Intuition What this page is for
The parent note built the formulas. Here we run them through every kind of situation an exam or the real world can throw at you — tiny data, degenerate data, shifting and scaling, samples vs populations, and word problems. If a case exists, it has a worked example below with a picture where geometry helps.
We only use two engines the parent gave us. Let me restate them in plain words so nothing here is a mystery symbol.
Below is the full list of "case classes." Every cell is covered by at least one example — the tag in each [!example] tells you which cell it hits.
Case class
What makes it tricky
Covered by
A. Definition, plain data
just follow steps
Ex 1
B. Shortcut on same data
must match A exactly
Ex 2
C. Degenerate — all equal
σ should be 0
Ex 3
D. Shift invariance (x + c )
variance must not move
Ex 4
E. Scaling (c x , incl. negative c )
variance × c 2 , SD $\times
c
F. Sample vs population (n vs n − 1 )
which divisor?
Ex 6
G. Frequency / grouped data
weights, not raw list
Ex 7
H. Word problem, real units
interpret SD in context
Ex 8
I. Exam twist — combine sets
derive from summary stats
Ex 9
J. Limiting behaviour (n → ∞ , one outlier grows)
how σ reacts
Ex 10
The figure above is the mental map: every example is a dart-scatter picture, and σ measures how wide the scatter is.
Worked example Example 1 — Case A (definition engine)
Data (a population ): 3 , 5 , 5 , 7 , 10 .
Forecast: guess whether σ is closer to 1 , 2.5 , or 5 before reading on.
Mean. x ˉ = 5 3 + 5 + 5 + 7 + 10 = 5 30 = 6 .
Why this step? Spread is measured from the centre , so we need the centre first.
Deviations d i = x i − x ˉ : − 3 , − 1 , − 1 , 1 , 4 .
Why? Raw signed distance of each point from the middle.
Square them: 9 , 1 , 1 , 1 , 16 .
Why? Squaring kills the signs (else they cancel to 0 ) and punishes far points more.
Sum & average: sum = 28 , so σ 2 = 28/5 = 5.6 .
Why? Variance is the average squared distance.
Square-root: σ = 5.6 ≈ 2.37 .
Why? Returns us to the data's own units.
Verify: deviations must sum to zero: − 3 − 1 − 1 + 1 + 4 = 0 ✓. Answer 2.37 matches the "closer to 2.5 " forecast.
Worked example Example 2 — Case B (shortcut engine, same data)
Same 3 , 5 , 5 , 7 , 10 , now via σ 2 = x 2 − ( x ˉ ) 2 .
Forecast: it must give 5.6 again — the algebra in the parent note proved the two engines are identical.
Squares of the data: 9 , 25 , 25 , 49 , 100 ; their sum = 208 .
Why? We need x 2 = mean of the squares.
x 2 = 208/5 = 41.6 .
( x ˉ ) 2 = 6 2 = 36 .
Why? Square-of-mean, the second term.
σ 2 = 41.6 − 36 = 5.6 .
Verify: matches Example 1 exactly ✓. Sanity check the ordering trap: x 2 = 41.6 ≥ ( x ˉ ) 2 = 36 , as it must, since their difference is a variance and variances are never negative.
Worked example Example 3 — Case C (zero spread)
Data: 8 , 8 , 8 , 8 .
Forecast: everyone's at the same spot. What should the spread be?
Mean = 32/4 = 8 .
Deviations all = 8 − 8 = 0 .
Why this matters? Every point is the centre.
Squares all 0 , sum 0 , σ 2 = 0/4 = 0 , σ = 0 .
Verify (shortcut): x 2 = 64 , ( x ˉ ) 2 = 64 , difference = 0 ✓. This is the only way σ 2 = 0 — the parent's non-negativity property: variance is zero iff all values are identical .
These two are the workhorses of exam questions: they let you re-use a variance you already computed.
Intuition Read the figure first
Sliding the whole scatter sideways (add c ) never changes the gaps between darts — so variance is unmoved. Stretching the scatter (multiply by c ) multiplies every gap by c ; squared gaps grow by c 2 .
Worked example Example 4 — Case D (shift invariance)
Take Example 1 data and add c = 100 to every value: 103 , 105 , 105 , 107 , 110 .
Forecast: will σ 2 stay 5.6 or jump to something huge?
New mean: old mean + 100 = 106 .
