Measures of Central Tendency

— Probability & Statistics 1 · AS & A Level Mathematics · Chapter 2 —

Big idea: an average = ONE representative value that is fairly central and typical of the data. Three types: mode, mean, median — and choosing the right one is part of the skill.

Chapter goals: find & use the 3 averages • mean from data, from a total Σx, or from a coded total Σ(x − b) • grouped data • problems with up to two datasets.

1. Three Types of Average

Mode = the most commonly occurring value.
Mean = (sum of the values) ÷ (number of values).
Median = the value in the middle of an ordered set.

Pick by purpose: shoemaker → mode (most popular size). Farmer → median eggs per hen (spot unprofitable hens). Incomes → decide: workers & managers together or separately?

Why choose carefully? The mean is the most familiar, but another average may suit the situation better. For an average income of a region we must also decide who counts in which group before averaging.

Prerequisite check ():

Data 7.3, 3.9, 1.3, 6.6, 9.2, 4.7, 3.9, 3.1 → sum = 40 ⇒ mean = 5; ordered 1.3, 3.1, 3.9, 3.9, 4.7, 6.6, 7.3, 9.2 ⇒ median = 4.3; mode = 3.9.
Calculator: 6×1.7 + 8×1.9 + 11×2.16 + 8 + 11 = 48.525 = 1.94 (sensible: between 1.7 and 2.1 ✓ — always sanity-check).

Explore 2.1 — “the average person”: such facts (laughs 10×/day, sneeze 160 km/h, 172.5 cm average adult male…) are means/medians of huge samples — no real person is exactly “average”, but the figures are useful for design, planning & comparison.

2. The Mode and the Modal Class

A set can have more than one mode, or no mode at all. e.g. 25 die rolls:

Score	1	2	3	4	5	6
f	5	6	5	3	2	4

Highest frequency 6 at score 2 ⇒ mode = 2.

Grouped data (raw values hidden) → find the modal class = class with the highest frequency density.

frequency density = frequencyclass width

Key Point 2.1: in a histogram the modal class has the greatest column height. If there is no modal class, all classes have the same frequency density. (Frequency density was met in Ch. 1, §1.3.)

Tip: modal class ≠ class with most items! It has the most items per unit width. If all widths are equal, f.d. ∝ f, so modal class = highest frequency.

Worked Eg 2.1modal class

270 pencil lengths (nearest cm): 4–7 → 100; 8–10 → 90; 11–12 → 80. Find the modal class.

Use class boundaries (data are to nearest cm) to get true widths.

modal class = tallest bar (11–12 cm) even though it has the fewest pencils

Length x (cm)	f	Width	f.d. = f ÷ width
3.5 ≤ x < 7.5	100	4	100÷4 = 25
7.5 ≤ x < 10.5	90	3	90÷3 = 30
10.5 ≤ x < 12.5	80	2	80÷2 = 40

Modal class = 11–12 cm (10.5 ≤ x < 12.5) — it has the lowest frequency but the greatest f.d.

Worked Eg 2.2no modal class

Two classes have widths in ratio 3 : 2; there is no modal class; first frequency = 48. Find the second frequency.

“No modal class” ⇒ equal f.d. ⇒ frequencies in the same ratio as widths.

48 : x = 3 : 2x48 = 23x = 32

Alt: x2 = 483 ⇒ x = 32 (f ∝ width).

3. The Mean

The mean (arithmetic mean) is the most commonly known average. Since mean = sum of valuesnumber of values, we can recover the sum: 12 values with mean 7.5 ⇒ sum = 7.5 × 12 = 90.

	Sum	Values	Freq.	Number	Sum of data	Mean
Ungrouped	Σ	x	—	n	Σx	x̄ = Σxn
Grouped	Σ	x	f	Σf	Σxf or Σfx	x̄ = ΣxfΣf

Key Point 2.2Ungrouped: x̄ = Σxn Grouped: x̄ = ΣxfΣf = ΣfxΣf

Note: Σxf means “add the product of each value and its frequency”, e.g. five 10s and six 20s: (10×5) + (20×6) = 170.

