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25 Quantitative Aptitude (Maths) — Common Notes 🕒 Updated 2026-10-17
Common Notes SSC CGL · CHSL · Police · Bank · Railway Chapter 25

📊 Statistics (Advanced)

Based on the SSC CGL Tier-II Paper-II (JSO / Statistical Investigator Grade-II) syllabus: mean, median and mode of grouped data, partition values, dispersion, moments, skewness and kurtosis, correlation and regression, Bayes' theorem, random variables and distributions, sampling, estimation, tests of significance, ANOVA, time series and index numbers. Basic statistics (mean, median, mode, SD) is in Chapter 23.

📌 How this chapter is arranged: the 12 official units of SSC CGL Tier 2 Paper II (JSO / SI Gr-II), in the same order — each unit has notes, 📘 formulas, ✏️ worked examples and ⚠️ exam traps, with unit-wise Practice MCQs at the end. Note: JSO (Paper II) PYQs are not yet available on SF Prep — every question here is a practice question prepared by SF Prep, not a PYQ.

📌 1. Unit 1 · Collection, Classification and Presentation of Data

Primary dataSecondary data
MeaningCollected first-hand by the investigator for the purpose in handAlready collected by someone else, published or unpublished
Examples / sourcesSurveys, interviews, questionnairesCensus reports, RBI bulletins, government publications, journals
Cost and timeMoreLess (but check suitability and reliability)
  • Methods of collecting primary data: direct personal investigation, indirect oral investigation, information from local sources/correspondents, mailed questionnaire (filled by the informant), schedules through enumerators (filled by the enumerator).
  • Census = every unit is studied; sample survey = only some units are studied.
  • Bases of classification: geographical (place), chronological (time), qualitative (attribute), quantitative (class intervals).
  • Class intervals: exclusive (10–20 excludes 20) and inclusive (10–19). Sturges’ rule: number of classes k = 1 + 3.322 log₁₀N.
  • Parts of a table: table number, title, head note, column headings (caption), row headings (stub), body, footnote, source note.
  • Presentation: histogram, frequency polygon, ogive (the ‘less than’ and ‘more than’ ogives meet at the median), pie chart (angle = value/total × 360°).

Frequency distributions and diagrams

  • Discrete distribution: separate values (e.g. number of children in a family); continuous distribution: class intervals (e.g. height 150–155 cm).
  • Cumulative frequency (cf): "less than" (add from the top) and "more than" (add from the bottom). Relative frequency = f/N.
  • A histogram is only for continuous data; the area of each rectangle is proportional to the frequency. With unequal class widths, height = frequency density = f ÷ class width.
  • One-dimensional diagrams: line and bar diagrams; two-dimensional: rectangles, squares, circles (pie). A Lorenz curve shows inequality of income or wealth.
✏️ Worked Example: A family spends ₹24,000 a month, ₹6,000 of it on food. What is the central angle for food in a pie chart?
→ Angle = (value ÷ total) × 360° = (6,000 ÷ 24,000) × 360°
→ = ¼ × 360° = 90°
✏️ Worked Example: Convert the inclusive classes 10–19, 20–29, 30–39 to exclusive classes and give the class mark of the first class.
→ Gap between classes = 20 − 19 = 1; half = 0.5. Subtract 0.5 from each lower limit and add 0.5 to each upper limit.
→ Exclusive classes: 9.5–19.5, 19.5–29.5, 29.5–39.5 (width 10).
→ Class mark = (10 + 19)/2 = (9.5 + 19.5)/2 = 14.5
✏️ Worked Example: By Sturges’ rule, how many classes suit N = 100 observations?
→ k = 1 + 3.322 log₁₀100 = 1 + 3.322 × 2 = 7.644
→ Round up → about 8 classes
⚠️ Exam Traps:
  • A questionnaire is filled by the informant; a schedule is filled by the enumerator.
  • Caption = column headings; stub = row headings.

📌 2. Unit 2 · Measures of Central Tendency and Partition Values

Data are classified into class intervals. Class mark (mid-value) = (lower limit + upper limit)/2. Presentation: histogram (locates mode), ogive (locates median and quartiles), frequency polygon, bar and pie diagrams.

📘 Formulas:
  • Mean (short-cut): x̄ = A + Σfd/N, where d = x − A
  • Step-deviation: x̄ = A + (Σfu/N) × h, where u = (x − A)/h
  • Median = L + [(N/2 − cf)/f] × h
  • Mode = L + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h
  • Q₁ = L + [(N/4 − cf)/f] × h; for Q₃ use 3N/4; for Dₖ use kN/10; for Pₖ use kN/100
  • Q₂ = D₅ = P₅₀ = Median
  • For two numbers: GM² = AM × HM; and AM ≥ GM ≥ HM
📘 Formulas:
  • Weighted mean = Σwx/Σw; geometric mean GM = antilog(Σf log x / N); harmonic mean HM = N/Σ(f/x)
  • Properties: Σ(x − x̄) = 0; Σ(x − A)² is least when A = x̄; Σ|x − A| is least when A = median
  • Empirical relation (moderately skewed data): Mode ≈ 3 Median − 2 Mean
  • Which average when: open-end classes → median; growth rates/ratios → GM; speeds/rates over equal distances → HM
✏️ Worked Example: Classes 0–10, 10–20, 20–30, 30–40, 40–50; frequencies 5, 8, 15, 7, 5 (N = 40). Find the mean by the step-deviation method.
→ Class marks x = 5, 15, 25, 35, 45; A = 25, h = 10 → u = −2, −1, 0, 1, 2
→ fu = −10, −8, 0, 7, 10 → Σfu = −1
→ x̄ = 25 + (−1/40) × 10 = 25 − 0.25 = 24.75
✏️ Worked Example: Find the median and the mode of the same distribution.
→ Cumulative frequencies: 5, 13, 28, 35, 40; N/2 = 20 → median class 20–30 (L = 20, cf = 13, f = 15)
→ Median = 20 + [(20 − 13)/15] × 10 = 20 + 4.67 = 24.67
→ Modal class 20–30: f₁ = 15, f₀ = 8, f₂ = 7 → Mode = 20 + [(15 − 8)/(30 − 8 − 7)] × 10 = 20 + 7/15 × 10 = 24.67
→ Mean ≈ median ≈ mode → the distribution is almost symmetric.
✏️ Worked Example: Find Q₁, Q₃ and D₇ of the same distribution.
→ Q₁: N/4 = 10 → class 10–20 (cf = 5, f = 8): Q₁ = 10 + (10 − 5)/8 × 10 = 16.25
→ Q₃: 3N/4 = 30 → class 30–40 (cf = 28, f = 7): Q₃ = 30 + (30 − 28)/7 × 10 = 32.86
→ D₇: 7N/10 = 28 → class 20–30 (cf = 13, f = 15): D₇ = 20 + (28 − 13)/15 × 10 = 30
✏️ Worked Example: A car covers 60 km at 30 km/h and returns at 60 km/h. What is the average speed?
→ Equal distances at different speeds → harmonic mean
→ HM = 2 × 30 × 60/(30 + 60) = 3600/90 = 40 km/h (the arithmetic mean 45 would be wrong)
⚠️ Exam Traps:
  • L = lower limit of the median/modal class; cf = cumulative frequency of the class BEFORE it (not of that class).
  • In the mode formula f₀ = frequency before the modal class, f₂ = frequency after it.

Partition values: deciles and percentiles

Deciles (D₁…D₉) split the data into 10 equal parts and percentiles (P₁…P₉₉) into 100 equal parts; quartiles (Q₁, Q₂, Q₃) into 4.

📘 Formulas:
  • Individual series (ascending order): Dₖ = k(n + 1)/10-th item; Pₖ = k(n + 1)/100-th item
  • Grouped: Dₖ = L + [(kN/10 − cf)/f] × h; Pₖ = L + [(kN/100 − cf)/f] × h
  • Relations: Q₁ = P₂₅, Q₂ = D₅ = P₅₀ = median, Q₃ = P₇₅, D₁ = P₁₀, D₉ = P₉₀
  • Percentile range = P₉₀ − P₁₀; Kelly’s coefficient of skewness = (P₉₀ + P₁₀ − 2P₅₀)/(P₉₀ − P₁₀)
  • Percentile rank 80 = 80% of the observations lie below that value
⚠️ Exam Traps:
  • In grouped data locate the class with kN/10 (not n + 1).