Why? Adding c to every point shifts the balance point by c .
New deviations = x i − 106 : − 3 , − 1 , − 1 , 1 , 4 — identical to before.
Why? The + 100 appears in both x i and x ˉ , so it cancels inside x i − x ˉ .
Therefore σ 2 = 5.6 unchanged.
Verify: Var ( x + c ) = Var ( x ) — matches Example 1's 5.6 ✓.
Worked example Example 5 — Case E (scaling, including a negative factor)
Multiply Example 1 data by c = − 2 : − 6 , − 10 , − 10 , − 14 , − 20 .
Forecast: the property says variance × c 2 and SD × ∣ c ∣ . With c = − 2 , that's variance × 4 , SD × 2 . Guess the numbers.
New mean: − 2 × 6 = − 12 .
Why? Multiplying every value by c multiplies the balance point by c .
New deviations: x i − x ˉ each become − 2 × (old deviation): 6 , 2 , 2 , − 2 , − 8 .
Why? c x i − c x ˉ = c ( x i − x ˉ ) .
Squares: 36 , 4 , 4 , 4 , 64 , sum = 112 , σ 2 = 112/5 = 22.4 .
σ = 22.4 ≈ 4.73 .
Verify: old σ 2 = 5.6 ; 5.6 × c 2 = 5.6 × 4 = 22.4 ✓. Old σ = 2.37 ; 2.37 × ∣ c ∣ = 2.37 × 2 = 4.73 ✓. The negative sign of c vanished because variance uses c 2 and SD uses ∣ c ∣ — spread has no direction.
Worked example Example 6 — Case F (which divisor?)
Three exam times (minutes) drawn as a sample to estimate a whole cohort: 10 , 12 , 14 .
Forecast: sample variance uses n − 1 , so it should be bigger than the population version. Guess by how much.
Mean = 36/3 = 12 .
Squared deviations: ( − 2 ) 2 , 0 2 , 2 2 = 4 , 0 , 4 ; sum = 8 .
If population: σ 2 = 8/3 ≈ 2.67 .
As a sample: s 2 = 8/ ( 3 − 1 ) = 8/2 = 4 .
Why divide by n − 1 = 2 ? Bessel's correction: the sample mean was tuned to these exact points, making deviations slightly too small, so we inflate by using a smaller divisor.
s = 4 = 2 minutes.
Verify: s 2 = 4 > σ 2 ≈ 2.67 ✓ (sample version is larger, as forecast). Ratio 4/ ( 8/3 ) = 1.5 = n − 1 n = 2 3 ✓ — exactly the correction factor.
Definition Frequency version of the engines
When a value x i occurs f i times, treat each as f i copies. With total count N = ∑ f i :
x ˉ = N 1 ∑ f i x i , σ 2 = N 1 ∑ f i ( x i − x ˉ ) 2 = N ∑ f i x i 2 − ( x ˉ ) 2 .
Same idea — each squared distance just gets weighted by how often it happens.
Worked example Example 7 — Case G (weighted data)
Marks and how many students got them:
mark x i
2
4
6
frequency f i
1
3
2
Forecast: most students sit at 4 ; guess whether σ is below or above 2 .
Total count N = 1 + 3 + 2 = 6 .
Weighted mean: ∑ f i x i = 1 ⋅ 2 + 3 ⋅ 4 + 2 ⋅ 6 = 2 + 12 + 12 = 26 , so x ˉ = 26/6 ≈ 4.33 .
Why weight? A mark that appears 3 times pulls the centre three times as hard.
Weighted mean of squares: ∑ f i x i 2 = 1 ⋅ 4 + 3 ⋅ 16 + 2 ⋅ 36 = 4 + 48 + 72 = 124 ; divide by N : 124/6 ≈ 20.67 .
Shortcut: σ 2 = 20.67 − ( 4.33 ) 2 = 20.67 − 18.78 ≈ 1.89 .
σ ≈ 1.37 .
Verify: exact fractions — x ˉ = 13/3 , x 2 = 62/3 , so σ 2 = 3 62 − 9 169 = 9 186 − 169 = 9 17 ≈ 1.89 ✓. Below 2 , as forecast, because most students cluster at 4 .
Worked example Example 8 — Case H (interpret the units)
A machine fills bottles. Five bottles measured (ml), treat as the population of this batch: 498 , 502 , 500 , 501 , 499 .
