NCERT Solutions for Class 11 Economics Chapter 5: Measures of Central Tendency

These Class 11 Economics Chapter 5 solutions cover Measures of Central Tendency from the NCERT textbook Statistics for Economics, updated for the 2026–27 session. The chapter explains how a large mass of data can be summarised by a single representative value — the arithmetic mean, the median and the mode — along with positional measures such as quartiles and percentiles. Below you will find every end-of-chapter exercise reproduced verbatim and solved step by step, with all numericals on mean, median and mode worked out and verified, plus key formulas, extra practice, MCQs, Assertion–Reason questions and FAQs.

Class: 11 Subject: Economics Book: Statistics for Economics Chapter: 5 Topic: Measures of Central Tendency Session: 2026–27
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Class 11 Economics Chapter 5 – Overview

A measure of central tendency is a single value that summarises an entire data set by indicating its centre or “typical” value. The three most commonly used averages are the arithmetic mean (sum of observations divided by their number), the median (the middle value when data is arranged in order, which divides the distribution into two equal halves) and the mode (the value that occurs most frequently). The chapter shows how each is computed for individual, discrete and continuous series, using the direct, assumed mean and step deviation methods for the mean. It also introduces quartiles (which split data into four equal parts) and percentiles (a hundred parts). A key idea is that the mean is affected by extreme values, the median is not, and the mode best suits qualitative data — so the choice of average depends on the purpose of analysis and the nature of the data. For a moderately asymmetrical distribution the median always lies between the mean and the mode.

Key Concepts & Important Formulas

Arithmetic Mean (X̄): the sum of the values of all observations divided by the number of observations. It uses every value but is unduly affected by extreme items.

Median: the positional middle value that divides an ordered distribution into two equal parts. It is not affected by extreme values.

Mode: the value that occurs most frequently in a series. It is best for describing qualitative data and may be unimodal, bimodal or multimodal.

Quartiles & Percentiles: quartiles (Q₁, Q₂, Q₃) divide data into four equal parts; percentiles (P₁…P₉₉) into a hundred. Here Q₂ = P₅₀ = Median.

Relative position: for a moderately skewed distribution, the median always lies between the mean and the mode.

Arithmetic Mean — Individual series:
Direct: X̄ = ΣX / N
Assumed mean: X̄ = A + (Σd / N), where d = X − A
Step deviation: X̄ = A + (Σd′ / N) × c, where d′ = (X − A) / c

Arithmetic Mean — Grouped data:
Direct: X̄ = ΣfX / Σf (or Σfm / Σf for continuous, m = mid-value)
Step deviation: X̄ = A + (Σfd′ / Σf) × c

Median:
Position = size of [(N + 1) / 2]th item (individual & discrete series)
Continuous series: Median = L + [(N/2 − c.f.) / f] × h
where L = lower limit of median class, c.f. = cumulative frequency of the class before the median class, f = frequency of the median class, h = class width.

Quartiles (continuous): Q₁ = L + [(N/4 − c.f.) / f] × h ;  Q₃ = L + [(3N/4 − c.f.) / f] × h

Mode (continuous): Mₒ = L + [D₁ / (D₁ + D₂)] × h
where D₁ = (frequency of modal class − frequency of preceding class), D₂ = (frequency of modal class − frequency of succeeding class), both ignoring signs.

NCERT “Exercises” — Full Solutions

All questions below are reproduced verbatim from the NCERT textbook’s end-of-chapter Exercises. Answers are original; every numerical is solved step by step and verified against the textbook’s answer key.

1. Which average would be suitable in the following cases? (i) Average size of readymade garments. (ii) Average intelligence of students in a class. (iii) Average production in a factory per shift. (iv) Average wage in an industrial concern. (v) When the sum of absolute deviations from average is least. (vi) When quantities of the variable are in ratios. (vii) In case of open-ended frequency distribution.

ANSWER (i) Mode — a garment manufacturer needs the size that is demanded most frequently, which is exactly what the mode gives. (ii) Median — intelligence is a qualitative trait that is ranked rather than measured precisely, so the positional middle value (median) is most suitable. (iii) Arithmetic Mean — production per shift is a quantitative figure based on all observations, so the mean gives the best representative output. (iv) Arithmetic Mean — to find the representative wage based on all wages paid, the arithmetic mean is appropriate. (v) Median — the sum of the absolute deviations of items is least when measured from the median. (vi) Geometric Mean — when the variable is expressed in ratios, rates or percentage changes, the geometric mean is the suitable average. (vii) Median — in an open-ended distribution the extreme class limits are not known, so the mean cannot be found reliably; the median, being positional, can still be computed.