📌 3. Unit 3 · Measures of Dispersion

Dispersion shows how widely values are spread around the average. Absolute measures (range, QD, MD, SD) carry the unit of the data; relative measures (coefficients, CV) are unit-free and are used to compare two series.

📘 Formulas:
  • Quartile deviation QD = (Q₃ − Q₁)/2; coefficient of QD = (Q₃ − Q₁)/(Q₃ + Q₁)
  • Mean deviation MD = Σ|x − A|/n; MD is least when taken about the median
  • Coefficient of range = (L − S)/(L + S)
  • Coefficient of variation CV = (σ/x̄) × 100%; lower CV = more consistent
  • Combined mean x̄ = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂)
  • Combined variance σ² = [n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)]/(n₁ + n₂), dᵢ = x̄ᵢ − x̄
  • For a normal distribution, roughly QD : MD : SD = 10 : 12 : 15
📘 Formulas:
  • Variance σ² = Σx²/n − x̄² (short form); grouped: σ² = Σfx²/N − x̄²
  • SD of the first n natural numbers = √[(n² − 1)/12]; SD of n terms of an AP with common difference d = d√[(n² − 1)/12]
  • SD of two numbers a, b = |a − b|/2; range ≥ SD
✏️ Worked Example: Find the standard deviation of 2, 4, 4, 4, 5, 5, 7, 9.
→ Mean = 40/8 = 5
→ (x − 5)² = 9, 1, 1, 1, 0, 0, 4, 16 → sum = 32
→ σ² = 32/8 = 4 → σ = 2
✏️ Worked Example: Mean deviation about the median of 3, 5, 8, 10, 14?
→ Median = 8; |x − 8| = 5, 3, 0, 2, 6 → sum 16
→ MD = 16/5 = 3.2; coefficient of MD = 3.2/8 = 0.4
✏️ Worked Example: Batsman A: average 50, SD 10; batsman B: average 40, SD 6. Who is more consistent?
→ CV(A) = 10/50 × 100 = 20%; CV(B) = 6/40 × 100 = 15%
→ Lower CV = more consistent → B (even though A has the higher average)
✏️ Worked Example: Group 1: n = 50, mean 60, SD 8; group 2: n = 50, mean 50, SD 6. Combined SD?
→ Combined mean = (50 × 60 + 50 × 50)/100 = 55; d₁ = 5, d₂ = −5
→ σ² = [50(64 + 25) + 50(36 + 25)]/100 = (4450 + 3050)/100 = 75
→ σ = √75 ≈ 8.66
⚠️ Exam Traps:
  • A change of origin (x + k) leaves SD, QD, MD unchanged; a change of scale (kx) multiplies them by |k|.

📌 4. Unit 4 · Moments, Skewness and Kurtosis

📘 Formulas:
  • Raw moments about A: μ′ᵣ = Σf(x − A)ʳ/N; mean = A + μ′₁
  • μ₂ = μ′₂ − μ′₁² (variance); μ₃ = μ′₃ − 3μ′₂μ′₁ + 2μ′₁³
  • Karl Pearson's coefficient Sₖ = (Mean − Mode)/σ = 3(Mean − Median)/σ
  • Bowley's coefficient = (Q₃ + Q₁ − 2Q₂)/(Q₃ − Q₁), lies between −1 and +1
  • β₁ = μ₃²/μ₂³; β₂ = μ₄/μ₂²
📘 Formulas:
  • μ₄ = μ′₄ − 4μ′₃μ′₁ + 6μ′₂μ′₁² − 3μ′₁⁴
  • About the mean: μ₀ = 1, μ₁ = 0, μ₂ = σ²; central moments do not change with a change of origin
  • γ₁ = √β₁ (sign of μ₃), γ₂ = β₂ − 3; for a symmetric distribution μ₃ = 0 and β₁ = 0
  • Karl Pearson’s coefficient usually lies between −3 and +3; Bowley’s between −1 and +1
Value of β₂Curve
β₂ = 3Mesokurtic — normal curve
β₂ > 3Leptokurtic — more peaked
β₂ < 3Platykurtic — flatter
✏️ Worked Example: The first four moments about A = 4 are 1, 4, 10, 45. Find the mean, variance, β₁ and β₂.
→ Mean = 4 + 1 = 5; μ₂ = 4 − 1² = 3
→ μ₃ = 10 − 3(4)(1) + 2(1)³ = 0 → β₁ = 0 (symmetric)
→ μ₄ = 45 − 4(10)(1) + 6(4)(1) − 3(1) = 26 → β₂ = 26/3² = 2.89 (< 3, slightly platykurtic)
✏️ Worked Example: Q₁ = 20, Q₂ = 30, Q₃ = 50. Bowley’s coefficient of skewness?
→ (Q₃ + Q₁ − 2Q₂)/(Q₃ − Q₁) = (50 + 20 − 60)/(50 − 20) = 10/30
→ = 0.33 → positive skewness
✏️ Worked Example: Mean 45, median 42, SD 9. Karl Pearson’s coefficient?
→ Sₖ = 3(Mean − Median)/σ = 3(45 − 42)/9
→ = 1 (positive skewness)
⚠️ Exam Traps:
  • Positive skew: Mean > Median > Mode (long right tail); reversed for negative skew.

📌 5. Unit 5 · Correlation and Regression

  • Scatter diagram: points rising → positive, falling → negative, no pattern → r ≈ 0.
  • Coefficient of determination r² = the share of the variance of y explained by x; probable error PE = 0.6745(1 − r²)/√n.
  • At r = ±1 the two regression lines coincide; at r = 0 they are perpendicular (x = x̄ and y = ȳ).
📘 Formulas:
  • r = Cov(x, y)/(σₓσᵧ), −1 ≤ r ≤ 1
  • Spearman's rank correlation: R = 1 − 6Σd²/[n(n² − 1)]
  • Regression of y on x: y − ȳ = bᵧₓ(x − x̄), bᵧₓ = r·σᵧ/σₓ; bₓᵧ = r·σₓ/σᵧ
  • r² = bᵧₓ × bₓᵧ; r, bᵧₓ and bₓᵧ always have the same sign
  • Both regression lines intersect at (x̄, ȳ)
  • If u = (x − a)/h, v = (y − b)/k, r is unchanged and bᵧₓ = (k/h)·bᵥᵤ
  • Partial correlation r₁₂.₃ = (r₁₂ − r₁₃r₂₃)/√[(1 − r₁₃²)(1 − r₂₃²)]
  • Association of attributes, Yule's Q = (ad − bc)/(ad + bc)

Multiple regression and partial/multiple correlation (three variables)

