Statistics

2.1 Core concepts

• Measures of central tendency (mean, median, mode) only locate the centre of data; they say nothing about how scattered the data are around that centre — hence the need for a single number called a measure of dispersion (NCERT §13.1, p. 258). Two distributions can share mean and median yet have very different spreads.
• The four measures of dispersion are Range, Quartile deviation, Mean deviation, and Standard deviation; Quartile deviation is excluded from study (NCERT §13.2, p. 259). Range is the simplest; SD is the most informative.
• Range of a series = Maximum value − Minimum value; it gives a rough idea of scatter but does not measure dispersion about a central tendency (NCERT §13.3, p. 259). Range is unstable: a single outlier can inflate it dramatically.
• The deviation of an observation x from a fixed value a is x − a; the sum of deviations from the mean is always zero, so the mean of signed deviations is useless as a dispersion measure (NCERT §13.4, p. 259). This is a key consequence of the definition x̄ = (Σxi)/n.
• To overcome cancellation of signs, we take absolute deviations; the mean of absolute deviations from a central value a is the mean deviation M.D.(a) = Σ|xi − a| / n (NCERT §13.4, p. 260). Mean deviation is intuitive but mathematically awkward (the absolute-value function is not smooth).
• For ungrouped data: M.D.(x̄) = (1/n) Σ|xi − x̄|; M.D.(M) = (1/n) Σ|xi − M|, where M is the median (NCERT §13.4.1, p. 260). M.D. about median is minimum: among all choices of central value a, M.D.(a) is least when a is the median.
• For a discrete frequency distribution with N = Σfi: M.D.(x̄) = (1/N) Σfi|xi − x̄|, and median is found by locating the observation whose cumulative frequency is equal to or just greater than N/2 (NCERT §13.4.2, pp. 262–264).
• For a continuous frequency distribution, mid-points xi of each class are used; the median class is the class whose c.f. is just greater than or equal to N/2, and Median = l + ((N/2 − C)/f) × h, where l, f, h are the lower limit, frequency and width of the median class and C is the c.f. of the preceding class (NCERT §13.4.2, p. 268).
• A step-deviation short-cut for mean is di = (xi − a)/h, giving x̄ = a + h·(Σfidi)/N (NCERT §13.4.2, p. 267). Choosing a near the centre of the data minimises arithmetic.
• Limitations of mean deviation — it uses absolute values so cannot be subjected to further algebraic treatment, and M.D. about median is unreliable when variability is high (NCERT §13.4.3, p. 271). This motivates moving to squared deviations.
• To avoid the absolute-value problem, square the deviations instead. The simple sum Σ(xi − x̄)² is not a good measure (it grows with n), so we take its mean: variance σ² = (1/n) Σ(xi − x̄)² for ungrouped data (NCERT §13.5, pp. 271–274).
• Because variance has the square of the observations, the proper dispersion measure is the positive square root, standard deviation σ = √[(1/n) Σ(xi − x̄)²] (NCERT §13.5.1, p. 274). SD is in the same units as the data, making it directly interpretable.
• For a discrete frequency distribution: σ = √[(1/N) Σfi(xi − x̄)²] (NCERT §13.5.2, p. 275).
• For a continuous frequency distribution (replacing each class by its mid-point): σ = √[(1/N) Σfi(xi − x̄)²], and an algebraically equivalent computational form is σ = (1/N) √[N Σfixi² − (Σfixi)²] (NCERT §13.5.3, pp. 276–277). The alternative form avoids computing x̄ first.
• Short-cut (step-deviation) method: with yi = (xi − A)/h, x̄ = A + h·ȳ and σx = h·σy, giving σ² = (h²/N²)[N Σfiyi² − (Σfiyi)²] (NCERT §13.5.4, pp. 279–281). The factor h² appears because variance is squared in unit; SD scales by |h|.
• Key results from Miscellaneous Examples — (a) multiplying each observation by k multiplies the variance by k² (so SD by |k|) (Example 13, p. 282); (b) adding a constant a to every observation leaves the variance unchanged (Example 15, p. 284); (c) given mean and SD with a wrong observation, the correct mean and SD are found by adjusting Σxi and Σxi² (Example 16, pp. 284–285). All three results recur on CUET papers.
• The coefficient of variation CV = (σ/x̄)·100 compares dispersions of data sets with different units or magnitudes; lower CV means greater consistency. CV is dimensionless, allowing comparison across units (e.g., height in cm vs weight in kg).
• Variance divides by n (not n − 1) — this is the "population" variance, appropriate when the data set is treated as exhaustive. The unbiased estimator using n − 1 is part of higher statistics, not Class XI.
• Worked tip for the alternative SD formula: build a small table of xi², multiply by fi, sum column-wise; one pass yields both Σfixi and Σfixi². This is more efficient than computing each (xi − x̄)² individually.
• Skewness comparison via Misc. Example 14 (p. 283): when one data set has a larger SD than another, the first is more dispersed; this is the basis of "consistency comparison" in cricket-batsman / quality-control problems on CUET.
• For comparing two series where the means are equal, the series with smaller SD is more consistent; where the means differ, the coefficient of variation provides a meaningful comparison.
• Practical computational warning: in step-deviation, choose A as one of the xi (often the middle observation) and h as the common difference of class widths so that the yi become small integers like −3, −2, …, 3. This dramatically simplifies arithmetic and reduces transcription errors.
• Moving from range → MD → SD trades more computation for better statistical behaviour (algebraic tractability, sensitivity to all observations, units consistency, invariance under translation). This is why higher statistics — regression, hypothesis testing, ANOVA — is built on variance and SD, not on range or mean deviation.