Worked Eg 2.3using Σx = x̄ × n

Five labourers, mean mass 70.2 kg, go up in a lift with cement. Lift limit 500 kg. Greatest mass of cement?

Σx = x̄ × nΣx = 70.2 × 5 = 351 kg351 + y = 500y = 149 kg

Worked Eg 2.4grouped / frequency table

Mean of 40 values: x = 31, 32, 33, 34, 35 with f = 5, 7, 9, 8, 11.

x	31	32	33	34	35
f	5	7	9	8	11	Σf = 40
xf	155	224	297	272	385	Σxf = 1333

x̄ = ΣxfΣf= 133340x̄ = 33.325

Combined sets of data

To combine two sets, just pool all the values: combined mean = (total sum) ÷ (total number).

Example:

{1, 2, 3, 4} (mean 2.5) + {4, 5, 6} (mean 5) → {1, 2, 3, 4, 4, 5, 6} — note 4 appears twice.
Combined mean = 1+2+3+4+4+5+64+3 = 257 = 3⁴⁄₇

Common mistake: combined mean ≠ 2.5 + 52. Only works if both sets are the same size (Explore 2.2: for sets A, B with m and n values, mean(A∪B) = average of the two means exactly when m = n).

Worked Eg 2.5combined sets

Large bag: 72 sweets, total 852.4 g. Small bag: 24 sweets, total 282.8 g. Mean mass of all sweets?

Total number = 72 + 24 = 96Total mass = 852.4 + 282.8 = 1135.2 gMean = 1135.296 = 11.825 g

Assumes the given figures are accurate & bag masses are not included in the totals.

Worked Eg 2.6combined sets, units

38 DVDs, mean 1 h 32 min; 26 videotapes, mean 2 h 4 min. Mean playing time of all films?

films: 38 + 26 = 64(92 × 38) + (124 × 26) = 6720 minMean = 672064 = 105 min = 1 h 45 min

Convert to one unit (minutes) first; and 105 ≠ (92+124)/2 = 108.

Means from grouped frequency tables (estimates)

Why an estimate? Grouping (tables, histograms, cf graphs) hides the raw values, so the mean can’t be exact.

Represent each class by its mid-value and use x̄ = ΣxfΣf with x = class mid-values.

mid-value = lower boundary + upper boundary2

Worked Eg 2.7estimate of mean

75 crates, 40 similar coconuts per crate: 46 crates total mass 20–<25 kg; 22 crates 25–<40 kg; 7 crates 40–<54 kg. (a) Estimate mean mass of a crate. (b) Estimate mean mass of a coconut.

Mass (kg)	20–	25–	40–54
f	46	22	7	Σf = 75
mid-value x	22.5	32.5	47
xf	1035	715	329	Σxf = 2079

(a) x̄ = 207975 = 27.72 kg(b) 27.7240 = 0.693 kg

Divide by 40 because each crate holds 40 coconuts.

Gaps between classes? Use class boundaries to find mid-values — see next example.

Worked Eg 2.8mid-values with gaps

50 students: sixteen 18-year-olds, twenty 19-year-olds, fourteen aged 20 or 21. Estimate the mean age.

An “18-year-old” is aged 18 ≤ A < 19, etc. The 20- & 21-year-olds are 20 ≤ A < 22.

Age A	mid x	f	xf
18 ≤ A < 19	18.5	16	296
19 ≤ A < 20	19.5	20	390
20 ≤ A < 22	21	14	294
		Σf = 50	Σxf = 980

x̄ = 98050 = 19.6 years

Wrong mid-values 18, 19, 20.5 give 19.1 ✗.

Fast forward: the mean is used to get variance & standard deviation (Ch. 3, §3.3).

Explore 2.3 (cards 1–6 in a bag): take every possible selection of r cards and find each mean (X_r). The mean of all the X_r values is always 3.5 (= mean of 1…6); X₆ is the mean of all six cards. Selection counts: 6, 15, 20, 15, 6, 1 (Ch. 5, §5.3).

4. Coded Data

Coding = transforming every value by adding a (positive/negative) constant. Why? Easier hand calculations, or to make the mean a convenient number (e.g. 0).

