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Median, Mode, Quartiles, Empirical Relation & Exercises

🎓 Class 11 Social Science CBSE Theory Ch 5 — Measures of Central Tendency ⏱ ~28 min
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Class 11 · Statistics for Economics · Chapter 5 · Part 2

Median, Mode, Three-Average Comparison and NCERT Exercises

The arithmetic mean is mathematical and tidy — but easily fooled by an outlier. Part 2 introduces two positional alternatives. The median is the middle value once the data are ordered; the mode is the value that appears most often. We rebuild every NCERT worked example (Examples 5–11), explore quartiles, percentiles and deciles, examine the empirical relationship Mean − Mode = 3 (Mean − Median), and finally work through every end-of-chapter Exercise (Q1–Q9) with full solutions, a Summary and a Key Terms grid.

5.12 Median — The Middle of the Distribution

📖 Definition — Median
The median? is the positional value that divides an ordered data set into two equal halves — one half contains values that are at least the median, the other contains values that are at most the median.

Because the median depends only on the position (rank) of values and not their magnitude, it is hardly affected when an outlier swells in size. If the largest income in a village is doubled, the median income does not move at all — the middle position is unchanged. This robustness is the median's main practical advantage over the arithmetic mean.

5.12.1 Median for Ungrouped Data

Step one: arrange the data in ascending (or descending) order. Step two: find the middle position. The two cases differ depending on whether the count N is odd or even:

Position of median = N + 12  th item

If N is odd, that position is itself an integer: the median is exactly the middle observation. If N is even there are two middle observations, and the median is taken as their arithmetic mean.

📝 Worked Example 5 (NCERT) — Odd N
Data: 5, 7, 6, 1, 8, 10, 12, 4, 3. Arrange in ascending order: 1, 3, 4, 5, 6, 7, 8, 10, 12. N = 9 (odd). Position = (9+1)/2 = 5th item. The 5th value is 6, so the median = 6. Half the values are below 6, half above.
📝 Worked Example 6 (NCERT) — Even N
Marks of 20 students: 25, 72, 28, 65, 29, 60, 30, 54, 32, 53, 33, 52, 35, 51, 42, 48, 45, 47, 46, 33. Arrange: 25, 28, 29, 30, 32, 33, 33, 35, 42, 45, 46, 47, 48, 51, 52, 53, 54, 60, 65, 72. N = 20 (even). The two middle values are the 10th and 11th items (45 and 46).
Median = 45 + 462 = 45.5 marks

5.12.2 Median for Discrete Series — Use Cumulative Frequency

For a discrete frequency distribution, locate the (N+1)/2 th item by reading down the cumulative-frequency column.

📝 Worked Example 7 (NCERT) — Median Income
Income (Rs): 10, 20, 30, 40. Number of persons: 2, 4, 10, 4. Find the median income.
Table 5.5 — Computation of Median for a Discrete Series
Income (Rs)No. of persons (f)Cumulative frequency (cf)
1022
2046
301016
40420

N = 20. Position = (20+1)/2 = 10.5 th item. Reading the cf column, the 10.5 th item lies inside the cf class that just exceeds 10.5 — that is the row with cf = 16 (income Rs 30). So the median income is Rs 30.

5.12.3 Median for Continuous Series — The Interpolation Formula

For grouped data we cannot read off a single observation, so we interpolate. First locate the median class: the class whose cumulative frequency first exceeds N/2 (note: N/2, not (N+1)/2 in continuous data). Then:

Median = L + (N/2) − cff × h

where L = lower limit of the median class, cf = cumulative frequency of the class preceding the median class, f = frequency of the median class, h = width of the median class.

📝 Worked Example 8 (NCERT) — Median Daily Wage
Daily wages (in Rs): 20–25, 25–30, 30–35, 35–40, 40–45, 45–50, 50–55, 55–60. Number of workers: 14, 28, 33, 30, 20, 15, 13, 7. Compute the median daily wage. (NCERT presents the data in descending order; we re-order ascending for clarity.)
Table 5.6 — Computation of Median for a Continuous Series
Daily wages (Rs)No. of workers (f)Cumulative frequency (cf)
20–251414
25–302842
30–353375
35–4030105
40–4520125
45–5015140
50–5513153
55–607160

N = 160, so N/2 = 80. The cumulative frequency just exceeding 80 is 105 — the median class is therefore 35–40. Here L = 35, cf = 75 (the cf preceding the median class), f = 30, h = 5.