Forecast: these are tightly bunched near 500 ; guess σ in ml.
Mean = 5 498 + 502 + 500 + 501 + 499 = 5 2500 = 500 ml.
Deviations: − 2 , 2 , 0 , 1 , − 1 ml.
Squares: 4 , 4 , 0 , 1 , 1 ; sum = 10 .
σ 2 = 10/5 = 2 ml² .
Why "ml²"? Squaring the deviations squared the units too.
σ = 2 ≈ 1.41 ml .
Why square-root? To report spread in real millilitres, not ml².
Verify: deviations sum − 2 + 2 + 0 + 1 − 1 = 0 ✓. Interpretation: typical bottle is about 1.41 ml off the 500 ml target — the machine is precise. Reporting "2 " without the square-root would be the classic forgot-to-root mistake, and it would be in ml², a nonsense unit for a bottle .
Sometimes you're not given the raw numbers, only summary sums. The shortcut engine shines here.
Worked example Example 9 — Case I (merge using
∑ x and ∑ x 2 )
Group P: n P = 3 values with ∑ x = 12 , ∑ x 2 = 56 .
Group Q: n Q = 2 values with ∑ x = 18 , ∑ x 2 = 164 .
Find the variance of the combined population.
Forecast: combining a low-mean group with a high-mean group usually increases spread. Watch for that.
Combined count N = 3 + 2 = 5 .
Combined mean: x ˉ = 5 12 + 18 = 5 30 = 6 .
Why add the ∑ x ? Total sum over total count is the overall balance point.
Combined mean of squares: x 2 = 5 56 + 164 = 5 220 = 44 .
Why add the ∑ x 2 ? ∑ x 2 is additive across disjoint groups, just like ∑ x .
Shortcut: σ 2 = 44 − 6 2 = 44 − 36 = 8 .
σ = 8 ≈ 2.83 .
Verify: consistent example — Group P could be 2 , 4 , 6 (∑ = 12 , ∑ x 2 = 4 + 16 + 36 = 56 ✓) and Group Q 8 , 10 (∑ = 18 , ∑ x 2 = 64 + 100 = 164 ✓). Merged list 2 , 4 , 6 , 8 , 10 : mean 6 , squared devs 16 , 4 , 0 , 4 , 16 sum 40 , σ 2 = 40/5 = 8 ✓.
Worked example Example 10 — Case J (one value runs away)
Start with 0 , 0 , 0 , 0 (population). One entry grows to x : data becomes x , 0 , 0 , 0 . How does σ 2 behave as x → ∞ ?
Forecast: one wild dart should blow the spread up without limit. Does it grow like x or like x 2 ?
Mean: x ˉ = x /4 .
Mean of squares: x 2 = 4 x 2 + 0 + 0 + 0 = 4 x 2 .
Shortcut: σ 2 = 4 x 2 − ( 4 x ) 2 = 4 x 2 − 16 x 2 = 16 3 x 2 .
Limit: as x → ∞ , σ 2 = 16 3 x 2 → ∞ , growing like x 2 (SD grows like x ).
Why x 2 , not x ? Squaring deviations means a far-off point contributes its distance squared — outliers dominate. This is exactly why the parent note warned squaring "penalises big deviations more."
Verify: plug x = 4 : data 4 , 0 , 0 , 0 , mean 1 , squared devs 9 , 1 , 1 , 1 sum 12 , σ 2 = 12/4 = 3 ; formula 16 3 ⋅ 16 = 3 ✓.
Recall Self-test: name the case, then solve
For each mini-prompt, first say which matrix cell (A–J) it is.
Data 5 , 5 , 5 population variance ::: Case C — degenerate, σ 2 = 0 .
You already found σ 2 = 9 ; now every value is multiplied by 3 . New variance? ::: Case E — × c 2 = 9 × 9 = 81 .
Same data, but now add 7 to every value. New variance? ::: Case D — unchanged, still 9 .
n = 3 sample, sum of squared deviations = 18 . Sample variance? ::: Case F — 18/ ( 3 − 1 ) = 9 .
Merge: N = 4 , ∑ x = 8 , ∑ x 2 = 24 . Variance? ::: Case I — 24/4 − ( 8/4 ) 2 = 6 − 4 = 2 .
Mnemonic One line to survive them all
Centre, subtract, square, average, root — and only two dials change it: shift does nothing, stretch squares it.