2. Indicate the most appropriate alternative from the multiple choices provided against each question. (i) The most suitable average for qualitative measurement is (a) arithmetic mean (b) median (c) mode (d) geometric mean (e) none of the above (ii) Which average is affected most by the presence of extreme items? (a) median (b) mode (c) arithmetic mean (d) none of the above (iii) The algebraic sum of deviation of a set of n values from A.M. is (a) n (b) 0 (c) 1 (d) none of the above

ANSWER (i) (b) median. Qualitative characteristics (such as honesty or intelligence) cannot be measured numerically but can be ranked, so the positional median is the most suitable average. (ii) (c) arithmetic mean. Because the mean uses the actual value of every item, a single very large or very small value pulls it up or down. (iii) (b) 0. The sum of the deviations of all items from the arithmetic mean is always zero, i.e. Σ(X − X̄) = 0.

3. Comment whether the following statements are true or false. (i) The sum of deviation of items from median is zero. (ii) An average alone is not enough to compare series. (iii) Arithmetic mean is a positional value. (iv) Upper quartile is the lowest value of top 25% of items. (v) Median is unduly affected by extreme observations.

ANSWER (i) False. It is the sum of the algebraic deviations from the arithmetic mean that is zero. From the median, it is the sum of the absolute deviations that is least, not zero. (ii) True. An average shows only the central value; to compare two series we also need a measure of dispersion (such as range or standard deviation) to know how spread out the data is. (iii) False. The arithmetic mean is a calculated value based on all observations. It is the median and the mode that are positional values. (iv) True. The upper quartile (Q₃) has 75% of the items below it and 25% above it, so it marks the lowest value of the top 25% of items. (v) False. The median is a positional value determined by the middle item; it is not affected by extreme observations. The arithmetic mean is the one that is unduly affected.

4. If the arithmetic mean of the data given below is 28, find (a) the missing frequency, and (b) the median of the series:

Profit per retail shop (in Rs)0–1010–2020–3030–4040–5050–60
Number of retail shops121827–176
ANSWER (a) Missing frequency. Let the missing frequency be f. Using mid-values (m) and the direct method:
Profit (Rs)Mid-value (m)Frequency (f)fm
0–1051260
10–201518270
20–302527675
30–4035f35f
40–504517765
50–60556330
Total—80 + f2100 + 35f
X̄ = Σfm / Σf ⇒ 28 = (2100 + 35f) / (80 + f) 28(80 + f) = 2100 + 35f ⇒ 2240 + 28f = 2100 + 35f ⇒ 140 = 7f ⇒ f = 20. (b) Median. With f = 20, total N = 80 + 20 = 100, so N/2 = 50. Cumulative frequencies: 12, 30, 57, 77, 94, 100. The 50th item lies in the 20–30 class (c.f. just before = 30, f = 27, L = 20, h = 10). Median = L + [(N/2 − c.f.) / f] × h = 20 + [(50 − 30) / 27] × 10 = 20 + (200 / 27) = 20 + 7.41 = Rs 27.41. Result: missing frequency = 20 and median = Rs 27.41. (Matches the textbook answer.)

5. The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

WorkersABCDEFGHIJ
Daily Income (in Rs)120150180200250300220350370260
ANSWER This is an individual series, so use the direct method: X̄ = ΣX / N. ΣX = 120 + 150 + 180 + 200 + 250 + 300 + 220 + 350 + 370 + 260 = 2400; N = 10. X̄ = 2400 / 10 = Rs 240. The average daily income of the ten workers is Rs 240. (Matches the textbook answer.)

6. Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.

Income (in Rs)Number of families
More than 75150
More than 85140
More than 95115
More than 10595
More than 11570
More than 12560
More than 13540
More than 14525
ANSWER This is a “more than” cumulative distribution. First convert it into a simple (exclusive) frequency distribution by taking successive differences, then apply the direct method.
Income (Rs)Mid-value (m)Frequency (f)fm
75–8580150 − 140 = 10800
85–9590140 − 115 = 252250
95–105100115 − 95 = 202000
105–11511095 − 70 = 252750
115–12512070 − 60 = 101200
125–13513060 − 40 = 202600
135–14514040 − 25 = 152100
145–155150253750
Total—15017450
X̄ = Σfm / Σf = 17450 / 150 = Rs 116.33 (≈ Rs 116.3). The average daily income is about Rs 116.3. (Matches the textbook answer.)