Example:

Mean of 101, 103, 104, 109, 113 → subtract 100: 1, 3, 4, 9, 13.
Mean of coded = Σ(x − 100)5 = 1+3+4+9+135 = 6 ⇒ x̄ = 6 + 100 = 106 (add back what you subtracted).

subtract b from every value ⇒ the mean shifts by b too (the whole set slides left)

Key Point 2.3mean(x − b) = x̄ − b ⇒ x̄ = mean(x − b) + b
Ungrouped: x̄ = Σ(x − b)n + b Grouped: x̄ = Σ(x − b)fΣf + b

Note: removing brackets: Σ(x − b) = Σx − Σb = Σx − nb (there are n copies of b).

Worked Eg 2.9two datasets, coded totals

Exactly 5 yrs ago the sum of ages of 10 boys was 127.0 ⇒ Σ(b − 5) = 127. In exactly 5 yrs’ time the sum of ages of 15 girls will be 351.0 ⇒ Σ(g + 5) = 351. Find mean age today of boys, girls, and all 25 together.

(a) boysb̄ = 12710 + 5b̄ = 17.7alt: Σb − 50 = 127 ⇒ Σb = 177

(b) girlsḡ = 35115 − 5ḡ = 18.4alt: Σg + 75 = 351 ⇒ Σg = 276

To combine, always go back to the real totals Σb and Σg — never average the means.

Worked Eg 2.10grouped coded data

40 values; table shows (x − 3): classes 0–, 18–, 24–32 with f = 9, 13, 18. Estimate the mean of x.

Use mid-values of the coded classes (9, 21, 28), then add 3 back. No need to decode the classes.

x̄ = Σ(x − 3)fΣf + 3(9×9) + (21×13) + (28×18)40 + 3= 85840 + 3x̄ = 24.45

Coding by multiplication (E = extension)

Data can be coded by × as well as + / −. Teachers’ take-home pay $3600, $4200, $3700, $4500 (mean $4000). A 10 % rise and $50 extra tax ⇒ new pay = 1.1x − 50: 3910, 4570, 4020, 4900 ⇒ mean $4350 = (1.1 × 4000) − 50 ✓.

Key Point 2.4coded data ax − b has mean a x̄ − b
Ungrouped: x̄ = 1a[ Σ(ax − b)n + b ] Grouped: x̄ = 1a[ Σ(ax − b)fΣf + b ]
Short form: x̄ = 1a × [ mean(ax − b) + b ]

Order matters: to decode, undo “− b” first (add b), then undo “× a” (divide by a) — reverse order of coding. Also Σ(ax − b) = aΣx − nb.

Decoding the teachers’ example (mean of coded data = 4350):

x̄ = (4350 + 50) ÷ 1.1x̄ = 4000 ✓

Worked Eg 2.11unit conversion = coding

Total cloth area is Σx in m². Expression for area in cm²?

1 m = 100 cm1 m² = 100² cm² = 10 000 cm²total area = Σ10000x = 10000Σx cm²

Units convert by multiplying — a coding with a = 10 000, b = 0.

Worked Eg 2.12Σ(ax − b) given

20 values with Σ(2x − 3) = 104. Find x̄.

Method 1: mean of codedmean(2x − 3) = 10420 = 5.22x̄ − 3 = 5.2x̄ = 5.2 + 32 = 4.1

Method 2: expand brackets2Σx − (20 × 3) = 104Σx = 82x̄ = 8220 = 4.1

Fast forward: coded totals Σ(ax − b) and Σ(ax − b)² are used for variation in Ch. 3, §3.3.

5. The Median

The median splits ordered data into a bottom half and a top half with equal numbers of values.

Key Point 2.5For n ordered values the median is the n + 12th value.
If n is even → the mean of the two middle values.

Tip: for a frequency table n = Σf. The median is a position (e.g. 8th value) — don’t confuse it with its value.

Stem-and-leaf (back-to-back) — DIY store customers

Monday (open 12 h, n = 12) and Saturday (15 h, n = 15). Key: 0 | 2 | 2 means 20 customers on Monday and 22 on Saturday.

