Chapter 14.2 — Exercise 14.1 — Finding the Mean
Problems based on finding the mean of grouped data. This is Lesson 2 of 5 in Chapter 14: Statistics.
Nine Datasets, Three Methods
Exercise 14.1 puts direct, assumed mean, and step deviation to work across nine real datasets, plus one problem that runs the mean formula in reverse to recover a missing frequency. Which method suits which dataset depends entirely on the class marks involved — nothing about the data itself demands one method over another.
Direct Method: Plants in 20 Houses
A survey of the number of plants in 20 houses, grouped into seven classes from 0–2 up to 12–14, uses the direct method exactly as introduced in the chapter opening — small class marks here, so there's no real arithmetic advantage to shifting them.
| Class (plants) | fᵢ (houses) | xᵢ | fᵢxᵢ |
|---|---|---|---|
| 0–2 | 1 | 1 | 1 |
| 2–4 | 2 | 3 | 6 |
| 4–6 | 1 | 5 | 5 |
| 6–8 | 5 | 7 | 35 |
| 8–10 | 6 | 9 | 54 |
| 10–12 | 2 | 11 | 22 |
| 12–14 | 3 | 13 | 39 |
| Total | 20 | 162 |
Mean = Σfᵢxᵢ / Σfᵢ = 162 / 20 = 8.1 plants per houseThis is the direct method at its most straightforward — seven small classes, single-digit-to-low-teens class marks, and a total that divides out cleanly. There's no real incentive here to shift to an assumed mean; the arithmetic is already about as light as it can get.
Assumed Mean Method: Daily Wages of 50 Workers
Fifty workers' daily wages, grouped into five Rs. 50-wide classes from Rs. 200–250 up to Rs. 400–450, use noticeably larger class marks — a good candidate for the assumed mean method, choosing the middle class's mark (325) as the reference point a.
| Class (Rs.) | fᵢ | xᵢ | dᵢ = xᵢ−325 | fᵢdᵢ |
|---|---|---|---|---|
| 200–250 | 12 | 225 | −100 | −1200 |
| 250–300 | 14 | 275 | −50 | −700 |
| 300–350 | 8 | 325 | 0 | 0 |
| 350–400 | 6 | 375 | 50 | 300 |
| 400–450 | 10 | 425 | 100 | 1000 |
| Total | 50 | −600 |
Mean = a + Σfᵢdᵢ/Σfᵢ = 325 + (−600/50) = 325 − 12 = Rs. 313Notice the negative running total — Σfᵢdᵢ comes out to −600, not a positive number, and that's entirely expected here rather than a sign of an error. It simply means the wages are, on balance, pulled below the assumed reference of 325 by the two lower classes' larger combined frequency (12+14=26 workers below 325, versus only 16 above it).
Running the Formula Backward: A Missing Frequency
A frequency table with one unknown frequency f, in the class 19–21, is given alongside the already-known mean, Rs. 18. Rather than computing a mean from a complete table, this problem instead solves an equation for the single missing piece that would produce that mean.
| Class (Rs.) | fᵢ | xᵢ | fᵢxᵢ |
|---|---|---|---|
| 11–13 | 7 | 12 | 84 |
| 13–15 | 6 | 14 | 84 |
| 15–17 | 9 | 16 | 144 |
| 17–19 | 13 | 18 | 234 |
| 19–21 | f | 20 | 20f |
| 21–23 | 5 | 22 | 110 |
| 23–25 | 4 | 24 | 96 |
| Total | 44+f | 752+20f |
18 = (752+20f) / (44+f)
18(44+f) = 752+20f
792 + 18f = 752 + 20f
40 = 2f
f = 20Every other problem in this exercise treats fᵢxᵢ/Σfᵢ as an arithmetic calculation with a numeric result waiting at the end. Here it's instead treated as an algebraic equation, with the known mean substituted in and the unknown frequency isolated — the exact same formula, used in the opposite direction from every other problem around it.
