MHT-CET Maths · Measures of Dispersion

Shift and Scale — How Adding and Multiplying Change Mean, Variance and SD

Add a constant: the mean shifts by it and the variance does not move. Multiply by λ: the mean multiplies by λ, the SD by |λ|, the variance by λ².

All Measures of Dispersion notes MHT-CET Maths strategy

Why this matters

10 PYQs at 30% HARD — the chapter's only HARD questions are here, though six of the ten are one-line applications of the two rules. The HARD three are the same rules used inside an algebraic identity: the mean of (x − 5)² from a known mean and variance (set twice), and the b = a + c identity whose stored options had lost a factor of 3 until this session restored them from the paper. Learn the rules as facts about the formula, not as slogans.

Concept 1 of 4

Adding a Constant: Mean Shifts, Variance Stays

Intuition

Adding

8

to every observation slides the whole data set along the line. Distances between values — and so the spread — are untouched; only the centre moves.

Definition

$y_i = x_i + c$ : $\bar y = \bar x + c$ , $\sigma_y^2 = \sigma_x^2$ , $\sigma_y = \sigma_x$ .
Variance $6$ , mean $10$ ; each observation increased by $8$ : new variance $6$ , new mean $18$ .
Why: $y_i - \bar y = (x_i + c) - (\bar x + c) = x_i - \bar x$ , so every squared deviation is the same.
This is the fact behind the assumed-mean method and behind 'the SD of $a + 2, b + 2, c + 2$ is $d$ ' meaning the SD of $a, b, c$ is $d$ .

Shift rule

\overline{x + c} = \bar x + c,\qquad \operatorname{Var}(x + c) = \operatorname{Var}(x)

Diagram · mean = the balance point

Treat each value as equal weight on a beam; the mean is the point where it balances. The pulls on the left (deviations −3, −1) exactly cancel those on the right (0, +4), which is the identity Σ(xᵢ − x̄) = 0. One extreme value drags the balance point toward it — why the mean is sensitive to outliers.

Worked example

The mean and variance of

12

observations are

25

and

9

. Each observation is decreased by

5

. Find the new mean and SD.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Measures of DispersionEASY

The variance and mean of 15 observations are respectively 6 and 10. If each observation is increased by 8 then the new variance and new mean of resulting observations are respectively

[Q119 · Shift 1 · 2022]

Adding the constant to the variance

New variance

14

is option (C) on the classic stem. Adding a constant moves nothing but the mean.

Concept 2 of 4

Multiplying by λ: Mean × λ, SD × |λ|, Variance × λ²

Intuition

Scaling every value by

3

stretches every deviation by

3

, so the SD triples and the variance — a square — grows nine-fold. A multiply-then-add stem uses both rules; only the multiplication touches the variance.

Definition

$y_i = \lambda x_i$ : $\bar y = \lambda\bar x$ , $\sigma_y = |\lambda|\sigma_x$ , $\sigma_y^2 = \lambda^2\sigma_x^2$ .
Variance $12$ , each value $\times 3$ : $108$ . Variance $16$ , $\times 3$ : $144$ . Variance $5$ , $\times 2$ : $20$ .
Variance $5$ , each value $\times 3$ THEN $+ 8$ : $9 \times 5 = 45$ ; the $+8$ does nothing to the variance.
The sign of $\lambda$ is lost in the variance and the SD ( $|\lambda|$ ) but kept in the mean; that matters when a stem lets $p = \pm\frac12$ .

Scale rule

\overline{\lambda x} = \lambda\bar x,\qquad \sigma_{\lambda x} = |\lambda|\,\sigma_x,\qquad \operatorname{Var}(\lambda x) = \lambda^2\operatorname{Var}(x)

Diagram · mean deviation = average distance from the centre

MD = (3 + 1 + 0 + 4) / 4 = 2

Take each value's distance from the centre (the red segments, signs dropped) and average them. Mean deviation can be taken about the mean or the median; about the median it is smallest.

Worked example

The variance of

10

observations is

7

. Each observation is multiplied by

4

and then

3

is subtracted. Find the new variance and the new SD.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Measures of DispersionEASY

The variance of 20 observations is 5. If each observation is multiplied by 3 and then 8 is added to each number, then variance of resulting observations is

[Q120 · 9th May Shift 2 · 2023]

Multiplying the variance by λ, not λ²

3 \times 12 = 36

is option (B). The variance carries a square, so it scales by

9

108

Drill 4 more on multiplying by λ: mean × λ, sd × |λ|, variance × λ²

Concept 3 of 4

y = px − q With Target Mean and SD: Two Equations, Watch the Sign of p

Intuition

A combined transformation gives one equation from the mean (

p\bar x - q

) and one from the SD (

|p|\sigma

). The SD equation fixes

|p|

, leaving two candidates; the mean equation, together with any condition like

q \ne 0

, picks the sign.

