7	21	45
   4	3	12	48
   5	2	10	50
   Total	50	93

2. Mode = 2 (highest frequency 18)
3. Median: (50 + 1) ÷ 2 = 25.5th item; c.f. reaches 38 at X = 2, so median = 2
4. Mean = 93 ÷ 50 = 1.86

Answer: Mode 2, median 2, mean 1.86. Interpretation: the typical family has 2 children; on average about 1.86.

Choosing an average and relative position of mean, median and mode

NCERT section 6

Average	Merits	Limits	Best used for
Mean	Uses every value; simple; suits further algebra	Pulled by extreme values; cannot be found for open-ended classes without assumptions; may not be an actual value	Average income, output, marks when no extremes
Median	Not affected by extremes; can be found for open-ended classes and graphically	Ignores most values; needs ordering	Income and wealth (skewed data); ranks
Mode	Easy; shows the most typical value; can be used for attributes	May not exist or may be more than one; unstable	Most popular size, style, brand

Relative position: in a symmetrical distribution, mean = median = mode. In a moderately skewed distribution, the median always lies between the mean and the mode: either Mean > Median > Mode (positive skew, a long tail of high values) or Mean < Median < Mode (negative skew).

For the 50 workers: mean ₹346 > median ₹344.44 > mode ₹342.86: a slight positive skew, with a few high earners pulling the mean up.

Worked example

Suppose the incomes of five families are ₹15,000, ₹18,000, ₹20,000, ₹22,000 and ₹2,25,000. Which average best describes a typical family and why?

Try it first, then show the solutionSolution

1. Mean = 3,00,000 ÷ 5 = ₹60,000, pulled up by one very high income
2. Median = ₹20,000, the middle value
3. Four of five families earn near ₹20,000

Answer: The median (₹20,000), because the mean is distorted by one extreme value.

Measures of Central Tendency 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: Direct, Assumed and Weighted
• Median, Quartiles and Mode

Measures of Central Tendency: key facts

Mean (individual)

\bar{X} = \frac{\sum X}{N} = A + \frac{\sum d}{N}

d = X − A.

Mean (discrete)

\bar{X} = \frac{\sum fX}{\sum f} = A + \frac{\sum fd}{\sum f}

Weights are frequencies.

Mean (continuous, step deviation)

\bar{X} = A + \frac{\sum fd'}{\sum f} \times h

d' = (m − A) ÷ h; m = mid-point.

Weighted mean

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

Used in index numbers.

Median (continuous)

Me = L + \frac{N/2 - c.f.}{f} \times h

c.f. of the class before the median class.

Median position (individual/discrete)

\left(\frac{N+1}{2}\right)^{th}\ item

Even N: average of the two middle items.

Quartiles

Q_1 = \left(\frac{N+1}{4}\right)^{th},\ Q_3 = \left(\frac{3(N+1)}{4}\right)^{th}\ item

Interpolate for fractional positions.

Mode (continuous)

Mo = L + \frac{D_1}{D_1 + D_2} \times h

D1 = f1 − f0, D2 = f1 − f2.

Combined mean

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

Mean of two groups together.

How the school exam asks Measures of Central Tendency

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

Tricks to remember Measures of Central Tendency

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

• Median formula

  "Lower limit + (half N minus cf before) over f, times h": L + (N/2 − cf) ÷ f × h.

  Use it when: Median and quartiles of grouped data.

• Mode formula

  D1 = this minus before, D2 = this minus after; L + D1 ÷ (D1 + D2) × h.

  Use it when: Mode of grouped data.

• Relative position

  Alphabetical order: MEan, MEdian, MOde: the median (middle letter group) sits in the middle.

  Use it when: Relative position of averages.

Measures of Central Tendency: 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 E 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 1 + 1 + 2 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 Tendency

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

• ✗Mistake: Using the c.f. of the median class instead of the class before

  The c.f. column is read at the wrong row.

  Wrong working

  Me = 300 + (25 − 35) ÷ 18 × 100

  Right working

  Me = 300 + (25 − 17) ÷ 18 × 100 = 344.44

  ✓Do this: c.f. in the formula is of the preceding class.

  Usually costs: 1-2 marks.

• ✗Mistake: Forgetting to multiply by h in the step deviation method

  The formula is long.

  Wrong working

  Mean = 350 + (−2 ÷ 50) = 349.96

  Right working

  Mean = 350 + (−2 ÷ 50) × 100 = 346

  ✓Do this: Write the formula fully: A + (Σfd' ÷ Σf) × h.

  Usually costs: 1 mark.

• ✗Mistake: Not arranging data before finding the median

  Picking the middle of the given list.

  ✓Do this: Always sort first; the median is positional.

  Usually costs: 1-2 marks.

• ✗Mistake: Finding the mode when classes are unequal or inclusive without adjustment

  Applying the formula directly.

  ✓Do this: Make classes continuous (exclusive) and equal around the modal class first.

  Usually costs: 1 mark.

• ✗Mistake: No economic interpretation

  Students stop at the number.

  ✓Do this: Add one line: "half the workers earn less than ₹344" or "the average family has about 2 children". The syllabus asks for it.

  Usually costs: ½-1 mark per numerical.

