Locating the middle class and interpolating across it

What to assume they know

• Running totals, and converting between the two cumulative tables — cumulative frequency, both accumulations, and differencing a cumulative table back into frequencies
• Grouping loses the raw values, so we stand the class mark in for them — class limits, class size, and what grouping discards
• The median of ungrouped data and the position rules for odd and even counts
• Dividing a segment in a given ratio, and converting a fraction of a length into a distance
• Solving a linear equation in one unknown, and a pair of simple simultaneous equations
• Converting inclusive classes into continuous ones

What they should be able to do

• Compute n ÷ 2 and identify the median class from a cumulative frequency column
• Explain what each of l, n, cf, f and h refers to in a specific table
• Derive the median formula as linear interpolation rather than quoting it
• Compute the median of a grouped distribution and state its meaning in context
• Compute a median from a table supplied only in cumulative form
• Use the formula in reverse to find one or two missing frequencies given the median and the total
• Apply the continuity correction before computing a median, and say what difference it makes

Where it usually goes wrong

• "The median is the class mark of the median class." In the chapter's worked case the class mark of 60–70 is 65 and the median is 66.43. They agree only when the middle position happens to sit centrally in the class.
• "cf is the cumulative frequency of the median class." It is the running total for the class before it. Using the median class's own value is the most frequent wrong answer on this topic, and it is self-diagnosing every single time: the median class is by definition the first whose running total has reached n ÷ 2, so its own total is never less than n ÷ 2 and the substitution always leaves a numerator of zero or below. A negative or vanishing numerator means this mistake and no other.
• "Use (n + 1) ÷ 2 as you did for ungrouped data." Grouped questions use n ÷ 2. On the chapter's own data the two give 66.43 and 67.14 — a visible gap.
• "Pick whichever class has a running total closest to n ÷ 2." Pick the first one that has reached it. A nearer value on the low side has not yet reached the middle observation, so the middle cannot be in it.
• "The median needs the frequencies, so a cumulative table is unusable." Difference it first. Example 7 and Exercise 13.3 Q3 both begin that way.
• "The continuity correction is bookkeeping that does not change anything." On Exercise 13.3 Q4 it moves the median from 147 mm to 146.75 mm. It changes the lower limit you start from and the width you multiply by.
• "Even spreading inside the class is a fact about the data." It is an assumption, and it is the same species of assumption as the class mark for the mean — a stand-in adopted because the real positions were discarded.
• "If the median formula gives a value outside the median class, that's fine." It cannot, if the working is right. The fraction is between 0 and 1, so the answer always lands between l and l + h — a free sanity check.

Questions to check understanding

• Identify the median class from a cumulative frequency column and justify the choice
• Compute the median of a grouped distribution, showing the cumulative column
• Compute a median from a table given in "less than" or "below" form
• Find one or two missing frequencies from a stated median and total — the form of Example 8 and Exercise 13.3 Q2
• Convert inclusive classes to continuous ones and then find the median
• Interpret a computed median in the context of the data
• Explain why the median formula assumes even spreading, and what would change if the data were not evenly spread

Examples worth working on the board

Values marked verified are worked out here on the chapter's printed data.