Step Deviation Method: Household Food Expenditure
Twenty-five households' daily food expenditure, grouped into five Rs. 50-wide classes, brings in the step deviation method's extra division by class size h — worthwhile here specifically because h=50 is large enough that dividing it out keeps every uᵢ down to a small, easy integer.
| Class (Rs.) | fᵢ | xᵢ | uᵢ=(xᵢ−225)/50 | fᵢuᵢ |
|---|---|---|---|---|
| 100–150 | 4 | 125 | −2 | −8 |
| 150–200 | 5 | 175 | −1 | −5 |
| 200–250 | 12 | 225 | 0 | 0 |
| 250–300 | 2 | 275 | 1 | 2 |
| 300–350 | 2 | 325 | 2 | 4 |
| Total | 25 | −7 |
Mean = a + (Σfᵢuᵢ/Σfᵢ)×h = 225 + (−7/25)×50 = 225 − 14 = Rs. 211Compare the size of the numbers actually being handled here against the assumed-mean wages problem above: dᵢ values there ran up to ±100, while uᵢ values here never exceed ±2. That compression is the entire point of dividing by h — every deviation shrinks to a manageable single digit before it ever gets multiplied by a frequency.
Assumed Mean Once More: Heart Beats per Minute
Thirty women's heart rates, grouped into seven classes of size 3 from 65–86 beats per minute, use the assumed mean method a second time — worth seeing again with fractional class marks, since the class boundaries here don't land on whole numbers.
| Class (bpm) | fᵢ | xᵢ | dᵢ=xᵢ−75.5 | fᵢdᵢ |
|---|---|---|---|---|
| 65–68 | 2 | 66.5 | −9 | −18 |
| 68–71 | 4 | 69.5 | −6 | −24 |
| 71–74 | 3 | 72.5 | −3 | −9 |
| 74–77 | 8 | 75.5 | 0 | 0 |
| 77–80 | 7 | 78.5 | 3 | 21 |
| 80–83 | 4 | 81.5 | 6 | 24 |
| 83–86 | 2 | 84.5 | 9 | 18 |
| Total | 30 | 12 |
Mean = 75.5 + 12/30 = 75.5 + 0.4 = 75.9 beats per minuteThe class marks here (66.5, 69.5, and so on) all carry a decimal point, since the class width of 3 is an odd number and a midpoint of an odd-width interval always lands on a half-unit — yet the assumed-mean arithmetic proceeds exactly as it did for the whole-number wages problem earlier, with dᵢ still coming out as clean integers because a itself was chosen as one of those same half-unit class marks.
Four More, Compressed
| Dataset | Method | Result |
|---|---|---|
| Oranges per basket, 400 baskets (5 classes, 10–34) | Step deviation, a=22, h=5 | 22.31 oranges |
| SO₂ concentration, 30 localities (6 classes, 0.00–0.24 ppm) | Direct — decimal class marks, no shortcut needed | 0.099 ppm |
| Attendance, 40 students out of 56 days (7 classes, 35–56) | Assumed mean, a=45.5 | ≈49 days |
| Literacy rate, 35 cities (5 classes, 45–95%) | Assumed mean, a=70 | 69.43% |
The SO₂ problem is worth a second glance for what it doesn't need: even though its class marks are small decimals (0.02, 0.06, 0.10, …) rather than large integers, the direct method still applies cleanly — the assumed-mean and step-deviation methods exist to simplify large or awkward numbers, not small ones, and 0.02–0.22 was never actually difficult arithmetic to begin with. The attendance problem is worth a second glance too, for the opposite reason: its precise mean works out to 49.025 days, and rounding that down to a whole "49 days" is a reasonable final step, since attendance is only ever counted in whole days regardless of what the raw division produces.
The literacy-rate and oranges problems both round to two decimal places rather than a whole number — a percentage and a count of fruit both tolerate a fractional final answer in a way that "days attended" doesn't, since nobody attends 49.025 days of school but a basket can meaningfully average 22.31 oranges across four hundred baskets. Deciding how far to carry a final decimal is a small judgement call worth making deliberately for every mean calculation, based on what the number is actually measuring, rather than mechanically copying however many decimal places the raw division happens to produce.
Onward to the Mode
Every problem in this exercise asked the identical underlying question — what single value best represents the centre of this dataset — and answered it with Σfᵢxᵢ/Σfᵢ in one of its three equivalent forms. Exercise 14.2 asks a related but genuinely different question: not what the data averages to, but which single value occurs most often — a measure the mean, as every calculation above shows, doesn't actually reveal on its own. Revisit the chapter introduction for the class-mark and three-method formulas this entire exercise leaned on throughout.