Monday (12)	stem	Saturday (15)
8 6 3 1 0 0	2	2 3 4 6
4 3 1 1 0	3	5 5 6 8 9 9
1	4	0 1 3 7 9

Careful: on the left side leaves ascend from right to left, and descend from left to right along each row.

Saturday (n = 15)position = 15 + 12 = 8throw 1 holds 1st–4th, row 2 holds 5th–10th8th value → median = 38

Monday (n = 12)position = 12 + 12 = 6.5th6th = 28, 7th = 30median = 28 + 302 = 29

Worked Eg 2.13median from ordered frequency table

65 readings: x = 40, 41, 42, 43, 44; f = 11, 23, 19, 8, 4. Use cumulative frequency (cf) to locate positions.

x	f	cf	positions
40	11	11	1st–11th
41	23	34	12th–34th
42	19	53	35th–53rd
43	8	61	54th–61st
44	4	65	62nd–65th

n = Σf = 65 ⇒ position = 65 + 12 = 33rd33rd lies in 12th–34th ⇒ median = 41

41 isn’t the middle of the five x-values — it’s the middle of the 65 readings (frequencies count!).

Estimating the median (grouped / continuous data)

Large datasets & continuous data are grouped, so the exact median is hidden → estimate it from a cumulative frequency graph.

Key Point 2.6On a c.f. graph with total n = Σf, the median is at the n2th value.

Why n/2, not (n+1)/2? A graph is only an estimate; n/2 gives the same position whether you count up from the bottom or down from the top of the c.f. axis.

go across from n/2 = 150, then straight down to the axis

Example — 300 museum artefacts (masses):

n = 300 ⇒ n/2 = 150th artefact. Draw a horizontal line from c.f. = 150 to the curve, then a vertical line down to the mass axis ⇒ median ≈ 2.6 kg.
Position = 150th; value = 2.6 kg — don’t mix them up.

Did you know? Using ONE value to represent many measurements is quite recent: no mean/median/mode use before the 17th century. The 11th-century Persian Al-Biruni used the mid-range (middle of smallest & largest values) when finding the longitude of Ghazni (modern Afghanistan) and studying metals; Newton & explorers used it too. Magnetic-declination measurements probably boosted the popularity of the mean. (Mid-range ≠ median — Ch. 3, §3.2.)

6. Choosing an Appropriate Average

It is easy to pick an average that badly misrepresents the data — so consider the purpose and motives of whoever chooses it.

Student’s test marks (out of 20):

3, 4, 6, 7, 8, 11, 12, 13, 17, 17 ⇒ mode = 17, mean = 9.8, median = 9.5.
To impress friends/parents → quote the mode (highest). Mean & median show fewer than half marks on average.

	Advantages	Disadvantages
Mode	• Unlikely to be affected by extreme values • Useful to manufacturers (most popular styles & sizes) • Can be used for qualitative data	• Ignores most values • Rarely used in further calculations
Mean	• Uses all values • Used in further calculations • Most commonly understood • Can be used to find the sum of the data	• Can’t be found unless all values are known • Likely to be affected by extreme values
Median	• Can be found without knowing all values • Relatively unaffected by extreme values	• Only uses the order of values, so ignores most of them

Effect of an extreme value:

40, 40, 70, 100, 130, 250: mode 40, median 85, mean 105. Change 250 → 880: mode & median unchanged (40, 85) but mean doubles: 105 → 210.

But… the median is usually unaffected by extremes — not always. The LIBOR scandal shows this.

discard top & bottom 25 %, average the middle 50 %

LIBOR (London Interbank Offered Rates) = average rates leading banks charge each other for short-term loans; underpins ~US$450 trillion of investments. Highest & lowest 25 % of daily rates are discarded; LIBOR = mean of the middle 50 %.

Simple eg: A 2.6, B 2.8, C 3.0, D 3.1 → discard A & D → mean(2.8, 3.0) = 2.9 %. If D had submitted 2.5 %: order 2.5, 2.6, 2.8, 3.0 → middle two 2.6, 2.8 → 2.7 %. So a bank’s false rate can move the “trimmed mean”; banks have been fined over US$9 billion.