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig 13.1 & 13.2 (p. 258): dot plots of batsman A's and batsman B's scores on a number line — visually shows that despite identical mean and median, A's data are scattered while B's cluster near the centre. Underscores why dispersion is needed.
• Fig 13.3 (p. 267): number line showing shifting of origin to assumed mean A in the step-deviation method.
• Fig 13.4 (p. 267): number line showing change of scale (division by common factor h) — together explain why di = (xi − A)/h works.
• Fig 13.5 & 13.6 (p. 273): dot plots for set A (6 observations: 5, 15, …, 55) and set B (31 observations from 15 to 45) — geometrical proof that Σ(xi − x̄)² alone is misleading and one must take its mean.
• Tables 13.1 – 13.11 (pp. 263–280): standardised tabular layouts for computing MD and variance — students should recall the column structure (xi, fi, fixi, |xi − x̄|, fi|xi − x̄|, (xi − x̄)², fi(xi − x̄)², and for short-cut: yi, yi², fiyi, fiyi²).
• Process — mean deviation about mean (ungrouped): (i) compute x̄ = (Σxi)/n; (ii) compute each |xi − x̄|; (iii) sum and divide by n.
• Process — mean deviation about median (ungrouped): (i) arrange observations in ascending order; (ii) identify median (middle term for odd n; mean of two middle terms for even n); (iii) compute |xi − M| for each; (iv) sum and divide by n.
• Process — variance and SD (ungrouped): (i) compute x̄; (ii) compute each (xi − x̄)²; (iii) sum and divide by n for σ²; (iv) take positive square root for σ.
• Process — alternative computational SD: (i) compute Σfixi and Σfixi² directly; (ii) apply σ = (1/N) √[N Σfixi² − (Σfixi)²].
• Process — step-deviation: (i) choose convenient A and h; (ii) compute yi = (xi − A)/h; (iii) compute Σfiyi and Σfiyi²; (iv) ȳ = Σfiyi/N; (v) x̄ = A + h·ȳ; (vi) σ = (h/N) √[N Σfiyi² − (Σfiyi)²].

2.4 Common confusions / NTA trap points

• Forgetting the absolute value in mean deviation — Σ(xi − x̄) is always zero, so a student who skips the modulus will write the answer as 0 (NCERT §13.4, p. 259). NTA frequently sets a distractor of 0 for this reason.
• Confusing variance and standard deviation — variance is σ² (units squared); SD is the positive square root σ. Many students stop at variance and report it as SD.
• Mixing the divisor in grouped data — for mean deviation/variance of a frequency distribution the divisor is N = Σfi, not n (the number of distinct values).
• Applying the wrong formula for median class — the c.f. just greater than or equal to N/2 (not the c.f. just less than) identifies the median class, then Median = l + ((N/2 − C)/f) × h.
• Effect of transformations — adding a constant does not change variance/SD, but multiplying every observation by k multiplies variance by k² (and SD by |k|). NTA mixes "added" and "multiplied" deliberately to trip students (Examples 13 & 15, pp. 282, 284).
• Step-deviation method is only a computational short-cut for mean and SD; the answer must come out the same as by the direct formula (NCERT §13.5.4, pp. 279–281).
• Treating range as a dispersion measure about the mean — range ignores the mean entirely.
• Reporting SD with a negative sign — SD is non-negative by definition.
• Forgetting that the mean shifts by the same constant when observations shift; failing to update x̄ leads to wrong (xi − x̄).
• Mis-identifying the median class as "the one containing N/2" rather than "the one whose c.f. first reaches or exceeds N/2".
• Forgetting to multiply the deviation factor h when converting step-deviation SD back to original units (σx = h·σy, not σy).
• Treating mean deviation as exchangeable with SD in numerical problems; they differ unless data are symmetric.

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT Example 1, p. 261). M.D. about mean of 6, 7, 10, 12, 13, 4, 8, 12.

Step 1 — mean: Σ = 72; n = 8; x̄ = 9. Step 2 — |deviations|: 3, 2, 1, 3, 4, 5, 1, 3; sum = 22. Step 3 — divide: M.D. = 22/8 = 2.75.

Example B (NCERT Example 3, p. 262). M.D. about median of 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.

Step 1 — sort: 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21; median = 6th term = 9. Step 2 — |xi − 9|: 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12; sum = 58. Step 3 — divide: M.D. = 58/11 ≈ 5.27.

Example C (NCERT Example 8, p. 275). Variance of 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.

Step 1 — mean: Σ = 150; n = 10; x̄ = 15. Step 2 — Σ(xi − x̄)²: 81 + 49 + 25 + 9 + 1 + 1 + 9 + 25 + 49 + 81 = 330. Step 3 — variance: 330/10 = 33 (SD = √33 ≈ 5.74).

Example D (NCERT Misc. Example 16, p. 285). Mean 40, SD 5.1 for n = 100. One value 50 was wrongly used instead of 40. Find corrected mean and SD.

Step 1 — correct Σxi: 100 · 40 = 4000; corrected = 4000 − 50 + 40 = 3990; mean = 39.9. Step 2 — original Σxi²: var = 5.1² = 26.01 ⇒ Σxi²/100 − 40² = 26.01 ⇒ Σxi² = 162601; corrected = 162601 − 50² + 40² = 161701. Step 3 — new variance and SD: var = 161701/100 − 39.9² = 1617.01 − 1592.01 = 25 ⇒ SD = 5. Corrected: mean 39.9, SD 5.

Example E (NCERT Misc. Example 15, p. 284). Show that adding constant a to every observation does not change variance.

Step 1 — set yi = xi + a: ȳ = x̄ + a. Step 2 — compute deviations: yi − ȳ = (xi + a) − (x̄ + a) = xi − x̄. Step 3 — variance unchanged: (1/n) Σ(yi − ȳ)² = (1/n) Σ(xi − x̄)² = var(x). QED.