• The chapter's worked case (Table 13.15 and pp. 192–193). Fifty-three students, classes 0–10 through 90–100 in tens, frequencies 5, 3, 4, 3, 3, 4, 7, 9, 7, 8, cumulative 5, 8, 12, 15, 18, 22, 29, 38, 45, 53. Verified: n ÷ 2 = 26.5; the running total first passes it at 29, so the median class is 60–70, with l = 60, cf = 22, f = 7 and h = 10; the median is 60 + 10 × (26.5 − 22) ÷ 7 = 60 + 45 ÷ 7 = 66.43 to two places. The chapter reports 66.4.
• The same case read as interpolation — sections 6 and 7, and the way. Verified: 22 students are already accounted for below 60, so the middle position, 26.5, is the 4.5th of the 7 students inside 60–70. That is 4.5 ÷ 7 = 0.643 of the way through the class, and 0.643 of a ten-mark class is 6.43 marks, landing at 66.43. No formula was quoted; the same number came out.
• The rule change worth flagging — section 3. For ungrouped data with an odd count the chapter uses the (n + 1) ÷ 2 th position (p. 188); for grouped data it uses n ÷ 2 flat (p. 193), and it does not comment on the switch. Here n = 53 is odd, so the ungrouped rule would name position 27. Verified: both 26.5 and 27 fall inside 60–70, so the median class is unaffected — but the answers differ, 60 + 10 × 5 ÷ 7 = 67.14 against 66.43. The difference is real and examinable, and grouped questions want n ÷ 2.
• Example 7 (pp. 194–195). Fifty-one girls' heights, given cumulatively and differenced in the previous topic into classes below 140, 140–145, 145–150, 150–155, 155–160, 160–165 with frequencies 4, 7, 18, 11, 6, 5 and cumulative 4, 11, 29, 40, 46, 51. Verified: n ÷ 2 = 25.5; the running total first passes it at 29, so the median class is 145–150, with l = 145, cf = 11, f = 18, h = 5; the median is 145 + 5 × (25.5 − 11) ÷ 18 = 145 + 72.5 ÷ 18 = 149.03 cm. The chapter reads this back as roughly half the girls being shorter than that height and half taller.
• Example 8, the formula reversed (pp. 195–196). Ten classes, 0–100 up to 900–1000 in hundreds, frequencies 2, 5, x, 12, 17, 20, y, 9, 7, 4, with the total given as 100 and the median as 525. Verified: accumulating downward, the running total is 2, then 7, then picks up x and climbs by 12, 17 and 20 to reach 56 + x at the top of the class 500–600; y enters next, and the last entry, larger than that by 9 + 7 + 4, has to equal 100 — which forces the first relation, x + y = 24. The median 525 sits in 500–600, giving a lower limit of 500, a frequency of 20, a preceding running total of 36 + x and a width of 100, so that 525 = 500 + 100 × (50 − 36 − x) ÷ 20, i.e. 25 = 5 × (14 − x), giving x = 9 and then y = 15.
• The check the chapter skips — worth showing. The solution assumes 500–600 is the median class before x is known. Verified after the fact: with x = 9 the cumulative column reads 2, 7, 16, 28, 45, 65, so the running total has reached 45 by the boundary at 500 and 65 by the boundary at 600. Fifty falls between those two, which is exactly what puts the median inside 500–600 and makes the assumption safe. Students should learn to close that loop.
• Exercise 13.3, worked inputs (pp. 198–200). All values below are added here.
  • Q1, sixty-eight consumers, classes 65–85 to 185–205, counts 4, 5, 13, 20, 14, 8, 4. Verified: cumulative 4, 9, 22, 42, 56, 64, 68; n ÷ 2 = 34; median class 125–145; median 125 + 20 × (34 − 22) ÷ 20 = 137 units exactly.
  • Q2, sixty observations, classes 0–10 to 50–60, counts 5, x, 20, 15, y, 5, median given as 28.5. Verified: the counts give x + y = 15; the median lies in 20–30, with lower limit 20, frequency 20, preceding running total 5 + x and width 10, so that 28.5 = 20 + 10 × (30 − 5 − x) ÷ 20 gives 8.5 = (25 − x) ÷ 2, so x = 8 and y = 7.
  • Q3, one hundred policy holders, differenced into 18–20, 20–25, 25–30, 30–35, 35–40, 40–45, 45–50, 50–55, 55–60 with counts 2, 4, 18, 21, 33, 11, 3, 6, 2. Verified: cumulative 2, 6, 24, 45, 78, 89, 92, 98, 100; n ÷ 2 = 50; median class 35–40; median 35 + 5 × (50 − 45) ÷ 33 = 35 + 25 ÷ 33 = 35.76 years.
  • Q4, forty leaves, classes printed 118–126, 127–135, 136–144, 145–153, 154–162, 163–171, 172–180 with counts 3, 5, 9, 12, 5, 4, 2. The question's own hint gives the corrected classes 117.5–126.5, 126.5–135.5 and so on. Verified: corrected width 9; cumulative 3, 8, 17, 29, 34, 38, 40; n ÷ 2 = 20; median class 144.5–153.5; median 144.5 + 9 × (20 − 17) ÷ 12 = 144.5 + 2.25 = 146.75 mm. This is section 12's payload: compute it uncorrected, treating the class as 145–153 of width 8, and you get 145 + 8 × 3 ÷ 12 = 147 mm. The correction moves the answer by a quarter of a millimetre.
  • Q5, four hundred neon lamps, classes 1500–2000 to 4500–5000 in five hundreds, counts 14, 56, 60, 86, 74, 62, 48. Verified: cumulative 14, 70, 130, 216, 290, 352, 400; n ÷ 2 = 200; median class 3000–3500; median 3000 + 500 × (200 − 130) ÷ 86 = 3000 + 35000 ÷ 86 = 3406.98 hours.
  • Q6, one hundred surnames by letter count, classes 1–4 to 16–19 in threes, counts 6, 30, 40, 16, 4, 4. Verified: cumulative 6, 36, 76, 92, 96, 100; n ÷ 2 = 50; median class 7–10; median 7 + 3 × (50 − 36) ÷ 40 = 8.05 letters.
  • Q7, thirty students by weight, classes 40–45 to 70–75 in fives, counts 2, 3, 8, 6, 6, 3, 2. Verified: cumulative 2, 5, 13, 19, 25, 28, 30; n ÷ 2 = 15; median class 55–60; median 55 + 5 × (15 − 13) ÷ 6 = 56.67 kg.

Figures to have open

• The median class drawn open: a horizontal band of width h with its f observations spaced evenly along it, the lower limit labelled l, the already counted cf shown accumulating to its left, and a marker sliding to the (n ÷ 2 − cf)-th position. Every symbol of the formula labels a part of this one picture. This is the topic's central image and the chapter has nothing like it — the chapter contains no diagrams at all. Standard schematic, must be built.
• A cumulative bar chart with the n ÷ 2 line drawn across it, for section 2. Standard schematic. If anyone wants to draw the ogive here, note that the chapter names ogives but never draws one, so it would be an addition made here — see Running totals, and converting between the two cumulative tables.
• A single class shown twice, printed and continuity-corrected, with the resulting medians 147 and 146.75 marked. Standard schematic.
• Nothing needs to come from the printed page.

Where this sits in the book

• NCERT Class 10 Mathematics, Chapter 13 "Statistics", §13.4, pp. 192–196.
• Table 13.15, the median class identified, the formula with its five symbols and the worked value 66.4, pp. 192–193.
• Example 7, Table 13.16 and the median 149.03 cm, pp. 194–195.
• Example 8 and the recovery of the two missing frequencies, pp. 195–196.
• The ungrouped position rules, p. 188 — cited here for the contrast with n ÷ 2.
• Exercise 13.3, questions 1–7, pp. 198–200, including the continuity hint given with question 4 on p. 199.
• A Note to the Reader, p. 201, which makes continuity a precondition.
• Remark 2 on p. 197, which declines to treat unequal class sizes.