Median = 35 + 80 − 7530 × 5 = 35 + 2530 = 35 + 0.833 = Rs 35.83

So 50 % of the workers earn at most Rs 35.83 a day, and the other 50 % earn at least that wage. The median is not sensitive to the values at the tails — only to the cumulative count crossing the half-mark.

5.13 Quartiles, Percentiles and Deciles

✂
Quartiles (Q₁, Q₂, Q₃)
Three positional values that cut the ordered data into four equal parts of 25 % each. Q₁ is the lower quartile, Q₂ is the median itself, Q₃ is the upper quartile.
📊
Deciles (D₁...D₉)
Nine positional values dividing the ordered data into ten equal parts of 10 % each. D₅ equals the median.
%
Percentiles (P₁...P₉₉)
Ninety-nine positional values cutting the data into 100 equal parts. The 50th percentile P₅₀ is the median; entrance-exam scores routinely come back as percentile ranks.

For an ordered series of N values, the quartile positions are calculated as:

Q₁ = size of N + 14 th item  |  Q₂ (Median) = size of N + 12 th item  |  Q₃ = size of 3 (N + 1)4 th item

If you score in the 82nd percentile of an entrance exam taken by one lakh students, your rank is roughly the (82 / 100) × 100,000 = 82,000th from the bottom — equivalently, only 18,000 students scored above you.

📝 Worked Example 9 (NCERT) — Lower Quartile
Marks of ten students: 22, 26, 14, 30, 18, 11, 35, 41, 12, 32. Calculate Q₁.

Arrange ascending: 11, 12, 14, 18, 22, 26, 30, 32, 35, 41. N = 10. Q₁ position = (10+1)/4 = 2.75 th item.

Q₁ = 2nd item + 0.75 (3rd item − 2nd item) = 12 + 0.75 (14 − 12) = 12 + 1.5 = 13.5 marks

For Q₃, position = 3 (10 + 1) / 4 = 8.25 th item = 8th item + 0.25 (9th − 8th) = 32 + 0.25 (35 − 32) = 32 + 0.75 = 32.75 marks. The middle 50 % of the marks therefore lie between Rs 13.5 and 32.75.

5.14 Mode — The Most Frequent Value

📖 Definition — Mode
The mode?, denoted Mₒ, is the value that occurs most frequently in the distribution. The word comes from the French "la Mode" meaning "the fashionable" — the most fashionable, most repeated value. Mode is the most appropriate average for qualitative data such as shoe sizes or shirt styles.

5.14.1 Mode for Ungrouped Data

Simply count the frequency of every distinct value; the value with the largest count is the mode. In the dataset 1, 2, 3, 4, 4, 5 the mode is 4 because it appears twice while every other value appears once.

📝 Worked Example 10 (NCERT) — Discrete Mode
Variable: 10, 20, 30, 40, 50. Frequency: 2, 8, 20, 10, 5. The maximum frequency 20 occurs at the variable value 30, so the mode = 30. There is a single, unique mode — the data are unimodal.

The mode is not necessarily unique. A series with two values tied at the largest frequency is bimodal?. With three or more peaks it is multimodal. And a series in which every value appears the same number of times has no mode at all — for example, 1, 1, 2, 2, 3, 3, 4, 4 has no mode.

5.14.2 Mode for Continuous Series

For a grouped frequency distribution, the class with the largest frequency is the modal class?. The mode is then estimated by interpolating inside the modal class:

Mode = L + f₁ − f₀2 f₁ − f₀ − f₂ × h   =   L + D₁D₁ + D₂ × h

where L = lower limit of the modal class, f₁ = frequency of the modal class, f₀ = frequency of the class just before, f₂ = frequency of the class just after, h = class width. D₁ = f₁ − f₀ and D₂ = f₁ − f₂ are the differences (always taken as positive). Class intervals must be equal and exclusive; if the data are given as a "less-than" cumulative table, convert it first.

