MHT-CET Maths · Measures of Dispersion

Standard Series and Missing Observations — First n Naturals, Evens, Primes and Two Unknowns

Variance of the first n natural numbers is (n² − 1)/12, and scaling gives the evens; and when two observations are missing, the mean gives x + y and the variance gives x² + y², from which xy and |x − y| follow.

All Measures of Dispersion notes MHT-CET Maths strategy

Why this matters

12 PYQs, none HARD — the chapter's largest page and its most repetitive: 'mean 8, variance 16, five of seven observations are 2, 4, 10, 12, 14' has been set THREE times (asking for the product, its square root, or the difference of the missing pair), and the first-n-naturals variance three times. Two closed results and one algebraic move cover the page.

Concept 1 of 3: Variance of the First n Natural Numbers, and of the Evens by Scaling

\dfrac{\sum k^2}{n} - \left(\dfrac{n+1}{2}\right)^2

with

\sum k^2 = \dfrac{n(n+1)(2n+1)}{6}

simplifies to

\dfrac{n^2 - 1}{12}

. The first

n

even numbers are twice the naturals, so their variance is

4

times that.

Definition

$\operatorname{Var}(1, 2, \dots, n) = \dfrac{(n+1)(2n+1)}{6} - \dfrac{(n+1)^2}{4} = \dfrac{n^2 - 1}{12}$ .
First $2n$ naturals: $\dfrac{4n^2 - 1}{12}$ . First $50$ evens: $4 \times \dfrac{50^2 - 1}{12} = 4 \times \dfrac{2499}{12} = 833$ .
Check by the sums: first $50$ evens have mean $51$ , $\sum x^2 = 4 \cdot \dfrac{50 \cdot 51 \cdot 101}{6} = 171{,}700$ , $\dfrac{171700}{50} - 2601 = 833$ .
Non-standard sets (six primes above $5$ ) are computed directly: $7, 11, 13, 17, 19, 23$ → mean $15$ , $\sum x^2 = 1518$ , variance $28$ .
Option lists always carry $\dfrac{(n+1)(n+5)}{12}$ -style near-misses; the true numerator is $n^2 - 1 = (n-1)(n+1)$ .

Standard results

\operatorname{Var}(1..n) = \frac{n^2 - 1}{12},\qquad \operatorname{Var}(2, 4, \dots, 2n) = \frac{n^2 - 1}{3},\qquad \operatorname{Var}(1..2n) = \frac{4n^2 - 1}{12}

Worked example

Find the variance of the first

10

natural numbers, and of the first

10

even natural numbers.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q111Moderate

Example 1 · Measures of Dispersion · Standard Series and Missing Observations — First n Naturals, Evens, Primes and Two Unknowns

The variance of first

n

natural numbers is

Doubling instead of quadrupling for the evens

Evens are

2 \times

naturals, so the variance is

4 \times

, not

2 \times

\dfrac{n^2 - 1}{6}

is the doubled distractor.

Drill 4 more on variance of the first n natural numbers, and of the evens by scaling

Concept 2 of 3: Two Missing Observations: x + y From the Mean, x² + y² From the Variance

The mean fixes the sum of the two unknowns; the variance fixes the sum of their squares. Then

2xy = (x + y)^2 - (x^2 + y^2)

and

(x - y)^2 = (x^2 + y^2) - 2xy

give whatever the stem asks — product, root of the product, or difference.

Definition

Seven observations, mean $8$ , variance $16$ , five known $2, 4, 10, 12, 14$ : $x + y = 56 - 42 = 14$ ; $\sum x^2 = 7(16 + 64) = 560$ , known $460$ , so $x^2 + y^2 = 100$ .
Then $xy = \dfrac{196 - 100}{2} = 48$ ; $\sqrt{xy} = 4\sqrt3$ ; $(x - y)^2 = 100 - 96 = 4$ , $|x - y| = 2$ ; the pair is $\{6, 8\}$ .
$a, b, 8, 5, 10$ with mean $6$ , variance $6.8$ : $a + b = 7$ , $a^2 + b^2 = 25$ , $ab = 12$ : $\{3, 4\}$ .
$3, 5, 7, a, b$ with mean $5$ , SD $2$ : $a + b = 10$ , $a^2 + b^2 = 62$ , $ab = 19$ : $a, b$ are the roots of $t^2 - 10t + 19 = 0$ .
The unknowns are the roots of $t^2 - (x + y)t + xy = 0$ ; write that quadratic when the stem asks for 'the equation whose roots are…'.

