MHT-CET Maths · Measures of Dispersion

Mean and Variance From Sums — Σx, Σx² and Deviations From an Assumed Mean

Variance is the mean of the squares minus the square of the mean: σ² = Σx²/n − x̄². Given any two of Σx, Σx², n, mean and variance, the rest follow — and deviations from an assumed mean plug into the same formula.

All Measures of Dispersion notes MHT-CET Maths strategy

Why this matters

9 PYQs, none HARD — the formula page. The stems hand you Σx and Σx² (or Σ(x − a) and Σ(x − a)²) and ask for the SD, or hand you the mean and SD and ask for Σx²; one replaces a wrongly recorded observation, one adds three observations without moving the mean, two are frequency tables. All of it is the single identity below, rearranged.

Concept 1 of 4

σ² = Σx²/n − x̄²: Recover Any One Quantity From the Others

Intuition

Variance is the average squared distance from the mean, and expanding

(x - \bar x)^2

turns that into the mean of the squares minus the square of the mean. Given the mean and SD,

\sum x^2 = n(\sigma^2 + \bar x^2)

; given

\sum x

and

\sum x^2

, the SD follows.

Definition

$\sigma^2 = \dfrac{\sum x_i^2}{n} - \bar x^2$ , $\bar x = \dfrac{\sum x_i}{n}$ , SD $= \sigma = \sqrt{\sigma^2}$ .
$n = 100$ , $\bar x = 50$ , $\sigma = 5$ : $\sum x^2 = 100(25 + 2500) = 252{,}500$ .
Variance zero means every observation equals the mean: $\bar x = 5$ , $\sigma^2 = 0$ , $\sum x^2 = 400 \Rightarrow 25n = 400 \Rightarrow n = 16$ .
Six primes $7, 11, 13, 17, 19, 23$ : $\bar x = 15$ , $\sum x^2 = 1518$ , $\sigma^2 = 253 - 225 = 28$ .
CET uses the population variance (divide by $n$ ), never $n - 1$ .

The identity

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

Visualization · move the points, watch the squared deviations

x1: 2.0x2: 4.0x3: 6.0x4: 8.0x5: 10.0

Variance σ² = 8.00SD σ = 2.83Σ(xᵢ − x̄)² = 40.00

Each red square has side |xᵢ − x̄|, so its AREA is the squared deviation. Variance is the AVERAGE area. Pull all points toward the mean — every square shrinks. Pull them apart — they grow.

Worked example

The mean of

40

observations is

12

and their SD is

3

. Find the sum of their squares.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Measures of DispersionEASY

The mean of 100 observations is 50 and their standard deviation is 5, then the sum of all squares of all the observations is

[Q145 · 2nd May Shift 1 · 2023]

Forgetting to add the mean squared

\sum x^2 = n\sigma^2

gives

2500

, not

252{,}500

. The mean of the squares is the variance PLUS the square of the mean.

Drill 1 more on σ² = σx²/n − x̄²: recover any one quantity from the others

Concept 2 of 4

Deviations From an Assumed Mean: Σ(x − a) and Σ(x − a)²

Intuition

Subtracting a constant

a

from every observation shifts the mean by

a

and leaves the variance alone. So with

d_i = x_i - a

\bar x = a + \dfrac{\sum d}{n}

and

\sigma^2 = \dfrac{\sum d^2}{n} - \left(\dfrac{\sum d}{n}\right)^2

— the same identity, applied to the deviations.

Definition

$n = 20$ , $\sum(x - 2) = 20$ , $\sum(x - 2)^2 = 100$ : $\sigma^2 = \dfrac{100}{20} - \left(\dfrac{20}{20}\right)^2 = 5 - 1 = 4$ , SD $2$ . (The mean is $2 + 1 = 3$ , not needed.)
$50$ observations with $\sum(x - 30) = 50$ : $\bar x = 30 + \dfrac{50}{50} = 31$ .
The correction term $\left(\dfrac{\sum d}{n}\right)^2$ vanishes only when $a$ IS the mean; otherwise it must be subtracted.
Choosing $a$ near the centre keeps the deviations small — the point of the method in hand calculation.

Assumed-mean formulas

\bar x = a + \frac{\sum d_i}{n},\qquad \sigma^2 = \frac{\sum d_i^2}{n} - \left(\frac{\sum d_i}{n}\right)^2,\qquad d_i = x_i - a

Worked example

For

10

observations,

\sum(x - 5) = 10

and

\sum(x - 5)^2 = 50

. Find the mean and the variance.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Measures of DispersionMODERATE

For 20 observations of variable

x

, if

\sum(x_i-2)=20

and

\sum(x_i-2)^2=100

, then the standard deviation of variable

x

[Q149 · 12th May Shift 2 · 2024]

Reading Σ(x − 2)²/n as the variance

\dfrac{100}{20} = 5

is the mean of the squared deviations from

2

, not from the mean. Subtract

\left(\dfrac{\sum d}{n}\right)^2 = 1

; the SD is

2

, not

\sqrt5

Drill 2 more on deviations from an assumed mean: σ(x − a) and σ(x − a)²

Concept 3 of 4

A Wrong Observation Replaced, or Observations Added: Fix the Sums First

Intuition

Everything runs through

\sum x

and

\sum x^2

. Replacing a value changes both sums by the difference of the old and new values (and of their squares); adding observations changes

n

and the sums, and 'the mean stays the same' is an equation in the unknown.