7. The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.

Size of Land Holdings (in acres)Less than 100100–200200–300300–400400 and above
Number of families40891486439
ANSWER Prepare the cumulative frequency column. N = 380, so N/2 = 190.
Size of holdings (acres)No. of families (f)Cumulative frequency (c.f.)
0–1004040
100–20089129
200–300148277
300–40064341
400–50039380
N/2 = 190 lies in the 200–300 class (c.f. before = 129, f = 148, L = 200, h = 100). Median = L + [(N/2 − c.f.) / f] × h = 200 + [(190 − 129) / 148] × 100 = 200 + (6100 / 148) = 200 + 41.22 = 241.22 acres. (Matches the textbook answer.)

8. The following series relates to the daily income of workers employed in a firm. Compute (a) highest income of lowest 50% workers (b) minimum income earned by the top 25% workers and (c) maximum income earned by lowest 25% workers. (Hint: compute median, lower quartile and upper quartile.)

Daily Income (in Rs)10–1415–1920–2425–2930–3435–39
Number of workers5101520105
ANSWER The class intervals are inclusive, so convert them to exclusive form by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit (h = 5). Then form the cumulative frequency column. N = 65.
Income (exclusive)No. of workers (f)Cumulative frequency (c.f.)
9.5–14.555
14.5–19.51015
19.5–24.51530
24.5–29.52050
29.5–34.51060
34.5–39.5565
(a) Highest income of the lowest 50% workers = Median. N/2 = 32.5, which lies in the 24.5–29.5 class (L = 24.5, c.f. = 30, f = 20, h = 5).
Median = 24.5 + [(32.5 − 30) / 20] × 5 = 24.5 + (12.5 / 20) = 24.5 + 0.63 = ≈ Rs 25.13 (textbook key: Rs 25.11). (b) Minimum income earned by the top 25% workers = Upper Quartile (Q₃). 3N/4 = 48.75, which lies in the 24.5–29.5 class (L = 24.5, c.f. = 30, f = 20, h = 5).
Q₃ = 24.5 + [(48.75 − 30) / 20] × 5 = 24.5 + (93.75 / 20) = 24.5 + 4.69 = Rs 29.19. (c) Maximum income earned by the lowest 25% workers = Lower Quartile (Q₁). N/4 = 16.25, which lies in the 14.5–19.5 class (L = 14.5, c.f. = 5, f = 10, h = 5).
Q₁ = 14.5 + [(16.25 − 5) / 10] × 5 = 14.5 + (56.25 / 10) = 14.5 + 5.63 = ≈ Rs 20.13 (textbook key: Rs 19.92). Note: the textbook’s printed key reads (a) Rs 25.11, (b) Rs 19.92, (c) Rs 29.19. The small numerical differences arise from rounding; in the textbook’s key the lower-quartile and upper-quartile values are listed against (b) and (c). The logical mapping is: lowest 50% → Median, top 25% minimum → Q₃, lowest 25% maximum → Q₁, as worked above.

9. The following table gives production yield in kg. per hectare of wheat of 150 farms in a village. Calculate the mean, median and mode values.

Production yield (kg/hectare)50–5353–5656–5959–6262–6565–6868–7171–7474–77
Number of farms381430362816105
ANSWER N = 150, class width h = 3. Build the computation table with mid-values (m) and cumulative frequencies (c.f.).
Yield (kg/ha)Mid-value (m)Frequency (f)fmc.f.
50–5351.53154.53
53–5654.58436.011
56–5957.514805.025
59–6260.5301815.055
62–6563.5362286.091
65–6866.5281862.0119
68–7169.5161112.0135
71–7472.510725.0145
74–7775.55377.5150
Total—1509573.0—
Mean: X̄ = Σfm / Σf = 9573 / 150 = 63.82 kg/hectare. Median: N/2 = 75, which lies in the 62–65 class (L = 62, c.f. = 55, f = 36, h = 3).
Median = 62 + [(75 − 55) / 36] × 3 = 62 + (60 / 36) = 62 + 1.67 = 63.67 kg/hectare. Mode: the modal class is 62–65 (highest frequency = 36). Here L = 62, D₁ = 36 − 30 = 6, D₂ = 36 − 28 = 8, h = 3.
Mode = L + [D₁ / (D₁ + D₂)] × h = 62 + [6 / (6 + 8)] × 3 = 62 + (18 / 14) = 62 + 1.29 = 63.29 kg/hectare. Result: mean = 63.82, median = 63.67, mode = 63.29 kg per hectare. (All match the textbook answers, confirming the relation Mode < Median < Mean for this right-skewed data.)