📘 Formulas:
  • Regression of X₁ on X₂ and X₃: X₁ = a + b₁₂.₃X₂ + b₁₃.₂X₃ (b₁₂.₃ = partial regression coefficient)
  • Multiple correlation: R²₁.₂₃ = (r₁₂² + r₁₃² − 2r₁₂r₁₃r₂₃)/(1 − r₂₃²); 0 ≤ R₁.₂₃ ≤ 1
  • Partial correlation: r₁₂.₃ = (r₁₂ − r₁₃r₂₃)/√[(1 − r₁₃²)(1 − r₂₃²)]; between −1 and +1
  • R₁.₂₃ is never less than |r₁₂| or |r₁₃|
✏️ Worked Example: x: 1, 2, 3, 4, 5 and y: 2, 4, 5, 4, 5. Find r and the regression of y on x.
→ x̄ = 3, ȳ = 4; dx = −2, −1, 0, 1, 2; dy = −2, 0, 1, 0, 1
→ Σdx·dy = 4 + 0 + 0 + 0 + 2 = 6; Σdx² = 10; Σdy² = 6
→ r = 6/√(10 × 6) = 6/7.75 ≈ 0.77
→ bᵧₓ = 6/10 = 0.6 → y − 4 = 0.6(x − 3) → y = 0.6x + 2.2; at x = 6, y = 5.8
✏️ Worked Example: The regression lines are 8x − 10y + 66 = 0 and 40x − 18y = 214. Find the means and r.
→ Solve together: first × 5 → 40x − 50y = −330; subtracting gives 32y = 544 → ȳ = 17, x̄ = 13
→ Take the first as y on x: y = 0.8x + 6.6 → bᵧₓ = 0.8; the second gives x = 0.45y + 5.35 → bₓᵧ = 0.45
→ r² = 0.8 × 0.45 = 0.36 → r = 0.6 (the other way round the product exceeds 1, which is impossible)
✏️ Worked Example: For 6 ranked items Σd² = 14. Spearman’s R?
→ R = 1 − 6 × 14/[6(36 − 1)] = 1 − 84/210
→ = 0.6
✏️ Worked Example: Association table of attributes: (AB) = 30, (Aβ) = 10, (αB) = 20, (αβ) = 40. Yule’s Q?
→ Q = (ad − bc)/(ad + bc) = (30 × 40 − 10 × 20)/(30 × 40 + 10 × 20)
→ = 1000/1400 ≈ 0.71 → positive association
✏️ Worked Example: r₁₂ = 0.7, r₁₃ = 0.5, r₂₃ = 0.5. Partial correlation r₁₂.₃?
→ r₁₂.₃ = (0.7 − 0.5 × 0.5)/√[(1 − 0.25)(1 − 0.25)] = 0.45/0.75
→ = 0.6
⚠️ Exam Traps:
  • bᵧₓ and bₓᵧ cannot both exceed 1 (their product r² ≤ 1).
  • r = 0 only means no LINEAR relation, not independence.
⚠️ Exam Traps:
  • Multiple correlation R is never negative; partial correlation can be negative.

📌 6. Unit 6 · Probability Theory

  • Random experiment, sample space S, event (a subset of S). Mutually exclusive: A ∩ B = ∅; exhaustive: A ∪ B ∪ … = S; equally likely: all have equal probability.
  • Definitions: classical P(A) = favourable/total equally likely outcomes; relative frequency (empirical) = limit of m/n over a very large number of trials; axiomatic (Kolmogorov): P(A) ≥ 0, P(S) = 1, and P(A ∪ B) = P(A) + P(B) for mutually exclusive events.
  • Independent events: one happening does not change the probability of the other; then A and B′, and A′ and B′, are also independent.
  • Odds in favour a : b → P = a/(a + b); odds against a : b → P = b/(a + b).
📘 Formulas:
  • Addition theorem: P(A ∪ B) = P(A) + P(B) − P(A ∩ B); three events: P(A ∪ B ∪ C) = ΣP(A) − ΣP(A ∩ B) + P(A ∩ B ∩ C)
  • Multiplication theorem: P(A ∩ B) = P(A) · P(B|A) = P(B) · P(A|B)
  • Conditional probability P(A|B) = P(A ∩ B)/P(B); if independent, P(A ∩ B) = P(A)P(B)
  • Total probability: P(A) = Σ P(Eᵢ)P(A|Eᵢ); Bayes’ theorem: P(Eᵢ|A) = P(Eᵢ)P(A|Eᵢ) / Σ P(Eⱼ)P(A|Eⱼ)
  • At least one: 1 − P(none) = 1 − (1 − p₁)(1 − p₂)…
✏️ Worked Example: P(A) = 0.5, P(B) = 0.4, P(A ∩ B) = 0.2. Find P(A ∪ B) and P(A|B); are A and B independent?
→ P(A ∪ B) = 0.5 + 0.4 − 0.2 = 0.7
→ P(A|B) = 0.2/0.4 = 0.5 = P(A)
→ P(A)P(B) = 0.2 = P(A ∩ B) → independent
✏️ Worked Example: Two cards are drawn without replacement from a pack of 52. Probability that both are aces?
→ P = P(first ace) × P(second ace | first ace) = 4/52 × 3/51
→ = 12/2652 = 1/221 (or ⁴C₂/⁵²C₂ = 6/1326)
✏️ Worked Example: Urn I has 3 white and 2 black balls; urn II has 2 white and 3 black. An urn is chosen at random and the ball drawn is white. Probability that it came from urn I?
→ P(I) = P(II) = 1/2; P(W|I) = 3/5, P(W|II) = 2/5
→ Bayes: P(I|W) = (½ × 3/5)/(½ × 3/5 + ½ × 2/5) = (3/10)/(5/10)
→ = 3/5
✏️ Worked Example: A, B and C solve a problem independently with probabilities 1/2, 1/3 and 1/4. Probability that the problem is solved?
→ P(nobody solves it) = (1/2)(2/3)(3/4) = 1/4
→ P(solved) = 1 − 1/4 = 3/4
⚠️ Exam Traps:
  • Mutually exclusive ≠ independent: if P(A), P(B) > 0 and A ∩ B = ∅, then P(A ∩ B) = 0 ≠ P(A)P(B) → such events are dependent.
  • P(A|B) and P(B|A) are different — in Bayes questions the "given" part goes in the denominator.

📌 7. Unit 7 · Random Variables and Probability Distributions

  • A random variable is a function that gives a number to every outcome of the sample space. Discrete (counts: 0, 1, 2…) → probability mass function p(x), Σp(x) = 1; continuous (measurements) → probability density function f(x), ∫f(x)dx = 1 and P(a < X < b) = area under the curve.
  • Distribution function F(x) = P(X ≤ x): non-decreasing, F(−∞) = 0, F(∞) = 1. For a continuous variable P(X = a) = 0.
📘 Formulas:
  • Conditional probability P(A|B) = P(A ∩ B)/P(B); independent events: P(A ∩ B) = P(A)P(B)
  • Bayes' theorem: P(Eᵢ|A) = P(Eᵢ)P(A|Eᵢ) / Σ P(Eⱼ)P(A|Eⱼ)
  • E(X) = Σx·p(x); Var(X) = E(X²) − [E(X)]²
  • E(aX + b) = aE(X) + b; Var(aX + b) = a²Var(X)
DistributionMeanVarianceKey point
Binomial B(n, p)npnpqMean > variance; P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ
Poisson P(λ)λλMean = variance; P(X = r) = e⁻λλʳ/r!
Normal N(μ, σ²)μσ²Mean = median = mode; Z = (X − μ)/σ
Exponential (λ)1/λ1/λ²f(x) = λe⁻λˣ, x ≥ 0; memoryless

The four main distributions — which one when

  • Binomial: n fixed independent trials, two outcomes each, constant success probability p. Symmetric when p = ½; positively skewed when p < ½. Variance npq ≤ n/4.
  • Poisson: rare events (misprints on a page, calls in an hour). Limit of the binomial when n is large, p small and np = λ. P(X = 0) = e⁻λ. The sum of two independent Poisson variables is Poisson (λ₁ + λ₂).
  • Normal: bell-shaped, symmetric (β₁ = 0, β₂ = 3); points of inflexion at μ ± σ; QD ≈ 0.6745σ (≈ ⅔σ), MD ≈ 0.7979σ (≈ ⅘σ). Standard normal Z = (X − μ)/σ ~ N(0, 1). Table: P(0 < Z < 1) = 0.3413, P(0 < Z < 1.96) = 0.4750, P(0 < Z < 2) = 0.4772.
  • Exponential: waiting time / lifetime. P(X > t) = e⁻λᵗ; mean 1/λ and SD also 1/λ. Memoryless: P(X > s + t | X > s) = P(X > t).
  • Continuous uniform U(a, b): mean (a + b)/2, variance (b − a)²/12.
✏️ Worked Example: X takes 0, 1, 2 with probabilities 1/4, 1/2, 1/4. Find E(X), Var(X) and Var(2X + 3).
→ E(X) = 0 + 1/2 + 2/4 = 1; E(X²) = 0 + 1/2 + 4/4 = 1.5
→ Var(X) = 1.5 − 1² = 0.5
→ Var(2X + 3) = 2² × 0.5 = 2 (the added 3 does not change the variance)
✏️ Worked Example: A fair coin is tossed 6 times. Probability of exactly 4 heads, and the mean and variance?
→ n = 6, p = q = ½: P(X = 4) = ⁶C₄ (½)⁶ = 15/64
→ Mean = np = 3; variance = npq = 1.5
✏️ Worked Example: Defects per metre of cloth follow a Poisson distribution with λ = 2. Probability of no defect, and of at most 1 defect, in a metre? (e⁻² ≈ 0.1353)
→ P(0) = e⁻² ≈ 0.135
→ P(1) = 2e⁻² → P(X ≤ 1) = e⁻²(1 + 2) = 3 × 0.1353 ≈ 0.406
✏️ Worked Example: Marks are distributed N(60, 10²). What proportion score above 80, and between 50 and 70?
→ X = 80 → Z = (80 − 60)/10 = 2 → P(Z > 2) = 0.5 − 0.4772 = 0.0228 (≈ 2.3%)
→ 50 to 70 → Z = −1 to 1 → 2 × 0.3413 = 0.6826 (≈ 68%)
✏️ Worked Example: Bulb life is exponential with mean 1000 hours. Probability that a bulb lasts more than 1000 hours? And that a bulb which has already run 500 hours lasts another 1000 hours?
→ λ = 1/1000 → P(X > 1000) = e⁻¹ ≈ 0.368
→ By the memoryless property the second answer is also e⁻¹ ≈ 0.368 — the first 500 hours do not matter.

