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

8 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

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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q145Easy

Example 1 · Measures of Dispersion · Mean and Variance From Sums — Σx, Σx² and Deviations From an Assumed Mean

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

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

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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 2 · Q149Moderate

Example 2 · Measures of Dispersion · Mean and Variance From Sums — Σx, Σx² and Deviations From an Assumed Mean

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

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 1 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

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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 1 · Q128Moderate

Example 3 · Measures of Dispersion · Mean and Variance From Sums — Σx, Σx² and Deviations From an Assumed Mean

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

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

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 1 · Q149Moderate

Example 4 · Measures of Dispersion · Mean and Variance From Sums — Σx, Σx² and Deviations From an Assumed Mean

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

f_i

: 2, 4, 6

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

Test yourself on Measures of Dispersion

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Related notes

Probability Distribution — the same mean and variance formulas for a random variable →