Definition

$\bar x = 20$ , $\sigma = 2$ ; $y = px - q$ with new mean $10$ and new SD $1$ : $|p| \cdot 2 = 1 \Rightarrow p = \pm\frac12$ ; $20p - q = 10$ . With $p = \frac12$ : $q = 0$ , excluded by $q \ne 0$ ; with $p = -\frac12$ : $-10 - q = 10$ , $q = -20$ .
Order of operations matters for the mean only: $px - q$ and $p(x - q)$ have the same SD but different means.
Always solve the SD equation first; it is the one with two solutions.

Combined rule

y = px - q:\quad \bar y = p\bar x - q,\qquad \sigma_y = |p|\,\sigma_x

Worked example

A data set has mean

30

and SD

4

. Each value is transformed to

y = px + q

so that the new mean is

15

and the new SD is

2

, with

q \ne 0

. Find

p

and

q

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Measures of DispersionMODERATE

The mean and the standard deviation of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by

p

and then reduced by

q

, where

p\neq0

and

q\neq0

. If the new mean and new standard deviation (s.d.) become half of the original values, then

q

is equal to

[Q134 · 9th May Shift 1 · 2023]

Taking p positive by default

p = \frac12

forces

q = 0

, which the stem forbids. The SD fixes only

|p|

; the remaining condition chooses the sign, and here it is negative.

Concept 4 of 4

Mean of (x − k)² and the b = a + c Identity: Expand, Then Use σ² + x̄²

Intuition

\dfrac{\sum (x - k)^2}{n} = \dfrac{\sum x^2}{n} - 2k\bar x + k^2

, and

\dfrac{\sum x^2}{n} = \sigma^2 + \bar x^2

. Any 'mean of squares of shifted values' collapses to

\sigma^2 + (\bar x - k)^2

Definition

$\bar x = 16$ , $\sigma^2 = 256$ , mean of $(x_i - 5)^2$ : $\sigma^2 + (\bar x - 5)^2 = 256 + 121 = 377$ . Same by expansion: $512 - 160 + 25$ .
$b = a + c$ identity: the SD of $a + 2, b + 2, c + 2$ is $d$ , so $d$ is the SD of $a, b, c$ ; mean $= \dfrac{2b}{3}$ ; $d^2 = \dfrac{a^2 + b^2 + c^2}{3} - \dfrac{4b^2}{9}$ , so $9d^2 = 3a^2 + 3c^2 - b^2$ , i.e. $b^2 = 3(a^2 + c^2) - 9d^2$ .
The three-observation identity is the same expansion with symbols; the shift rule is what lets the $+2$ be ignored.
$\sigma^2 + (\bar x - k)^2$ is minimised at $k = \bar x$ — the mean is the value about which the mean squared deviation is least.

Shifted squares

\frac{\sum (x_i - k)^2}{n} = \sigma^2 + (\bar x - k)^2

Worked example

The mean and variance of

20

observations are

10

and

4

. Find the mean of

(x_i - 7)^2

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Measures of DispersionHARD

If both mean and variance of 50 observations

x_1, x_2, \ldots, x_{50}

are equal to 16 and 256 respectively, then mean of

(x_1-5)^2, (x_2-5)^2, \ldots, (x_{50}-5)^2

[Q149 · 14th May Shift 2 · 2024]

Using Σx²/n = σ² in the expansion

The mean of the squares is

\sigma^2 + \bar x^2 = 512

, not

256

. Dropping

\bar x^2

gives

121

, which is not offered — the offered distractors come from arithmetic slips in

512 - 160 + 25

Drill 2 more on mean of (x − k)² and the b = a + c identity: expand, then use σ² + x̄²

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Adding a Constant: Mean Shifts, Variance Stays
Shift rule
$\overline{x + c} = \bar x + c,\qquad \operatorname{Var}(x + c) = \operatorname{Var}(x)$
Multiplying by λ: Mean × λ, SD × |λ|, Variance × λ²
Scale rule
$\overline{\lambda x} = \lambda\bar x,\qquad \sigma_{\lambda x} = |\lambda|\,\sigma_x,\qquad \operatorname{Var}(\lambda x) = \lambda^2\operatorname{Var}(x)$
y = px − q With Target Mean and SD: Two Equations, Watch the Sign of p
Combined rule
$y = px - q:\quad \bar y = p\bar x - q,\qquad \sigma_y = |p|\,\sigma_x$
Mean of (x − k)² and the b = a + c Identity: Expand, Then Use σ² + x̄²
Shifted squares
$\frac{\sum (x_i - k)^2}{n} = \sigma^2 + (\bar x - k)^2$

Watch out for (4)

Adding the constant to the variance
→ Adding a Constant: Mean Shifts, Variance Stays
Multiplying the variance by λ, not λ²
→ Multiplying by λ: Mean × λ, SD × |λ|, Variance × λ²
Taking p positive by default
→ y = px − q With Target Mean and SD: Two Equations, Watch the Sign of p
Using Σx²/n = σ² in the expansion
→ Mean of (x − k)² and the b = a + c Identity: Expand, Then Use σ² + x̄²

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