Joint distribution (two discrete random variables)

📘 Formulas:
  • Joint pmf p(x, y) = P(X = x, Y = y); p(x, y) ≥ 0 and ΣΣp(x, y) = 1
  • Marginal: P(X = x) = Σᵧ p(x, y); conditional: P(X = x | Y = y) = p(x, y)/P(Y = y)
  • Independent only if p(x, y) = P(X = x) × P(Y = y) for every (x, y)
  • Cov(X, Y) = E(XY) − E(X)E(Y); Var(aX ± bY) = a²Var X + b²Var Y ± 2ab Cov(X, Y)
  • For independent variables E(XY) = E(X)E(Y) and Cov = 0 (but Cov = 0 does not prove independence)
✏️ Worked Example: p(0, 0) = 0.2, p(0, 1) = 0.3, p(1, 0) = 0.1, p(1, 1) = 0.4. Are X and Y independent? Find Cov(X, Y).
→ Marginals: P(X = 1) = 0.1 + 0.4 = 0.5; P(Y = 1) = 0.3 + 0.4 = 0.7
→ p(1, 1) = 0.4 ≠ 0.5 × 0.7 = 0.35 → not independent
→ E(XY) = 1 × 1 × 0.4 = 0.4 → Cov = 0.4 − 0.5 × 0.7 = 0.05
💡 Shortcut Trick: In a normal curve ≈ 68.27% of values lie in μ ± σ, ≈ 95.45% in μ ± 2σ, ≈ 99.73% in μ ± 3σ.
⚠️ Exam Traps:
  • Binomial: mean > variance; Poisson: mean = variance — a binomial with "mean 4, variance 5" is impossible.
  • In a normal distribution divide by σ, not σ², when finding Z.

📌 8. Unit 8 · Sampling Theory

  • Parameter = measure of the population (μ, σ); statistic = measure of the sample (x̄, s).
  • Probability sampling: simple random, stratified, systematic, cluster, multistage, multiphase.
  • Non-probability sampling: purposive, convenience, quota.
  • Sampling error arises from taking a sample (falls as n grows); non-sampling error (wrong responses, processing mistakes) occurs even in a census.
  • Properties of a good estimator: unbiased, consistent, efficient, sufficient.
  • Methods of estimation: method of moments, maximum likelihood (MLE), least squares.
  • Population: finite or infinite; sampling frame = the list of units from which the sample is drawn; sampling unit = the unit chosen (person, household, village).
  • Advantages of a sample survey: lower cost, less time, more detailed study, and the only way for destructive testing (e.g. bulb life). Census: information on every unit, but costly.
  • SRSWR (with replacement): Nⁿ possible samples; SRSWOR (without replacement): ᴺCₙ. Select by the lottery method or a random number table.
  • Systematic: interval k = N/n, first unit random from 1 to k, then every k-th. Stratified: proportional allocation nₕ = n × Nₕ/N; Neyman allocation nₕ ∝ Nₕσₕ. Cluster: whole groups (e.g. villages) are chosen. Multistage: first districts, then villages, then households.
  • Central limit theorem: for large n (above about 30) the sampling distribution of x̄ is approximately normal N(μ, σ²/n), whatever the population distribution.
📘 Formulas:
  • Standard error SE(x̄) = σ/√n
  • 95% confidence interval for μ = x̄ ± 1.96·σ/√n; use 2.58 for 99%
  • s² = Σ(x − x̄)²/(n − 1) is an unbiased estimator of the population variance
📘 Formulas:
  • Finite population (SRSWOR): SE(x̄) = (σ/√n) × √[(N − n)/(N − 1)] — finite population correction
  • Standard error of a proportion SE(p) = √(PQ/n)
  • Sample size to estimate a mean n = (Zσ/E)²; for a proportion n = Z²PQ/E² (E = permissible error)
✏️ Worked Example: σ = 12, n = 36. Standard error of x̄?
→ SE = σ/√n = 12/6 = 2
✏️ Worked Example: σ = 10. How large a sample is needed to estimate the mean within ±2 with 95% confidence?
→ n = (1.96 × 10/2)² = 9.8² = 96.04
→ Always round up → 97
✏️ Worked Example: N = 1000 has three strata of 600, 300 and 100. Proportional allocation of n = 50?
→ nₕ = 50 × Nₕ/1000 → 50 × 0.6, 50 × 0.3, 50 × 0.1
→ = 30, 15, 5
✏️ Worked Example: A systematic sample of n = 25 is to be drawn from N = 500; the random start is 7. Which units are chosen?
→ k = 500/25 = 20
→ Units 7, 27, 47, 67, … , 487 (7 + 24 × 20)
✏️ Worked Example: N = 100, n = 25 (without replacement), σ = 10. SE(x̄)?
→ σ/√n = 10/5 = 2; correction = √(75/99) ≈ 0.870
→ SE ≈ 2 × 0.870 = 1.74
⚠️ Exam Traps:
  • Sampling error falls as n grows; non-sampling error (wrong answers, enumerator mistakes, non-response) occurs even in a census and can grow with n.
  • To halve the SE, n must be made four times as large (SE ∝ 1/√n).

📌 9. Unit 9 · Statistical Inference: Estimation and Testing of Hypotheses

Estimation

  • Point estimate = a single number (e.g. x̄ = 52); interval estimate = a range that contains the parameter with a stated confidence (e.g. a 95% confidence interval).
  • Unbiased: E(T) = θ. x̄ is an unbiased estimator of the population mean μ; Σ(x − x̄)²/n is biased, Σ(x − x̄)²/(n − 1) is unbiased. Consistent: T → θ as n grows. Efficient: least variance among unbiased estimators. Sufficient: keeps all the information in the sample.
  • Maximum likelihood estimator (MLE): the value that maximises the likelihood L(θ). Examples: binomial p̂ = x/n; Poisson λ̂ = x̄; normal μ̂ = x̄, σ̂² = Σ(x − x̄)²/n (biased but consistent).