Extra Practice Questions

Short Answer Type Questions

Q1. Define arithmetic mean and state its main drawback.

ANSWERThe arithmetic mean is the sum of the values of all observations divided by the number of observations (X̄ = ΣX / N). Its main drawback is that it is unduly affected by extreme values — a single very large or very small item can pull the mean up or down, making it less representative.

Q2. Why is the median preferred over the mean in an open-ended distribution?

ANSWERIn an open-ended distribution the first or last class limit is not known (e.g. “below 100” or “400 and above”), so mid-values cannot be found and the mean cannot be computed accurately. The median is a positional value located by cumulative frequency, so it can still be calculated without the exact extreme limits.

Q3. Find the median of the data: 17, 19, 21, 13, 16, 18, 24.

ANSWERArrange in ascending order: 13, 16, 17, 18, 19, 21, 24. Here N = 7, so the median is the (N+1)/2 = 4th item = 18.

Q4. State the two important mathematical properties of the arithmetic mean.

ANSWER(i) The algebraic sum of the deviations of all items from the arithmetic mean is always zero, i.e. Σ(X − X̄) = 0. (ii) The arithmetic mean is affected by extreme values; any large value on either end can push it up or down.

Q5. A data set is 1, 1, 2, 2, 3, 3. What is its mode? Explain.

ANSWERThere is no mode. The mode is the value that occurs most frequently, but here every value (1, 2 and 3) appears the same number of times, so no value is more frequent than the others.

Long Answer Type Questions

Q1. Explain the relative position of the arithmetic mean, median and mode for a moderately skewed distribution.

ANSWERFor a perfectly symmetrical (normal) distribution, the mean, median and mode coincide at the same value. For a moderately asymmetrical (skewed) distribution they separate, but the median always lies between the mean and the mode. If we denote Mean = Me, Median = Mi and Mode = Mo, the relative magnitudes appear in alphabetical order — either Me > Mi > Mo (positively/right skewed) or Me < Mi < Mo (negatively/left skewed). In a right-skewed distribution a few very large values pull the mean towards the higher end, so Mean > Median > Mode (as seen in Exercise 9, where 63.82 > 63.67 > 63.29). This empirical relationship is often written as Mode = 3 Median − 2 Mean.

Q2. Distinguish between the direct, assumed mean and step deviation methods of computing the arithmetic mean.

ANSWERAll three methods give the same arithmetic mean; they differ in convenience. In the direct method the mean is ΣfX/Σf (or Σfm/Σf for continuous data) — simple, but tedious when figures are large. In the assumed mean method we choose any value A as a working mean, take deviations d = X − A, and compute X̄ = A + (Σfd/Σf); this avoids multiplying large numbers. The step deviation method goes further by dividing each deviation by a common factor c, so d′ = (X − A)/c, and X̄ = A + (Σfd′/Σf) × c; it is the easiest when class intervals are equal because it keeps the figures small. The choice depends on the size of the data and the figures involved.

Q3. Compare the arithmetic mean, median and mode as measures of central tendency, stating where each is most useful.

ANSWERThe arithmetic mean is the most commonly used average because it is simple to calculate and is based on all observations; its weakness is that it is unduly affected by extreme items and cannot be found for open-ended classes, so it suits homogeneous quantitative data such as average wages or production. The median is the positional middle value; it is not affected by extreme values and can be computed for open-ended distributions and qualitative-ranked data, making it useful for incomes or intelligence where extremes exist. The mode is the most frequent value; it best describes qualitative data and is used by manufacturers to find the most demanded size or style. Median and mode can both be located graphically. Thus the appropriate average must be selected according to the purpose of analysis and the nature of the distribution.

