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MHT-CET Maths · Formula sheet

Measures of Dispersion formulas

11 formulas and 11 common traps for MHT-CET Maths Measures of Dispersion, grouped by subtopic.

Full notes with worked examples

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

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

The identity

σ2=∑xi2n−xˉ2∑xi2=n(σ2+xˉ2)\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

xˉ=a+∑din,σ2=∑di2n−(∑din)2,di=xi−a\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

∑xnew=∑x−xwrong+xright,∑xnew2=∑x2−xwrong2+xright2\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

xˉ=∑fixiN,σ2=∑fixi2N−xˉ2,N=∑fi\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

Common traps

Forgetting to add the mean squared

∑x2=nσ2\sum x^2 = n\sigma^2 gives 25002500, not 252,500252{,}500. The mean of the squares is the variance PLUS the square of the mean.

Reading Σ(x − 2)²/n as the variance

10020=5\dfrac{100}{20} = 5 is the mean of the squared deviations from 22, not from the mean. Subtract (∑dn)2=1\left(\dfrac{\sum d}{n}\right)^2 = 1; the SD is 22, not 5\sqrt5.

Keeping the old mean after correcting the sum

333015−(17015)2\dfrac{3330}{15} - \left(\dfrac{170}{15}\right)^2 is wrong; the corrected mean is 1212, giving 222−144=78222 - 144 = 78.

Using class limits instead of midpoints

The class 66–1212 contributes at x=9x = 9. Using 66 or 1212 shifts every sum and lands on a distractor.

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

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Adding a Constant: Mean Shifts, Variance Stays

Shift rule

x+c‾=xˉ+c,Var⁡(x+c)=Var⁡(x)\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

λx‾=λxˉ,σλx=∣λ∣ σx,Var⁡(λx)=λ2Var⁡(x)\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:yˉ=pxˉ−q,σy=∣p∣ σxy = 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

∑(xi−k)2n=σ2+(xˉ−k)2\frac{\sum (x_i - k)^2}{n} = \sigma^2 + (\bar x - k)^2

Common traps

Adding the constant to the variance

New variance 1414 is option (C) on the classic stem. Adding a constant moves nothing but the mean.

Multiplying the variance by λ, not λ²

3×12=363 \times 12 = 36 is option (B). The variance carries a square, so it scales by 99: 108108.

Taking p positive by default

p=12p = \frac12 forces q=0q = 0, which the stem forbids. The SD fixes only ∣p∣|p|; the remaining condition chooses the sign, and here it is negative.

Using Σx²/n = σ² in the expansion

The mean of the squares is σ2+xˉ2=512\sigma^2 + \bar x^2 = 512, not 256256. Dropping xˉ2\bar x^2 gives 121121, which is not offered — the offered distractors come from arithmetic slips in 512−160+25512 - 160 + 25.

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

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Variance of the First n Natural Numbers, and of the Evens by Scaling

Standard results

Var⁡(1..n)=n2−112,Var⁡(2,4,…,2n)=n2−13,Var⁡(1..2n)=4n2−112\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, x2+y2=Q ⇒ xy=S2−Q2,(x−y)2=2Q−S2x + 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

∑xi2(k)n−xˉ(k)2=σ2 ⇒ solve for k\frac{\sum x_i^2(k)}{n} - \bar x(k)^2 = \sigma^2 \ \Rightarrow\ \text{solve for } k

Common traps

Doubling instead of quadrupling for the evens

Evens are 2×2 \times naturals, so the variance is 4×4 \times, not 2×2 \times. n2−16\dfrac{n^2 - 1}{6} is the doubled distractor.

Dividing by n − 1

∑x2=7(16+64)\sum x^2 = 7(16 + 64) uses the population variance. Sample variance gives x2+y2=96+…x^2 + y^2 = 96 + \dots, no integer pair, and a wrong product.

Forgetting the mean also contains k

2+k24=5\dfrac{2 + k^2}{4} = 5 ignores xˉ=k4\bar x = \dfrac{k}{4} and gives k=18k = \sqrt{18}. The mean moves with the unknown; subtract its square.

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