Testing of hypotheses

  • Steps: (1) state the null hypothesis H₀ and the alternative H₁; (2) choose the level of significance α; (3) compute the test statistic; (4) compare with the critical value; (5) reject H₀ if |calculated| > critical.
  • H₁: μ ≠ μ₀ → two-tailed; H₁: μ > μ₀ or μ < μ₀ → one-tailed. Reject H₀ if the p-value < α.
  • Type-I error: rejecting a true H₀ (probability α = level of significance); Type-II error: accepting a false H₀ (β); power of a test = 1 − β.
  • Large sample (n > 30) or known σ → Z-test; small sample with unknown σ → t-test; comparing variances → F; independence / goodness of fit → χ².
  • At the 5% level the two-tailed critical Z = 1.96; one-tailed 1.645.
TestWhenStatisticDegrees of freedom (df)
Z (mean)Large sample or σ knownZ = (x̄ − μ₀)/(σ/√n)—
Z (proportion)Large sampleZ = (p − P)/√(PQ/n)—
t (one mean)Small sample, σ unknownt = (x̄ − μ₀)/(s/√n)n − 1
t (two independent means)Two small sampleswith the pooled sn₁ + n₂ − 2
Paired tBefore–after (same units)t = d̄/(s_d/√n)n − 1
χ²Goodness of fit / independence / varianceΣ(O − E)²/Ek − 1 − (parameters estimated); (r − 1)(c − 1)
FComparing two variances, ANOVAF = s₁²/s₂²(n₁ − 1, n₂ − 1)
Level of significanceTwo-tailed ZOne-tailed Z
5%1.961.645
1%2.582.33
📘 Formulas:
  • Z = (x̄ − μ₀)/(σ/√n)
  • χ² = Σ(O − E)²/E; for an r × c table df = (r − 1)(c − 1)
  • F = s₁²/s₂² (larger variance in the numerator)
  • One-way ANOVA (k treatments, N observations): df treatments = k − 1, error = N − k, total = N − 1
  • Two-way ANOVA (r rows, c columns): error df = (r − 1)(c − 1)
  • χ²(n) distribution: mean = n, variance = 2n
✏️ Worked Example: x̄ = 50, σ = 8, n = 64. 95% confidence interval for μ?
→ SE = 8/√64 = 1
→ 50 ± 1.96 × 1 = (48.04, 51.96)
✏️ Worked Example: Claim μ = 100. n = 36, x̄ = 103, σ = 9. Test at the 5% and 1% levels (two-tailed).
→ Z = (103 − 100)/(9/6) = 3/1.5 = 2
→ 5%: 2 > 1.96 → reject H₀; 1%: 2 < 2.58 → do not reject H₀
✏️ Worked Example: n = 16, x̄ = 52, s = 8; H₀: μ = 50. Value of t and decision (t₀.₀₂₅,₁₅ = 2.131)?
→ t = (52 − 50)/(8/4) = 1; df = 15
→ 1 < 2.131 → do not reject H₀
✏️ Worked Example: Claim: 50% of people favour a scheme. In a sample of 400, 220 are in favour. Test at the 5% level.
→ p = 0.55; SE = √(0.5 × 0.5/400) = 0.025
→ Z = 0.05/0.025 = 2 > 1.96 → claim rejected
✏️ Worked Example: A die thrown 60 times gives 8, 12, 9, 11, 6, 14. Is it fair? (χ²₀.₀₅,₅ = 11.07)
→ E = 10 for each face; (O − E)² = 4, 4, 1, 1, 16, 16 → sum 42
→ χ² = 42/10 = 4.2; df = 6 − 1 = 5
→ 4.2 < 11.07 → the die is taken as fair
✏️ Worked Example: 7 heads in 10 tosses. MLE of p?
→ For the binomial p̂ = x/n = 7/10 = 0.7
⚠️ Exam Traps:
  • "Accept H₀" means "not enough evidence to reject", not that H₀ is proved.
  • Lowering α (5% → 1%) reduces the Type-I error but raises the Type-II error (for the same n).

📌 10. Unit 10 · Analysis of Variance (ANOVA)

ANOVA (R. A. Fisher) tests the means of three or more groups at once. H₀: μ₁ = μ₂ = … = μₖ. The total variation is split into "between groups" and "within groups (error)" parts and F = MS(between)/MS(within) is computed; a large F means the means are not all equal.

  • Assumptions: independent observations, each group from a normal population, equal variances in all groups (homoscedasticity), additive effects. Equal group sizes are not required.
  • Link with design of experiments: completely randomised design (CRD) → one-way; randomised block design (RBD) → two-way; in an m × m Latin square (LSD) the error df = (m − 1)(m − 2).
📘 Formulas:
  • Correction factor CF = G²/N (G = total of all observations)
  • Total SS (TSS) = Σx² − CF; treatment SS (SST) = Σ(Tᵢ²/nᵢ) − CF; error SS (SSE) = TSS − SST
  • One-way df: treatments k − 1, error N − k, total N − 1; MST = SST/(k − 1), MSE = SSE/(N − k), F = MST/MSE
  • Two-way (one observation per cell, r rows × c columns): TSS = SSR + SSC + SSE; df: (r − 1) + (c − 1) + (r − 1)(c − 1) = rc − 1
Source of variationSSdfMSF
Treatments (between groups)SSTk − 1MST = SST/(k − 1)MST/MSE
Error (within groups)SSEN − kMSE = SSE/(N − k)
TotalTSSN − 1
✏️ Worked Example: Marks of students taught by three methods — A: 2, 4, 6; B: 4, 6, 8; C: 6, 8, 10. Carry out a one-way ANOVA (F₀.₀₅(2, 6) = 5.14).
→ Totals: T_A = 12, T_B = 18, T_C = 24; G = 54, N = 9 → CF = 54²/9 = 324
→ Σx² = 56 + 116 + 200 = 372 → TSS = 372 − 324 = 48
→ SST = (144 + 324 + 576)/3 − 324 = 348 − 324 = 24; SSE = 48 − 24 = 24
→ MST = 24/2 = 12; MSE = 24/6 = 4 → F = 12/4 = 3
→ 3 < 5.14 → do not reject H₀: no significant difference between the method means
✏️ Worked Example: 4 treatments with 5 observations each. SST = 60, SSE = 80. Value of F and decision (F₀.₀₅(3, 16) = 3.24)?
→ N = 20 → df: treatments 3, error 16, total 19
→ MST = 60/3 = 20; MSE = 80/16 = 5 → F = 4
→ 4 > 3.24 → reject H₀: the treatment means are not all equal
✏️ Worked Example: Two-way ANOVA: 4 blocks (rows) × 5 treatments (columns), one observation per cell. All the df?
→ Rows 4 − 1 = 3; columns 5 − 1 = 4
→ Error 3 × 4 = 12; total 20 − 1 = 19 (3 + 4 + 12 = 19 ✓)
⚠️ Exam Traps:
  • F is always ≥ 0; the ANOVA F-test is always right-tailed (one-tailed).
  • Mind the df in MST/MSE: F(k − 1, N − k) — numerator df first.

📌 11. Unit 11 · Time Series

  • Components of a time series: secular trend, seasonal (within a year), cyclical (business cycle, several years), irregular (floods, wars).
  • Measuring trend: free-hand curve, semi-averages, moving averages, least squares.
  • Models: additive Y = T + S + C + I; multiplicative Y = T × S × C × I (more common in practice).
  • Examples: sales rising at festivals → seasonal; boom and recession → cyclical; steady population growth → secular trend; earthquake/strike → irregular.
  • Moving averages: an odd period (3, 5 years) is placed against the middle year; an even period (4 years) must be centred. If the period equals the cycle length, the fluctuations are removed. Some values at the ends are lost.
  • Least squares method: Y = a + bX; code X so that ΣX = 0, then a = ΣY/n and b = ΣXY/ΣX².
  • Seasonal indices: simple averages, ratio-to-trend, ratio-to-moving-average, link relatives. Quarterly indices add up to 400, monthly ones to 1200. Deseasonalised value = actual ÷ seasonal index × 100.
✏️ Worked Example: The values for 2019–2023 are 10, 14, 12, 16, 20. Find the 3-year moving averages.
→ 2020: (10 + 14 + 12)/3 = 12; 2021: (14 + 12 + 16)/3 = 14; 2022: (12 + 16 + 20)/3 = 16
→ Moving averages 12, 14, 16 (none for 2019 and 2023)
✏️ Worked Example: Output for 2019–2023 is 10, 12, 15, 17, 21. Fit a straight-line trend by least squares and estimate 2025.
→ X = −2, −1, 0, 1, 2 (origin 2021); ΣY = 75, ΣXY = −20 − 12 + 0 + 17 + 42 = 27, ΣX² = 10
→ a = 75/5 = 15, b = 27/10 = 2.7 → Y = 15 + 2.7X
→ 2025 → X = 4: Y = 15 + 10.8 = 25.8
✏️ Worked Example: The averages of the four quarters are 40, 50, 60 and 50. Find the seasonal indices, and the deseasonalised value of third-quarter sales of 66.
→ Grand average = 200/4 = 50 → index = quarter average/50 × 100 = 80, 100, 120, 100 (sum 400 ✓)
→ Deseasonalised = 66/120 × 100 = 55
✏️ Worked Example: Values for 6 years: 20, 24, 22, 28, 32, 30. Trend increase per year by the method of semi-averages?
→ First half: (20 + 24 + 22)/3 = 22 (against year 2); second: (28 + 32 + 30)/3 = 30 (against year 5)
→ Increase = (30 − 22)/(5 − 2) = 8/3 ≈ 2.67 per year
⚠️ Exam Traps:
  • Seasonal swings repeat within a year; cyclical ones over several years — "every summer" = seasonal, "every 7–8 years" = cyclical.