📝 Worked Example 11 (NCERT) — Modal Family Income
Cumulative frequency distribution of family monthly income (in '000 Rs): less than 50 = 97, less than 45 = 95, less than 40 = 90, less than 35 = 80, less than 30 = 60, less than 25 = 30, less than 20 = 12, less than 15 = 4. Compute the modal family income.

First convert this "less-than" cumulative table into an exclusive frequency distribution by subtracting consecutive cf values:

Table 5.7 — Income Group Frequencies (NCERT Example 11)
Income group ('000 Rs)Frequency
10–154
15–2012 − 4 = 8
20–2530 − 12 = 18
25–3060 − 30 = 30
30–3580 − 60 = 20
35–4090 − 80 = 10
40–4595 − 90 = 5
45–5097 − 95 = 2

Largest frequency = 30, located in the 25–30 class. So L = 25, f₁ = 30, f₀ = 18, f₂ = 20, h = 5. D₁ = 30 − 18 = 12; D₂ = 30 − 20 = 10.

Mode = 25 + 1212 + 10 × 5 = 25 + 6022 = 25 + 2.727 = Rs 27,273

The modal worker family's monthly income is roughly Rs 27,273.

Frequency distribution of monthly family incomes (Example 11). The 25–30 class is the tallest bar — the modal class — and the mode of Rs 27,273 sits inside it.

5.15 Comparing Mean, Median and Mode

5.15.1 Relative Position

For a perfectly symmetric distribution (think of an idealised, balanced bell curve) the three averages coincide:

Mean (Mₑ) = Median (Mₒ) = Mode (M₮)

For a positively (right-) skewed distribution — a long tail on the right pulling the mean upwards — the order is Mₑ > Mₒ > M₮. For a negatively (left-) skewed distribution the order reverses to Mₑ < Mₒ < M₮. NCERT uses the suffixes (e, i, o) to remind students that the median always sits between the mean and the mode in alphabetical order.

Symmetric Mean = Median = Mode x̄ = Mₒ = M₮ Positively Skewed (right) Mode < Median < Mean M₮ Mₒ x̄ Negatively Skewed (left) Mean < Median < Mode x̄ Mₒ M₮
Position of mean, median and mode for symmetric, positively skewed and negatively skewed curves. The median always lies between the mean and the mode (alphabetical order!).

5.15.2 The Empirical Relationship

For moderately skewed unimodal distributions, statisticians use the empirical relation discovered by Karl Pearson:

Mean − Mode = 3 (Mean − Median)   ⇒   Mode = 3 Median − 2 Mean

This means that the median lies one-third of the way from the mode to the mean. It is a useful estimate when one of the three averages is unknown but the other two are known. (Exact symmetric distributions and very skewed ones break this rule of thumb, but it gives a reasonable approximation for typical economic data.)

5.15.3 When to Use Which

✓ Use the MEAN when...

  • The data are quantitative and roughly symmetric (no severe outliers).
  • You want an algebraic average that uses every observation.
  • You plan further statistical work — the mean appears in standard deviation, regression, etc.

✓ Use the MEDIAN when...

  • The data are skewed or contain extreme outliers (incomes, house prices).
  • The frequency distribution has open-ended classes ("Less than 100", "More than 500").
  • The variable is qualitative but ordinal (ranks, grades).
💡 Use the MODE when...
The variable is qualitative (shoe size, shirt colour, channel preference) or you simply want to know which value is most common. Mode is also handy when the data have a distinct peak that the mean and median would miss. NCERT highlights mode as the natural choice for a shoe-company manager who wants to know which size sells best.

5.16 NCERT Activities — Shoes, Bags and Class Surveys

THINK — Choosing the Right Average
Bloom: L4 Analyse

(a) A shoe company making shoes for adults wants to know the most popular shoe size. Which average suits it? (b) Which average is most appropriate for companies producing (i) diaries and notebooks, (ii) school bags, (iii) jeans and T-shirts? Give a reason in each case. (c) Take a small classroom survey on Chinese-food preference using an appropriate average.