MCQs & Assertion–Reason

1. The arithmetic mean of 5, 10, 15, 20, 25 is:

(a) 12    (b) 15    (c) 18    (d) 20

2. The sum of deviations of all observations from the arithmetic mean is always:

(a) 1    (b) equal to N    (c) zero    (d) maximum

3. Which average is the most suitable for measuring qualitative data?

(a) Arithmetic mean    (b) Median    (c) Mode    (d) Geometric mean

4. The value that divides a distribution into two equal halves is the:

(a) mean    (b) mode    (c) median    (d) range

5. The mode of the series 3, 4, 4, 5, 6, 4, 7 is:

(a) 3    (b) 4    (c) 5    (d) 7

6. Which measure of central tendency is most affected by extreme values?

(a) Median    (b) Mode    (c) Arithmetic mean    (d) Quartile

7. The position of the median in an individual or discrete series is given by:

(a) N/2    (b) (N+1)/2    (c) N/4    (d) 3(N+1)/4

8. The upper quartile (Q₃) has below it:

(a) 25% of items    (b) 50% of items    (c) 75% of items    (d) 100% of items

9. In a continuous series, the modal class is the class with the:

(a) lowest frequency    (b) largest frequency    (c) lowest mid-value    (d) highest cumulative frequency

10. The 50th percentile (P₅₀) is the same as the:

(a) lower quartile    (b) upper quartile    (c) median    (d) mode

Answer key: 1-(b), 2-(c), 3-(c), 4-(c), 5-(b), 6-(c), 7-(b), 8-(c), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: The arithmetic mean is unduly affected by extreme values.

Reason: The arithmetic mean is calculated using the actual value of every observation in the series.

A-R 2. Assertion: The median is a positional average.

Reason: The median is determined by the position of the middle item and not by the magnitude of all items.

A-R 3. Assertion: The mode is the most suitable average for qualitative data.

Reason: The mode is the value that occurs least frequently in a series.

A-R 4. Assertion: In a continuous series the median is located using N/2, not (N+1)/2.

Reason: In a continuous frequency distribution the data is treated as continuous, so N/2 gives the correct median position.

A-R 5. Assertion: For a moderately skewed distribution the median lies between the mean and the mode.

Reason: For such a distribution the three averages are related approximately as Mode = 3 Median − 2 Mean.

Answer key: 1-(A), 2-(A), 3-(C), 4-(A), 5-(A).

Exam Tips & Common Mistakes

How to score full marks in this chapter

Memorise the median and mode formulas exactly, including what each symbol (L, c.f., f, h, D₁, D₂) stands for, and always show the computation table. For continuous-series median and quartiles use N/2, N/4 and 3N/4 (not N+1); for the median of an individual or discrete series use (N+1)/2. Convert “more than / less than” cumulative tables and inclusive class intervals before calculating. State the units (Rs, acres, kg/hectare) in the final answer, and remember the relation Mode = 3 Median − 2 Mean for cross-checking.

Common mistakes to avoid

  • Using (N+1)/2 instead of N/2 to locate the median or quartile class in a continuous series.
  • Forgetting to convert inclusive class intervals (e.g. 10–14) into exclusive form (9.5–14.5) before finding the median or mode.
  • Not converting “more than” or “less than” cumulative frequencies into simple frequencies before computing the mean.
  • Taking the wrong cumulative frequency (use the c.f. of the class just before the median class).
  • Ignoring signs incorrectly in D₁ and D₂ for the mode (both differences are taken as positive).
  • Calling the arithmetic mean a positional value — it is a calculated value; the median and mode are positional.

Frequently Asked Questions

What is Chapter 5 of Class 11 Statistics for Economics about?

Chapter 5, Measures of Central Tendency, explains how a data set can be summarised by a single representative value — the arithmetic mean, the median and the mode — and how each is computed for individual, discrete and continuous series, along with positional measures such as quartiles and percentiles.

What is the difference between mean, median and mode?

The arithmetic mean is the sum of all values divided by their number; it uses every item but is affected by extreme values. The median is the middle value of an ordered data set and is not affected by extremes. The mode is the most frequently occurring value and is best for qualitative data.

Which average is best for open-ended frequency distributions?

The median is best for open-ended distributions because it is a positional value located through cumulative frequency and does not require the exact limits of the first or last class, unlike the arithmetic mean.

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