📌 12. Unit 12 · Index Numbers

  • An index number measures the relative change in a group (prices, quantities) over time or place; base year = 100. Price, quantity and value (value = Σp₁q₁/Σp₀q₀ × 100) indices.
  • Simple aggregative Σp₁/Σp₀ × 100; simple average of relatives Σ(p₁/p₀ × 100)/n; weighted: Laspeyres, Paasche, Fisher, Marshall–Edgeworth, Dorbish–Bowley (L + P)/2, Kelly (fixed weights).
  • Tests: unit test; time reversal P₀₁ × P₁₀ = 1; factor reversal P₀₁ × Q₀₁ = V₀₁; circular test (met by the simple aggregative and fixed-weight indices).
  • Chain base: each year uses the previous year = 100 (link relatives); multiplying them gives a fixed-base series.
📘 Formulas:
  • Laspeyres P₀₁ = Σp₁q₀/Σp₀q₀ × 100 (base-year quantities)
  • Paasche P₀₁ = Σp₁q₁/Σp₀q₁ × 100 (current-year quantities)
  • Fisher's ideal index = √(L × P)
  • Marshall–Edgeworth = Σp₁(q₀ + q₁)/Σp₀(q₀ + q₁) × 100
⚠️ Exam Traps:
  • Fisher's index satisfies both the Time Reversal Test and the Factor Reversal Test, hence "ideal".
  • Usually L > P (Laspeyres has an upward bias, Paasche a downward bias).
📘 Formulas:
  • Base shifting: new index = (old index ÷ old index of the new base year) × 100
  • Splicing: new series to the old base = new index × (old index of the link year/100); old series to the new base = old × 100/(old index of the link year)
  • Deflating: real wage = money wage ÷ price index × 100
  • Purchasing power of money = 100/CPI (in rupees)
Method of cost of living (consumer price) indexFormula
Aggregate expenditure methodΣp₁q₀/Σp₀q₀ × 100 (= Laspeyres)
Family budget methodΣPW/ΣW; P = p₁/p₀ × 100, W = p₀q₀ (base-year expenditure)

Both methods give the same answer. A CPI measures the change in the cost of living of a particular group (e.g. industrial workers).

✏️ Worked Example: Item A: p₀ = 10, q₀ = 8, p₁ = 12, q₁ = 10; item B: p₀ = 20, q₀ = 6, p₁ = 25, q₁ = 5. Laspeyres, Paasche and Fisher?
→ Σp₁q₀ = 96 + 150 = 246; Σp₀q₀ = 80 + 120 = 200 → L = 123
→ Σp₁q₁ = 120 + 125 = 245; Σp₀q₁ = 100 + 100 = 200 → P = 122.5
→ Fisher = √(123 × 122.5) ≈ 122.75
✏️ Worked Example: With 2015 = 100 the index is 125 in 2020 and 150 in 2023. With the base shifted to 2020, the 2023 index?
→ New = 150/125 × 100 = 120
✏️ Worked Example: Salary ₹10,000 in 2015 (CPI 100) and ₹16,000 in 2024 (CPI 160). Has the real salary risen?
→ Real salary in 2024 = 16,000/160 × 100 = ₹10,000
→ The same as in 2015 → no real increase
✏️ Worked Example: Family budget method: food P = 150, W = 50; rent P = 120, W = 20; others P = 110, W = 30. CPI?
→ ΣPW = 7500 + 2400 + 3300 = 13,200; ΣW = 100
→ CPI = 13,200/100 = 132
⚠️ Exam Traps:
  • In base shifting the new base year becomes 100 — divide by its old index.