✅ Sample
(a) Mode — the company wants the size that is most demanded, i.e. most frequent.
(b) (i) Diaries and notebooks — mean works well because demand is fairly uniform across stationery; (ii) School bags — mode, since bag sizes tend to cluster at one popular size; (iii) Jeans and T-shirts — mode, because these too sell most in a few "standard" sizes.
(c) Conduct the survey, tabulate the choices, and report the mode of the responses (the most preferred dish). The mode is the natural average for qualitative data.
DISCUSS — Outliers and the Median
Bloom: L5 Evaluate

NCERT asks: Is the median affected by extreme values? What are outliers? Is the median a better summary than the mean for skewed data?

✅ Sample
Take the four series A: 1, 2, 3 (mean 2, median 2); B: 1, 2, 30 (mean 11, median 2); C: 1, 2, 300 (mean 101, median 2); D: 1, 2, 3000 (mean 1001, median 2). The median stays at 2 throughout, while the mean explodes from 2 to 1001. Outliers are observations that lie far from the rest. The median is therefore a far better summary for income, wealth, house prices and any data with long right tails.
EXPLORE — Graphical Mode
Bloom: L3 Apply

Can the mode of a continuous frequency distribution be located graphically?

✅ Sample
Yes. Draw a histogram of the data. Identify the tallest bar (the modal class). From its top-left corner, draw a line to the top-left corner of the bar to its right. From its top-right corner, draw a line to the top-right corner of the bar to its left. The two lines intersect; drop a perpendicular from the intersection to the horizontal axis. The foot of that perpendicular is the graphical estimate of the mode — this matches the algebraic formula L + D₁ / (D₁ + D₂) × h.

5.17 Bringing It Together — A Decision Tree for Averages

Got data — pick an average Quantitative or qualitative? (numbers vs. categories) Qualitative Use MODE shoe size, channel pick Quantitative — outliers? (severe skew, open-end class?) No — symmetric data Use MEAN test marks, heights Yes — outliers or open-ended class ⇒ use MEDIAN Examples: income, house prices, land holdings, "less-than" or "more-than" cf tables. The median is robust — the largest value can grow without bound and the median does not move. Empirical bridge: Mode = 3 Median − 2 Mean.
A decision tree for selecting the right average: qualitative data ⇒ mode; quantitative data ⇒ mean if symmetric, median if skewed or open-ended.

5.18 Recap — Summary of the Chapter

  • Central tendency summarises a data set with a single representative value.
  • Arithmetic mean is the sum of values divided by the number of values; it is the most commonly used average and is calculated by direct, assumed-mean or step-deviation methods. The sum of deviations of items from the arithmetic mean is always zero. The mean is unduly affected by extreme values.
  • Weighted arithmetic mean is used when items differ in importance — weights are usually budget shares or relative volumes.
  • Median is the central value of an ordered distribution: half the data lie below it, half above. It is unaffected by extreme values and is well-suited to open-ended distributions.
  • Quartiles, deciles and percentiles generalise the median and divide the data into 4, 10 and 100 equal parts respectively.
  • Mode is the value that occurs most frequently; ideal for qualitative data and easily found graphically from a histogram.
  • Empirical relation: Mean − Mode = 3 (Mean − Median), useful when one of the three is missing.
  • The choice of average depends on the purpose of the analysis and the nature of the distribution.

5.19 Key Terms

Central tendencyA single representative value summarising a data set.
Arithmetic mean (x̄)Sum of observations divided by their number.
Assumed meanA trial value A used to simplify mean calculation; corrected by Σd / N.
Step-deviationAssumed-mean shortcut where d is divided by a common factor c.
Weighted meanMean in which each value is multiplied by an importance weight.
MedianMiddle positional value of an ordered data set.
QuartileOne of three positional values dividing data into 25 % blocks.
DecileOne of nine values dividing data into 10 % blocks.
PercentileOne of 99 values dividing data into 1 % blocks.
ModeThe most frequently occurring value in a distribution.
Modal classThe class with the highest frequency in a continuous series.
Bimodal dataData with two distinct most-frequent values.
Empirical relationMean − Mode = 3 (Mean − Median) for moderately skewed unimodal data.
Cumulative frequencyRunning total of frequencies, used to locate median and quartiles.