NCERT exercise guide: Measures of Central Tendency

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Measures of Central Tendency

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

   The value that occurs most often in a series is the:
   (a) mean
   (b) median
   (c) mode
   (d) quartile

   Hint

   Most fashionable.

   Answer

   (c) mode

2. Q2 · 1 mark · MCQExam level

   In a moderately skewed distribution:
   (a) the mean always lies between median and mode
   (b) the median always lies between mean and mode
   (c) the mode always lies between mean and median
   (d) all three are equal

   Hint

   NCERT relative position.

   Answer

   (b)

3. Q3 · 1 mark · Very short answerExam level

   What percentage of items lie below the upper quartile?

   Hint

   Q3.

   Answer

   75%.

4. Q4 · 3 marks · Short answerExam level

   Find the mean by the assumed mean method: 45, 52, 38, 61, 49, 55 (take A = 50).

   Hint

   Deviations from 50.

   Answer

   d: −5, 2, −12, 11, −1, 5; Σd = 0; mean = 50 + 0 ÷ 6 = 50.

5. Q5 · 3 marks · Short answerHarder exam level

   Find the median: 34, 28, 45, 19, 52, 41, 37, 30, 26.

   Hint

   Order first; N = 9.

   Answer

   Ordered: 19, 26, 28, 30, 34, 37, 41, 45, 52; median = 5th item = 34.

6. Q6 · 4 marks · Short answerHarder exam level

   Find the mean of: marks 10-20: 3; 20-30: 7; 30-40: 15; 40-50: 10; 50-60: 5, by the step deviation method.

   Hint

   A = 35, h = 10.

   Answer

   d': −2, −1, 0, 1, 2; fd': −6, −7, 0, 10, 10; Σfd' = 7; N = 40; mean = 35 + 7/40 × 10 = 36.75 marks.

7. Q7 · 4 marks · Short answerHarder exam level

   For the same distribution, find the median and mode.

   Hint

   N/2 = 20.

   Answer

   c.f.: 3, 10, 25, 35, 40; median class 30-40; Me = 30 + (20 − 10) ÷ 15 × 10 = 36.67. Modal class 30-40; D1 = 8, D2 = 5; Mo = 30 + 8/13 × 10 = 36.15.

8. Q8 · 3 marks · Short answerHarder exam level

   The mean marks of 25 boys are 62 and of 15 girls 70. Find the mean of the class.

   Hint

   Combined mean.

   Answer

   (25 × 62 + 15 × 70) ÷ 40 = (1,550 + 1,050) ÷ 40 = 65.

9. Q9 · 3 marks · Short answerHarder exam level

   Find Q1 and Q3: 5, 9, 12, 15, 18, 21, 24 (ordered).

   Hint

   N = 7.

   Answer

   Q1 = (7 + 1)/4 = 2nd item = 9; Q3 = 3 × 8/4 = 6th item = 21.

10. Q10 · 4 marks · Long answerHarder exam level

    Give two merits and two limits each of the median and the mode.

    Hint

    Extremes, open classes, existence.

    Answer

    Median: merits: not affected by extremes; usable with open-ended classes and graphically. Limits: ignores most values; needs ordering. Mode: merits: shows the most typical value; easy and usable for attributes. Limits: may not exist or may be multiple; unstable with small changes in data.

11. Q11 · 3 marks · Short answerHarder exam level

    The mean of 10 observations is 42. Later it is found that one value was read as 38 instead of 83. Find the correct mean.

    Hint

    Correct the total.

    Answer

    Wrong total 420; correct total = 420 − 38 + 83 = 465; correct mean = 46.5.

12. Q12 · 4 marks · Long answerHOTS / competency

    Suppose in a town the mean household income is ₹48,000 and the median ₹31,000. What does this tell you about income distribution, and which figure should the government use to describe a "typical" family?

    Hint

    Mean > median.

    Answer

    Mean much higher than median means a positively skewed distribution: a few very high incomes pull the mean up while most families earn much less. The median (₹31,000) better describes a typical family; using the mean would overstate how well-off most families are.

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 for mean by step deviation?Show answer

   A + (Σfd' ÷ Σf) × h.

2. 2.What is the sum of deviations from the mean?Show answer

   Zero.

3. 3.Position of the median in an individual series?Show answer

   (N + 1) ÷ 2th item.

4. 4.What is Q1?Show answer

   The value below which 25% of items lie.

5. 5.What is a bimodal series?Show answer

   One with two modes.

6. 6.Which average suits open-ended classes?Show answer

   Median (or mode).

7. 7.Where does the median lie relative to mean and mode?Show answer

   Between them.

Quick revision checklist

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

• ☐Can you find the mean by three methods and get the same answer?
• ☐Can you compute a weighted mean and say why it differs from the simple mean?
• ☐Can you find a missing frequency from a given mean?
• ☐Can you find the median of individual, discrete and continuous series?
• ☐Can you find Q1 and Q3 with interpolation?
• ☐Can you find the mode by formula and from a histogram?
• ☐Can you choose the right average for a situation?
• ☐Can you state the relative position of mean, median and mode?
• ☐Can you add a one-line economic interpretation to each answer?

Measures of Central Tendency: frequently asked questions