📌 13. Practice MCQs

Questions by unit: Unit 1: Q1–Q4; Unit 2: Q5–Q9; Unit 3: Q10–Q13; Unit 4: Q14–Q17; Unit 5: Q18–Q23; Unit 6: Q24–Q29; Unit 7: Q30–Q38; Unit 8: Q39–Q43; Unit 9: Q44–Q50; Unit 10: Q51–Q55; Unit 11: Q56–Q60; Unit 12: Q61–Q64.
📝 Practice MCQs (prepared by SF Prep — not PYQs)
Q1. Which graph is used to locate the median of a distribution graphically? Practice
(A) Ogive
(B) Histogram
(C) Pie chart
(D) Scatter diagram
Correct answer: (A)
Solution: The x-coordinate where the 'less than' and 'more than' ogives intersect is the median. A histogram locates the mode.
Q2. In a statistical table, what are the column headings called? Practice
(A) Stub
(B) Body
(C) Caption
(D) Head note
Correct answer: (C)
Solution: Caption = column headings; stub = row headings; the body holds the figures.
Q3. A researcher uses figures from the Census of India reports in a study. For the researcher these data are: Practice
(A) Secondary data
(B) Primary data
(C) Experimental data
(D) Qualitative data
Correct answer: (A)
Solution: The data were collected by someone else (the Census office), so for the researcher they are secondary.
Q4. A family’s monthly budget is ₹40,000, of which rent is ₹10,000. What is the central angle for rent in a pie chart? Practice
(A) 72°
(B) 90°
(C) 100°
(D) 120°
Correct answer: (B)
Solution: Angle = (10,000/40,000) × 360° = ¼ × 360° = 90°.
Q5. The arithmetic mean of the following frequency distribution is — Class: 20-30, 30-40, 40-50, 50-60, 60-70; Frequency: 4, 13, 9, 4, 10 Practice
(A) 40.75
(B) 45.75
(C) 50.75
(D) 45
Correct answer: (B)
Solution: Take class marks: Σfx = 1830, N = 40 → mean = 1830/40 = 45.75.
Q6. The median of the following distribution is — Class: 100-110, 110-120, 120-130, 130-140, 140-150; Frequency: 11, 12, 8, 2, 13 Practice
(A) 115
(B) 110
(C) 120
(D) 775/6
Correct answer: (C)
Solution: N/2 = 23; median class 110-120 (cf = 11, f = 12) → 110 + [(23 − 11)/12] × 10 = 120.
Q7. The mode of the following distribution is — Class: 10-20, 20-30, 30-40, 40-50, 50-60; Frequency: 8, 9, 8, 14, 10 Practice
(A) 45
(B) 42.5
(C) 44
(D) 46
Correct answer: (D)
Solution: Modal class 40-50: f₁ = 14, f₀ = 8, f₂ = 10 → 40 + [6/10] × 10 = 46.
Q8. Which of the following equals the third quartile (Q₃)? Practice
(A) D₇
(B) P₇₀
(C) D₃
(D) P₇₅
Correct answer: (D)
Solution: 75% of the observations lie below Q₃ → Q₃ = P₇₅.
Q9. In a distribution with N = 50, the class 20–30 has frequency 10 and the cumulative frequency before it is 15. Find the fourth decile D₄. Practice
(A) 22.5
(B) 25
(C) 27.5
(D) 30
Correct answer: (B)
Solution: 4N/10 = 20 → class 20–30 (cf 15 to 25). D₄ = 20 + [(20 − 15)/10] × 10 = 25.
Q10. In a distribution Q₁ = 15 and Q₃ = 26. The coefficient of quartile deviation is: Practice
(A) 5.5
(B) 11/41
(C) 41/11
(D) 11/82
Correct answer: (B)
Solution: (Q₃ − Q₁)/(Q₃ + Q₁) = (26 − 15)/(26 + 15) = 11/41.
Q11. A series has mean 120 and standard deviation 6. Its coefficient of variation is: Practice
(A) 10%
(B) 30%
(C) 5%
(D) 2.5%
Correct answer: (C)
Solution: CV = (σ/x̄) × 100 = (6/120) × 100 = 5%.
Q12. Two groups: n₁ = 30, x̄₁ = 59, σ₁ = 8; n₂ = 30, x̄₂ = 47, σ₂ = 8. The combined standard deviation is: Practice
(A) 8
(B) 100
(C) 12
(D) 10
Correct answer: (D)
Solution: Combined mean = 53; d₁ = 6, d₂ = −6; σ² = [30(64 + 36) + 30(64 + 36)]/60 = 100 → σ = 10.
Q13. The mean deviation about the mean of 4, 7, 8, 9, 12 is: Practice
(A) 2.5
(B) 1.6
(C) 10
(D) 2
Correct answer: (D)
Solution: Mean = 40/5 = 8; |x − 8| = 4, 1, 0, 1, 4 → sum 10; MD = 10/5 = 2.
Q14. A distribution has mean 58, mode 54 and standard deviation 8. Karl Pearson's coefficient of skewness is: Practice
(A) 1/2
(B) −1/2
(C) 1.5
(D) 2
Correct answer: (A)
Solution: Sₖ = (Mean − Mode)/σ = (58 − 54)/8 = 1/2.
Q15. In a positively skewed distribution, the correct order is: Practice
(A) Mode > Median > Mean
(B) Mean > Median > Mode
(C) Mean = Median = Mode
(D) Median > Mean > Mode
Correct answer: (B)
Solution: With positive skew the right tail is long and pulls the mean furthest right.
Q16. For a distribution μ₂ = 4 and μ₄ = 40. β₂ and the type of curve are: Practice
(A) β₂ = 2.5, leptokurtic
(B) β₂ = 2.5, mesokurtic
(C) β₂ = 2.5, platykurtic
(D) β₂ = 10, platykurtic
Correct answer: (C)
Solution: β₂ = μ₄/μ₂² = 40/16 = 2.5; β₂ < 3 → platykurtic.
Q17. The first two moments about A = 5 are μ′₁ = 2 and μ′₂ = 20. The mean and variance are: Practice
(A) Mean 7, variance 16
(B) Mean 3, variance 16
(C) Mean 7, variance 20
(D) Mean 7, variance 24
Correct answer: (A)
Solution: Mean = A + μ′₁ = 5 + 2 = 7; variance μ₂ = μ′₂ − μ′₁² = 20 − 4 = 16.
Q18. For two variables x and y, Cov(x, y) = 6, σₓ = 3 and σᵧ = 4. The correlation coefficient r is: Practice
(A) −1/2
(B) 2/3
(C) 6/7
(D) 1/2
Correct answer: (D)
Solution: r = Cov/(σₓσᵧ) = 6/(3 × 4) = 1/2.
Q19. 5 students were ranked by two judges; the sum of squares of rank differences Σd² = 36. Spearman's rank correlation coefficient is: Practice
(A) −4/5
(B) 1.8
(C) −43/65
(D) 7/10
Correct answer: (A)
Solution: R = 1 − 6Σd²/[n(n² − 1)] = 1 − 216/120 = −4/5.
Q20. The two regression lines are 3x + 4y = 40 and 6x + 5y = 68. The means (x̄, ȳ) are: Practice
(A) (4, 8)
(B) (8, 4)
(C) (9, 3)
(D) (8, 6)
Correct answer: (B)
Solution: Both lines pass through (x̄, ȳ); solving gives x = 8, y = 4 (check: 3·8 + 4·4 = 40, 6·8 + 5·4 = 68).
Q21. If r₁₂ = 0.6, r₁₃ = 0.8 and r₂₃ = 0, the multiple correlation coefficient R₁.₂₃ is: Practice
(A) 0.7
(B) 0.48
(C) 1.4
(D) 1
Correct answer: (D)
Solution: R² = (0.36 + 0.64 − 0)/(1 − 0) = 1 → R = 1. (R is never above 1.)
Q22. If bᵧₓ = 0.8 and bₓᵧ = 0.2, the correlation coefficient r is: Practice
(A) 0.16
(B) 0.5
(C) 0.4
(D) 1
Correct answer: (C)
Solution: r² = bᵧₓ × bₓᵧ = 0.16 → r = 0.4 (both coefficients are positive, so r is positive).
Q23. In an attribute table (AB) = 40, (Aβ) = 10, (αB) = 10, (αβ) = 40. Yule’s coefficient of association Q is: Practice
(A) 3/5
(B) 16/17
(C) 15/17
(D) 1/17
Correct answer: (C)
Solution: Q = (ad − bc)/(ad + bc) = (1600 − 100)/(1600 + 100) = 1500/1700 = 15/17.
Q24. In a factory, machines produce A 30%, B 30%, C 40% of the output; their defective rates are 4%, 2%, 5% respectively. An item chosen at random is defective. The probability that it was made by machine A is: Practice
(A) 6/19
(B) 3/10
(C) 4/11
(D) 3/250
Correct answer: (A)
Solution: Bayes: P(A|D) = (30 × 4)/(30 × 4 + 30 × 2 + 40 × 5) = 120/380 = 6/19.
Q25. P(A) = 0.6, P(B) = 0.5 and P(A ∪ B) = 0.8. What is P(A|B)? Practice
(A) 0.6
(B) 0.3
(C) 0.5
(D) 0.375
Correct answer: (A)
Solution: P(A ∩ B) = 0.6 + 0.5 − 0.8 = 0.3; P(A|B) = 0.3/0.5 = 0.6.
Q26. A and B are independent events with P(A) = 0.4 and P(B) = 0.5. P(A ∪ B) is: Practice
(A) 0.9
(B) 0.2
(C) 0.5
(D) 0.7
Correct answer: (D)
Solution: Independent → P(A ∩ B) = 0.4 × 0.5 = 0.2; P(A ∪ B) = 0.4 + 0.5 − 0.2 = 0.7.
Q27. Two cards are drawn without replacement from a pack of 52. The probability that both are hearts is: Practice
(A) 1/16
(B) 1/17
(C) 1/4
(D) 3/52
Correct answer: (B)
Solution: P = 13/52 × 12/51 = 156/2652 = 1/17. (1/16 would be the answer with replacement.)
Q28. 40% of a class are girls. 10% of the girls and 5% of the boys fail. A student chosen at random has failed; the probability that it is a girl is: Practice
(A) 2/5
(B) 3/7
(C) 2/3
(D) 4/7
Correct answer: (D)
Solution: Bayes: P(girl ∩ fail) = 0.4 × 0.1 = 0.04; P(boy ∩ fail) = 0.6 × 0.05 = 0.03; P = 0.04/(0.04 + 0.03) = 4/7.