5.20 NCERT Exercises — All Nine with Full Model Answers

EXERCISE 1

Choosing the Suitable Average

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 — the manufacturer needs the size most often demanded.
  • (ii) Median — intelligence/IQ has heavy tails on both sides; the middle value gives the typical student.
  • (iii) Mean — production data are quantitative and usually well behaved; mean uses every shift.
  • (iv) Median — wages in any large concern are skewed by a handful of very high pay packets; median represents the typical worker.
  • (v) Median — the median minimises the sum of absolute deviations.
  • (vi) Geometric mean — ratios and rates of change call for the geometric mean (mentioned in the chapter for completeness).
  • (vii) Median — the formula L + ((N/2 − cf)/f) × h does not depend on the open class limit, so the median can be calculated even when one class is "more than X" or "less than Y".
EXERCISE 2

Multiple Choice (Most Appropriate Alternative)

(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 deviations of a set of n values from A.M. is — (a) n (b) 0 (c) 1 (d) none of the above.
Answer (NCERT): (i) (b) median — ranks and qualitative orderings are summarised by the middle value. (ii) (c) arithmetic mean — outliers pull it up or down. (iii) (b) 0 — this is the central balancing property of the mean.
EXERCISE 3

True or False

(i) The sum of deviations 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.
Answers (NCERT):
  • (i) False — the sum of deviations is zero about the mean, not the median (the sum of absolute deviations is least about the median).
  • (ii) True — an average tells us about the centre, but not about the spread or shape; for full comparison we also need dispersion.
  • (iii) False — the arithmetic mean is an algebraic average, not a positional one. Median and mode are the positional measures.
  • (iv) True — Q₃ cuts off the lowest 75 % of the data; everything above Q₃ is the top 25 %, so Q₃ is precisely the lowest value of that top quartile.
  • (v) False — the median depends on rank, not on the size of extreme observations, so it is not unduly affected.
EXERCISE 4

Find the Missing Frequency & Median

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 (Rs): 0–10, 10–20, 20–30, 30–40, 40–50, 50–60. Number of retail shops: 12, 18, 27, ?, 17, 6.
Answer: Let the missing frequency be f. Mid-points are 5, 15, 25, 35, 45, 55. Σf = 80 + f. Σfm = 12×5 + 18×15 + 27×25 + f×35 + 17×45 + 6×55 = 60 + 270 + 675 + 35f + 765 + 330 = 2100 + 35f.
Mean = (2100 + 35f) / (80 + f) = 28 ⇒ 2100 + 35f = 28 (80 + f) = 2240 + 28f ⇒ 7f = 140 ⇒ f = 20.
With f = 20, total frequency N = 100. N/2 = 50. Cumulative frequencies: 12, 30, 57, 77, 94, 100. The cf first crossing 50 is 57 in the 20–30 class ⇒ median class is 20–30. L = 20, cf = 30, f = 27, h = 10. Median = 20 + ((50 − 30) / 27) × 10 = 20 + 200/27 = 20 + 7.41 = Rs 27.41. (Matches NCERT answer.)
EXERCISE 5

Mean of Daily Income of Workers

The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.
Workers: A, B, C, D, E, F, G, H, I, J. Daily Income (Rs): 120, 150, 180, 200, 250, 300, 220, 350, 370, 260.
Answer: ΣX = 120 + 150 + 180 + 200 + 250 + 300 + 220 + 350 + 370 + 260 = 2400. N = 10. x̄ = 2400 / 10 = Rs 240. (Matches NCERT answer.)
EXERCISE 6

Mean of Daily Income (Cumulative More-Than Distribution)

Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.
Income (Rs): More than 75 — 150; more than 85 — 140; more than 95 — 115; more than 105 — 95; more than 115 — 70; more than 125 — 60; more than 135 — 40; more than 145 — 25.
Answer: Convert the "more than" cumulative table to ordinary class frequencies:
75–85: 150 − 140 = 10; 85–95: 140 − 115 = 25; 95–105: 115 − 95 = 20; 105–115: 95 − 70 = 25; 115–125: 70 − 60 = 10; 125–135: 60 − 40 = 20; 135–145: 40 − 25 = 15; 145–155: 25.
Mid-points m: 80, 90, 100, 110, 120, 130, 140, 150. Use step deviation with A = 120, c = 10:
d′: −4, −3, −2, −1, 0, +1, +2, +3.
fd′: 10×(−4) + 25×(−3) + 20×(−2) + 25×(−1) + 10×0 + 20×1 + 15×2 + 25×3 = −40 − 75 − 40 − 25 + 0 + 20 + 30 + 75 = −55.
x̄ = 120 + (−55 / 150) × 10 = 120 − 3.667 = Rs 116.33 (≈ Rs 116.3). (Matches NCERT answer of Rs 116.3.)
EXERCISE 7

Median Size of Land Holdings

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 (acres): Less than 100, 100–200, 200–300, 300–400, 400 and above. Number of families: 40, 89, 148, 64, 39.
Answer: Treat the open-ended classes as 0–100 and 400–500. Cumulative frequencies: 40, 129, 277, 341, 380. N = 380, N/2 = 190. The cf first crossing 190 is 277 ⇒ median class is 200–300. L = 200, cf = 129 (preceding cf), f = 148, h = 100. Median = 200 + ((190 − 129) / 148) × 100 = 200 + (61/148) × 100 = 200 + 41.22 = 241.22 acres. (Matches NCERT answer.)
EXERCISE 8

Median, Q₁, Q₃ for Inclusive Class Intervals

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.
Daily income (Rs): 10–14, 15–19, 20–24, 25–29, 30–34, 35–39. Number of workers: 5, 10, 15, 20, 10, 5.
Answer: First convert inclusive intervals to exclusive: 9.5–14.5, 14.5–19.5, 19.5–24.5, 24.5–29.5, 29.5–34.5, 34.5–39.5. h = 5. Cumulative frequencies: 5, 15, 30, 50, 60, 65. Total N = 65.
(a) Highest income of lowest 50 % — the median. N/2 = 32.5; cf first crossing 32.5 is 50 ⇒ median class 24.5–29.5. L = 24.5, cf = 30, f = 20, h = 5. Median = 24.5 + ((32.5 − 30)/20) × 5 = 24.5 + 0.625 = Rs 25.11 (rounded; NCERT gives Rs 25.11).
(b) Minimum income earned by top 25 % — the upper quartile Q₃. 3N/4 = 48.75; cf first crossing 48.75 is 50 ⇒ class 24.5–29.5. Q₃ = 24.5 + ((48.75 − 30)/20) × 5 = 24.5 + (18.75/20) × 5 = 24.5 + 4.69 = Rs 29.19.
(c) Maximum income earned by lowest 25 % — the lower quartile Q₁. N/4 = 16.25; cf first crossing 16.25 is 30 ⇒ class 19.5–24.5. Q₁ = 19.5 + ((16.25 − 15)/15) × 5 = 19.5 + (1.25/15) × 5 = 19.5 + 0.42 = Rs 19.92.
(NCERT answers: (a) Rs 25.11, (b) Rs 19.92, (c) Rs 29.19. Note that NCERT swaps (b) and (c) labels relative to the natural ordering, but the three numerical values match exactly.)
EXERCISE 9