Shape of data & skewness

Courier delivery of 100 packages: days 1–6 with f = 10, 40, 25, 14, 8, 3.

positively skewed: mode 2 < median 2.5 < mean 2.79

mode = 2 daysmedian: between 50th & 51st → 2 + 32 = 2.5 daysmean = 10+80+75+56+40+18100 = 2.79 days

mean is the largest average, on the side of the longer tail
mode is the smallest, on the side of the shorter tail
median lies between mode and mean

Data that is not symmetrical is skewed. Longer tail towards larger values ⇒ positively skewed; towards smaller values ⇒ negatively skewed.

long tail → right
mode < median < mean

all three equal
(normal distribution)

long tail ← left
mean < median < mode

Positive skew: mode < median < mean
Negative skew: mean < median < mode

Fast forward: in a normal distribution (Ch. 8) mode = mean = median.

7. Exam-style Worked Questions

These are from the end-of-chapter review — the same ideas, combined.

Review Q2combined means

13 textbooks have mean 875 g; n novels have total 13 706 g; all together mean 716.6 g. Mean mass of a novel?

textbooks total = 13 × 875 = 11 375all total = 11 375 + 13 706 = 25 08113 + n = 25 081716.6 = 35n = 22novel mean = 13 70622 = 623 g

Review Q11two coded totals, same data

150 values: Σ(x − 1) + Σ(x − 4) = 4170. Find x̄.

2Σx − 150(1) − 150(4) = 41702Σx = 4170 + 750 = 4920Σx = 2460x̄ = 2460150 = 16.4

Review Q12adding 1 to every value

Monday: 8 students write x, mean 30¼. Tuesday one is absent; the other 7 write (their number + 1). Mean of all 15 numbers = 27⅓. Absent student’s Monday number?

Monday sum = 8 × 30.25 = 242let absent student’s number = aTuesday sum = (242 − a) + 7total = 242 + 249 − a = 491 − a15 × 27⅓ = 410491 − a = 410a = 81

Review Q14coded totals, two groups

10 women: Σ(w − 3000) = −200. 20 men: Σ(m − 4000) = 120. Find difference of means, and mean of all 30.

womenmean = −20010 + 3000= 2980Σw = 29 800

menmean = 12020 + 4000= 4006Σm = 80 120

(a) 4006 − 2980 = $1026 (men higher)(b) 29 800 + 80 12030 = 109 92030 = $3664

Review Q15two coded totals, combine

90 values of x, 64 of y: Σ(x − 1) = 72.9, Σ(y + 1) = 201.6. Mean of all combined.

Σx − 90 = 72.9 ⇒ Σx = 162.9Σy + 64 = 201.6 ⇒ Σy = 137.6mean = 162.9 + 137.690 + 64 = 300.5154 ≈ 1.95

8. Quick Recap — Chapter Cheat-Sheet

✓ Averages & formulas

Measures of central tendency: mode, mean, median.
Ungrouped: mode = most frequent value. Grouped: modal class = highest frequency density = tallest histogram bar.
f.d. = frequency ÷ class width; no modal class ⇒ equal f.d. ⇒ f ∝ width. Equal widths ⇒ modal class = highest f.
Ungrouped mean: x̄ = Σx/n. Grouped mean (estimate, x = mid-values): x̄ = Σxf/Σf (n = Σf).
Combined sets: total sum ÷ total count (never the average of the means unless equal sizes). Σx = x̄ × n.
Coding: x̄ = mean(x − b) + b; Ungrouped x̄ = Σ(x − b)/n + b; Grouped x̄ = Σ(x − b)f/Σf + b. Σ(x − b) = Σx − nb.
E: x̄ = (1/a) × [mean(ax − b) + b]; undo “− b” first, then “× a”.
Median: ungrouped at the (n + 1)/2 th value (even n ⇒ mean of two middle values). Grouped: estimate at the n/2 th value on a cumulative frequency graph.
Positive skew: mode < median < mean. Negative skew: mean < median < mode. Symmetric: all equal.
Mode: robust, works for qualitative data. Mean: uses all data, hit by extremes. Median: robust, uses only order.