Definition

$n = 15$ , $\sum x = 170$ , $\sum x^2 = 2830$ ; $20$ replaced by $30$ : $\sum x = 180$ , $\sum x^2 = 2830 - 400 + 900 = 3330$ ; $\sigma^2 = 222 - 144 = 78$ .
Mean $\bar x$ of $n$ values; add $n + 1$ , $n - 1$ , $2n - 1$ (sum $4n - 1$ ) with the mean unchanged: $\dfrac{n\bar x + 4n - 1}{n + 3} = \bar x \Rightarrow 4n - 1 = 3\bar x \Rightarrow n = \dfrac{3\bar x + 1}{4}$ .
A wrongly recorded value that is REMOVED (not replaced) also reduces $n$ by one.
Recompute the mean from the corrected sum before the variance; using the old mean with the new sum of squares is the standard slip.

Correcting sums

\sum x_{\text{new}} = \sum x - x_{\text{wrong}} + x_{\text{right}},\qquad \sum x^2_{\text{new}} = \sum x^2 - x_{\text{wrong}}^2 + x_{\text{right}}^2

Worked example

The mean of

10

observations is

20

and their variance is

9

. One observation, recorded as

15

, should have been

25

. Find the corrected variance.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Measures of DispersionMODERATE

In an experiment with 15 observations for x, the following results were available

\sum x^2 = 2830

\sum x = 170

. One observation 20 was found to be wrong and was replaced by the correct value 30. Then the corrected variance is

[Q128 · 11th May Shift 1 · 2023]

Keeping the old mean after correcting the sum

\dfrac{3330}{15} - \left(\dfrac{170}{15}\right)^2

is wrong; the corrected mean is

12

, giving

222 - 144 = 78

Drill 1 more on a wrong observation replaced, or observations added: fix the sums first

Concept 4 of 4

Grouped Data: Midpoints, Σfx and Σfx²

Intuition

With frequencies, every sum is weighted:

\bar x = \dfrac{\sum f x}{N}

\sigma^2 = \dfrac{\sum f x^2}{N} - \bar x^2

, where

N = \sum f

and a class interval is represented by its midpoint.

Definition

Classes $0$ – $6$ , $6$ – $12$ , $12$ – $18$ with $f = 2, 4, 6$ : midpoints $3, 9, 15$ ; $N = 12$ ; $\sum fx = 6 + 36 + 90 = 132$ , $\bar x = 11$ ; $\sum fx^2 = 18 + 324 + 1350 = 1692$ ; $\sigma^2 = 141 - 121 = 20$ , SD $2\sqrt5$ .
Frequencies in a parameter: marks $2, 3, 5, 7$ with $f = (x+1)^2, 2x - 5, x^2 - 3x, x$ and $N = 20$ : $2x^2 + 2x - 4 = 20 \Rightarrow x = 3$ ; frequencies $16, 1, 0, 3$ ; $\sum fx = 32 + 3 + 0 + 21 = 56$ ; mean $2.8$ .
A frequency must be a whole number $\ge 0$ ; that is what rejects the negative root of the quadratic in $x$ .
$\sqrt{20} = 2\sqrt5$ : options list both forms, and both are the same answer.

Weighted sums

\bar x = \frac{\sum f_i x_i}{N},\qquad \sigma^2 = \frac{\sum f_i x_i^2}{N} - \bar x^2,\qquad N = \sum f_i

Visualization · change the bin width, watch the shape change

Bin width:

Same 30 data points each time — only the bin width changes. At width 50 the shape looks almost uniform; at width 5 it looks jagged. Choosing bin width is part of the analysis, not the data.

Worked example

Find the SD of the distribution: values

1, 2, 3

with frequencies

1, 2, 1

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Measures of DispersionMODERATE

The standard deviation of the following distribution: C.I.: 0–6, 6–12, 12–18;

f_i

: 2, 4, 6

[Q149 · 9th May Shift 1 · 2024]

Using class limits instead of midpoints

The class

6

–

12

contributes at

x = 9

. Using

6

12

shifts every sum and lands on a distractor.

Drill 1 more on grouped data: midpoints, σfx and σfx²

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)

σ² = Σx²/n − x̄²: Recover Any One Quantity From the Others
The identity
$\sigma^2 = \frac{\sum x_i^2}{n} - \bar x^2 \qquad \sum x_i^2 = n\left(\sigma^2 + \bar x^2\right)$
Deviations From an Assumed Mean: Σ(x − a) and Σ(x − a)²
Assumed-mean formulas
$\bar x = a + \frac{\sum d_i}{n},\qquad \sigma^2 = \frac{\sum d_i^2}{n} - \left(\frac{\sum d_i}{n}\right)^2,\qquad d_i = x_i - a$
A Wrong Observation Replaced, or Observations Added: Fix the Sums First
Correcting sums
$\sum x_{\text{new}} = \sum x - x_{\text{wrong}} + x_{\text{right}},\qquad \sum x^2_{\text{new}} = \sum x^2 - x_{\text{wrong}}^2 + x_{\text{right}}^2$
Grouped Data: Midpoints, Σfx and Σfx²
Weighted sums
$\bar x = \frac{\sum f_i x_i}{N},\qquad \sigma^2 = \frac{\sum f_i x_i^2}{N} - \bar x^2,\qquad N = \sum f_i$

Watch out for (4)

Forgetting to add the mean squared
→ σ² = Σx²/n − x̄²: Recover Any One Quantity From the Others
Reading Σ(x − 2)²/n as the variance
→ Deviations From an Assumed Mean: Σ(x − a) and Σ(x − a)²
Keeping the old mean after correcting the sum
→ A Wrong Observation Replaced, or Observations Added: Fix the Sums First
Using class limits instead of midpoints
→ Grouped Data: Midpoints, Σfx and Σfx²

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