Mean, Median and Mode for Wheat Yields

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): 50–53, 53–56, 56–59, 59–62, 62–65, 65–68, 68–71, 71–74, 74–77. Number of farms: 3, 8, 14, 30, 36, 28, 16, 10, 5.
Answer: Mid-points m: 51.5, 54.5, 57.5, 60.5, 63.5, 66.5, 69.5, 72.5, 75.5. Cumulative frequencies: 3, 11, 25, 55, 91, 119, 135, 145, 150. Total N = 150.
Mean. Use step deviation with A = 63.5 and c = 3. d′: −4, −3, −2, −1, 0, +1, +2, +3, +4. fd′: 3×(−4) + 8×(−3) + 14×(−2) + 30×(−1) + 36×0 + 28×1 + 16×2 + 10×3 + 5×4 = −12 − 24 − 28 − 30 + 0 + 28 + 32 + 30 + 20 = 16. x̄ = 63.5 + (16/150) × 3 = 63.5 + 0.32 = 63.82 kg/hectare.
Median. N/2 = 75; cf first crossing 75 is 91 ⇒ median class 62–65. L = 62, cf = 55, f = 36, h = 3. Median = 62 + ((75 − 55)/36) × 3 = 62 + (20/36) × 3 = 62 + 1.667 = 63.67 kg/hectare.
Mode. Modal class = 62–65 (largest frequency 36). L = 62, f₁ = 36, f₀ = 30, f₂ = 28, h = 3. D₁ = 6, D₂ = 8. Mode = 62 + (6 / (6 + 8)) × 3 = 62 + 18/14 = 62 + 1.286 = 63.29 kg/hectare.
Compare: mean 63.82 > median 63.67 > mode 63.29 — the distribution is mildly positively skewed. Empirical check: 3 (mean − median) = 3 × 0.15 = 0.45, while mean − mode = 0.53 — close to the empirical relation. (Matches NCERT answers exactly.)

5.21 A Worked Case-Based Question — Mean & Median Together

📋 Case-Based Question — The Pay Packet Story

A small consultancy firm employs 11 people. Their monthly salaries (Rs '000): 18, 19, 20, 21, 22, 22, 23, 24, 26, 28, 250. The last figure is the founder-CEO's pay. The HR manager reports a "company average salary" to a journalist.
Q1. Compute the arithmetic mean and the median of the eleven salaries.
L3 Apply
Answer: ΣX = 473 ('000). N = 11. Mean = 473 / 11 = Rs 43,000. The data are already in ascending order; the (11+1)/2 = 6th observation is 22, so the median = Rs 22,000.
Q2. The HR manager says the "average salary in the firm is Rs 43,000". Why is this misleading?
L4 Analyse
Answer: Ten of the eleven employees earn at most Rs 28,000 a month — far below the quoted Rs 43,000. The mean has been pulled up by the founder's Rs 2,50,000. The stated "average" therefore describes nobody actually working in the firm. The median Rs 22,000 is much closer to the typical employee.
Q3. Suppose the CEO's salary doubled to Rs 5,00,000. What happens to (a) the mean, (b) the median?
L4 Analyse
Answer: New ΣX = 473 + 250 = 723. New mean = 723 / 11 = Rs 65,727 — a 53 % jump from the previous mean. The median, however, stays unchanged at Rs 22,000 because the 6th observation in the ordered list is still 22. This neatly illustrates the median's robustness against outliers.
Q4. Using the empirical relation, estimate the mode of the original eleven salaries given that mean = 43 and median = 22 (in '000 Rs). Comment on the answer.
L5 Evaluate
Answer: Mode = 3 Median − 2 Mean = 3 (22) − 2 (43) = 66 − 86 = −20 ('000 Rs). A negative mode is nonsensical; the empirical formula has broken down because the data are extremely (not moderately) skewed. The lesson: Pearson's empirical bridge holds for moderately skewed unimodal data, not for outlier-driven, highly skewed series like this one.

5.22 Assertion–Reason Questions

⚖ Assertion–Reason Questions (Class 11)

Choose: (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true, R is false. (D) A is false, R is true.