Two unknowns

x + y = S,\ x^2 + y^2 = Q \ \Rightarrow\ xy = \frac{S^2 - Q}{2},\quad (x - y)^2 = 2Q - S^2

Worked example

Six observations have mean

5

and variance

\dfrac{10}{3}

. Four of them are

2, 4, 6, 8

. Find the product of the other two.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q118Moderate

Example 2 · Measures of Dispersion · Standard Series and Missing Observations — First n Naturals, Evens, Primes and Two Unknowns

The mean and variance of seven observations are 8 and 16 respectively. If five of the observations are 2, 4, 10, 12, 14, then the product of remaining two observations is

Dividing by n − 1

\sum x^2 = 7(16 + 64)

uses the population variance. Sample variance gives

x^2 + y^2 = 96 + \dots

, no integer pair, and a wrong product.

Drill 4 more on two missing observations: x + y from the mean, x² + y² from the variance

Concept 3 of 3: One Unknown Value: Write the Variance in Terms of It and Solve

With one unknown

k

in the data, the mean and the sum of squares are both expressions in

k

; the variance equation is then a quadratic in

k

(or in

k^2

), and a sign condition picks the root.

Definition

$-1, 0, 1, k$ with variance $5$ , $k > 0$ : $\bar x = \dfrac{k}{4}$ , $\dfrac{2 + k^2}{4} - \dfrac{k^2}{16} = 5 \Rightarrow 8 + 4k^2 - k^2 = 80 \Rightarrow k^2 = 24 \Rightarrow k = 2\sqrt6$ .
A missing score fixed by the mean first: five tests $54, 45, 41, 43, 57$ and a mean of $48$ over six: sixth $= 288 - 240 = 48$ . Then deviations from $48$ : $6, -3, -7, -5, 9, 0$ ; $\sigma^2 = \dfrac{36 + 9 + 49 + 25 + 81}{6} = \dfrac{100}{3}$ , SD $\dfrac{10}{\sqrt3}$ .
When the mean is a clean number, work with deviations $x_i - \bar x$ directly — smaller squares, no correction term.
Multiply through by $16$ (or the denominators present) before collecting; the coefficient of $k^2$ comes out as $3$ , from $4 - 1$ .

One unknown

\frac{\sum x_i^2(k)}{n} - \bar x(k)^2 = \sigma^2 \ \Rightarrow\ \text{solve for } k

Worked example

The variance of

0, 2, 4, k

5

and

k > 4

. Find

k

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 1 · Q118Moderate

Example 3 · Measures of Dispersion · Standard Series and Missing Observations — First n Naturals, Evens, Primes and Two Unknowns

If the variance of the numbers

-1,0,1,k

is 5, where

k>0

, then

k

is equal to

Forgetting the mean also contains k

\dfrac{2 + k^2}{4} = 5

ignores

\bar x = \dfrac{k}{4}

and gives

k = \sqrt{18}

. The mean moves with the unknown; subtract its square.

Drill 1 more on one unknown value: write the variance in terms of it and solve

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 (3)

Variance of the First n Natural Numbers, and of the Evens by Scaling
Standard results
$\operatorname{Var}(1..n) = \frac{n^2 - 1}{12},\qquad \operatorname{Var}(2, 4, \dots, 2n) = \frac{n^2 - 1}{3},\qquad \operatorname{Var}(1..2n) = \frac{4n^2 - 1}{12}$
Two Missing Observations: x + y From the Mean, x² + y² From the Variance
Two unknowns
$x + y = S,\ x^2 + y^2 = Q \ \Rightarrow\ xy = \frac{S^2 - Q}{2},\quad (x - y)^2 = 2Q - S^2$
One Unknown Value: Write the Variance in Terms of It and Solve
One unknown
$\frac{\sum x_i^2(k)}{n} - \bar x(k)^2 = \sigma^2 \ \Rightarrow\ \text{solve for } k$

Watch out for (3)

Doubling instead of quadrupling for the evens
→ Variance of the First n Natural Numbers, and of the Evens by Scaling
Dividing by n − 1
→ Two Missing Observations: x + y From the Mean, x² + y² From the Variance
Forgetting the mean also contains k
→ One Unknown Value: Write the Variance in Terms of It and Solve

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Related notes

Mean and Variance From Sums — the identity behind x² + y² →