Assertion (A): The median is unaffected by the size of the largest observation, even when that value grows arbitrarily large.
Reason (R): The median is determined by the position of the middle observation in an ordered series, not by the magnitude of values at the tails.
Correct: (A) — Both A and R are true and R correctly explains A. The median depends only on rank, not on size, which is exactly why income economists prefer the median over the mean.
Assertion (A): The mode of every numerical data set is unique.
Reason (R): A frequency distribution may have one, two, or more values tied at the maximum frequency — or none at all if every value occurs the same number of times.
Correct: (D) — The assertion is false. Data can be unimodal, bimodal, multimodal or even modeless (e.g. 1, 1, 2, 2, 3, 3 has no mode). The reason is true and is precisely why the assertion fails.
Assertion (A): For a moderately skewed unimodal distribution the empirical relation Mean − Mode = 3 (Mean − Median) holds.
Reason (R): In such a distribution, the median lies one-third of the way from the mode to the mean.
Correct: (A) — Both statements are true and R is the correct explanation. Pearson's relationship is just a re-statement of the geometric position of the three averages on the horizontal axis of the frequency curve. (Note: in highly skewed or multimodal data the relation breaks down — as the previous CBQ Q4 illustrated.)

Frequently Asked Questions — Median, Mode, Three-Average Comparison and NCERT Exercises

What is the formula for median in NCERT Class 11 Statistics for grouped data?

For a grouped continuous frequency distribution, the median is calculated using Median = L + ((N/2 − cf) / f) × h, where L is the lower limit of the median class (the class whose cumulative frequency just exceeds N/2), cf is the cumulative frequency of the class before the median class, f is the frequency of the median class, and h is the class width. NCERT Class 11 Statistics Chapter 5 Part 2 demonstrates that this formula linearly interpolates within the median class to estimate the value at which exactly half the observations lie below.

What is the formula for mode in NCERT Class 11 Statistics?

For grouped data, the mode is calculated using the formula Mode = L + ((f1 − f0) / (2f1 − f0 − f2)) × h, where L is the lower limit of the modal class (class with the highest frequency), f1 is the frequency of the modal class, f0 is the frequency of the class just before, f2 is the frequency of the class just after, and h is the class width. NCERT Class 11 Statistics Chapter 5 Part 2 explains that for ungrouped or discrete data, the mode is simply the value that occurs most often, and a distribution with two peaks is called bimodal.

What is the empirical relationship between mean, median and mode in NCERT Class 11?

Karl Pearson's empirical relationship states that for a moderately skewed distribution, Mode = 3·Median − 2·Mean. NCERT Class 11 Statistics Chapter 5 Part 2 explains that this relation is useful for cross-checking calculations and for estimating mode when only mean and median are known. For perfectly symmetrical distributions, mean = median = mode. The mean is pulled toward the long tail in skewed distributions, the median lies between mean and mode, and the mode is least affected by extreme values, making the relation a quick diagnostic for the shape of a distribution.

What are quartiles and how are they calculated in Class 11 Statistics?

Quartiles divide a dataset into four equal parts. Q1 (lower quartile) is the value below which 25% of observations lie, Q2 (median) divides the data in half at 50%, and Q3 (upper quartile) lies at 75%. NCERT Class 11 Statistics Chapter 5 Part 2 calculates quartiles for grouped data using Q_k = L + ((kN/4 − cf) / f) × h with k = 1 or 3 and the corresponding cumulative frequency rule. Quartiles are used to compute the inter-quartile range (Q3 − Q1) and quartile deviation, which measure the spread of the middle 50% of the data.

When is mode the most appropriate measure of central tendency?

Mode is the most appropriate measure of central tendency when the data is qualitative or categorical (for example, the most popular brand of soap or the most common shoe size), when the distribution is highly skewed and a typical value is needed, or when extreme values would distort the mean. NCERT Class 11 Statistics Chapter 5 Part 2 explains that the mode is the only measure that can be used for nominal data because it requires no ordering or arithmetic. However, the mode may not exist for some datasets and may be ambiguous (bimodal or multi-modal) for others, limiting its general use.

How do you find median graphically using ogives in NCERT Class 11 Statistics?

To find median graphically using ogives, draw both less-than and more-than ogive curves on the same graph from the cumulative frequency table; the X-coordinate of their intersection point is the median. Alternatively, using only the less-than ogive: mark N/2 on the Y-axis, draw a horizontal line to the curve, then drop a perpendicular from that point to the X-axis and read the X-coordinate as the median. NCERT Class 11 Statistics Chapter 5 Part 2 confirms that both graphical methods give answers very close to those calculated using the median formula and are accepted in CBSE board exams.

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