Q29. If two events A and B are mutually exclusive and both have probability greater than zero, then they are: Practice
(A) Always independent
(B) Always dependent (not independent)
(C) Exhaustive
(D) Equally likely
Correct answer: (B)
Solution: A ∩ B = ∅ → P(A ∩ B) = 0, while P(A)P(B) > 0; so the condition for independence fails — the events are dependent.
Q30. A binomial distribution has mean 12 and variance 3. n and p are: Practice
(A) n = 16, p = 1/4
(B) n = 15, p = 3/4
(C) n = 16, p = 3/4
(D) n = 32, p = 3/8
Correct answer: (C)
Solution: q = variance/mean = 3/12 = 1/4 → p = 3/4; n = mean ÷ p = 12 ÷ 3/4 = 16.
Q31. In which distribution are the mean and variance always equal? Practice
(A) Binomial
(B) Normal
(C) Exponential
(D) Poisson
Correct answer: (D)
Solution: For Poisson, mean = variance = λ. Binomial has mean > variance; exponential has mean 1/λ, variance 1/λ².
Q32. Joint distribution of X, Y ∈ {0, 1}: p(0, 0) = 0.1, p(0, 1) = 0.3, p(1, 0) = 0.2, p(1, 1) = k. Find P(X = 1). Practice
(A) 0.4
(B) 0.6
(C) 0.5
(D) 0.2
Correct answer: (B)
Solution: Total is 1 → k = 0.4. P(X = 1) = p(1, 0) + p(1, 1) = 0.2 + 0.4 = 0.6.
Q33. A fair coin is tossed 4 times. The probability of exactly 2 heads is: Practice
(A) 1/4
(B) 1/2
(C) 3/8
(D) 5/16
Correct answer: (C)
Solution: Binomial, n = 4, p = ½: P(X = 2) = ⁴C₂ (½)⁴ = 6/16 = 3/8.
Q34. For a Poisson variable P(X = 1) = P(X = 2). Its mean λ is: Practice
(A) 2
(B) 1
(C) 4
(D) 1/2
Correct answer: (A)
Solution: λe⁻λ = λ²e⁻λ/2 → 1 = λ/2 → λ = 2.
Q35. X ~ N(50, 5²). The probability that X exceeds 60 is (P(0 < Z < 2) = 0.4772): Practice
(A) 0.0228
(B) 0.4772
(C) 0.9772
(D) 0.1587
Correct answer: (A)
Solution: Z = (60 − 50)/5 = 2; P(Z > 2) = 0.5 − 0.4772 = 0.0228.
Q36. The time until a machine fails is exponential with mean 5 years. The probability that it works for more than 10 years is: Practice
(A) e⁻⁰·⁵
(B) 1 − e⁻²
(C) e⁻¹⁰
(D) e⁻²
Correct answer: (D)
Solution: λ = 1/5 = 0.2; P(X > 10) = e⁻λᵗ = e⁻⁰·² ˣ ¹⁰ = e⁻².
Q37. X takes the values 1, 2, 3 with probabilities 0.2, 0.5, 0.3. E(X) is: Practice
(A) 2
(B) 2.1
(C) 1.9
(D) 2.3
Correct answer: (B)
Solution: E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1 + 0.9 = 2.1.
Q38. If Var(X) = 4, then Var(3X − 2) is: Practice
(A) 10
(B) 12
(C) 36
(D) 34
Correct answer: (C)
Solution: Var(aX + b) = a²Var(X) = 9 × 4 = 36; the −2 does not change the variance.
Q39. Dividing a population into homogeneous sub-groups (e.g. age groups) and drawing a random sample from each is called: Practice
(A) Cluster sampling
(B) Stratified sampling
(C) Systematic sampling
(D) Quota sampling
Correct answer: (B)
Solution: Strata are internally homogeneous and each is sampled at random — stratified sampling. In cluster sampling whole clusters are chosen.
Q40. The population standard deviation is 15 and the sample size is 25. The standard error of the sample mean is: Practice
(A) 0.6
(B) 3
(C) 5
(D) 9
Correct answer: (B)
Solution: SE = σ/√n = 15/√25 = 15/5 = 3.
Q41. σ = 20. The minimum sample size needed to estimate the mean within ±4 with 95% confidence is: Practice
(A) 97
(B) 96
(C) 49
(D) 10
Correct answer: (A)
Solution: n = (Zσ/E)² = (1.96 × 20/4)² = 9.8² = 96.04 → round up to 97.
Q42. Three strata contain 2000, 3000 and 5000 units. Proportional allocation of a sample of 200 is: Practice
(A) 67, 67, 66
(B) 20, 30, 50
(C) 40, 60, 100
(D) 50, 60, 90
Correct answer: (C)
Solution: N = 10,000; nₕ = 200 × Nₕ/N → 200 × 0.2 = 40, 200 × 0.3 = 60, 200 × 0.5 = 100.
Q43. A systematic sample of 60 units is to be drawn from a list of 1200 units. The sampling interval k is: Practice
(A) 60
(B) 12
(C) 72
(D) 20
Correct answer: (D)
Solution: k = N/n = 1200/60 = 20; the first unit is random from 1–20, then every 20th.
Q44. A coin tossed 50 times shows heads 37 times. H₀: the coin is fair. The value of χ² is: Practice
(A) 5.76
(B) 24/25
(C) 11.52
(D) 288/37
Correct answer: (C)
Solution: E = 25 (heads), 25 (tails); χ² = (37 − 25)²/25 + (13 − 25)²/25 = 5.76 + 5.76 = 11.52.
Q45. A Type-I error means: Practice
(A) Accepting H₀ when it is false
(B) Accepting H₀ when it is true
(C) Always accepting H₁
(D) Rejecting H₀ when it is true
Correct answer: (D)
Solution: The probability of a Type-I error is α, the level of significance. Accepting a false H₀ is a Type-II error (β).
Q46. x̄ = 80, σ = 10, n = 100. The 95% confidence interval for the population mean is: Practice
(A) (79.02, 80.98)
(B) (77.42, 82.58)
(C) (60.4, 99.6)
(D) (78.04, 81.96)
Correct answer: (D)
Solution: SE = 10/√100 = 1; 80 ± 1.96 × 1 = (78.04, 81.96). (2.58 gives the 99% interval.)
Q47. H₀: μ = 50; n = 25, x̄ = 52, σ = 5. The value of the test statistic Z is: Practice
(A) 0.4
(B) 10
(C) 2
(D) 1
Correct answer: (C)
Solution: Z = (52 − 50)/(5/√25) = 2/1 = 2.
Q48. In a t-test comparing the means of two independent small samples of sizes 12 and 10, the degrees of freedom are: Practice
(A) 20
(B) 22
(C) 21
(D) 9
Correct answer: (A)
Solution: df = n₁ + n₂ − 2 = 12 + 10 − 2 = 20.
Q49. A sample 2, 4, 3, 5, 1 is drawn from a Poisson population. The maximum likelihood estimate of λ is: Practice
(A) 15
(B) 3
(C) 2.5
(D) √3
Correct answer: (B)
Solution: For the Poisson λ̂ = x̄ = (2 + 4 + 3 + 5 + 1)/5 = 15/5 = 3.
Q50. Which of these is an unbiased estimator of the population variance σ²? Practice
(A) Σ(x − x̄)²/(n − 1)
(B) Σ(x − x̄)²/n
(C) Σ(x − x̄)²/(n + 1)
(D) Σ(x − x̄)/n
Correct answer: (A)
Solution: E[Σ(x − x̄)²] = (n − 1)σ², so dividing by n − 1 makes the estimator unbiased; dividing by n underestimates σ² (biased).
Q51. A one-way ANOVA has 5 treatments and 30 observations in all. The error (within-groups) degrees of freedom are: Practice
(A) 4
(B) 29
(C) 25
(D) 24
Correct answer: (C)
Solution: Error df = N − k = 30 − 5 = 25 (treatment df = 4, total df = 29).
Q52. One-way ANOVA: k = 4 treatments, N = 24, SST = 48, SSE = 60. The value of F is: Practice
(A) 0.8
(B) 16/3
(C) 4
(D) 5/4
Correct answer: (B)
Solution: MST = 48/(4 − 1) = 16; MSE = 60/(24 − 4) = 3; F = 16/3 ≈ 5.33.
Q53. In an ANOVA the total sum of squares is 120 and the between-groups sum of squares is 45. The error sum of squares is: Practice
(A) 165
(B) 90
(C) 60
(D) 75
Correct answer: (D)
Solution: TSS = SST + SSE → SSE = 120 − 45 = 75.
Q54. In a two-way ANOVA with 5 rows and 4 columns (one observation per cell) the error degrees of freedom are: Practice
(A) 19
(B) 20
(C) 12
(D) 7
Correct answer: (C)
Solution: Error df = (r − 1)(c − 1) = 4 × 3 = 12; total df = 20 − 1 = 19.
Q55. Which of these is not an assumption of ANOVA? Practice
(A) All groups have equal size
(B) Observations are independent
(C) Each group comes from a normal population
(D) All groups have equal variance
Correct answer: (A)
Solution: The assumptions are independence, normality and equal variances; a one-way ANOVA can be done with unequal group sizes too.
Q56. In the series 8, 12, 13, 17, 20, the 3-year moving average placed against the third term is: Practice
(A) 13
(B) 15.5
(C) 11
(D) 14
Correct answer: (D)
Solution: Against the third term: (12 + 13 + 17)/3 = 42/3 = 14. (The first moving average, (8 + 12 + 13)/3 = 11, goes against the second term.)
Q57. Ice-cream sales rise every summer. Which component of a time series is this? Practice
(A) Cyclical variation
(B) Seasonal variation
(C) Secular trend
(D) Irregular variation
Correct answer: (B)
Solution: A fluctuation that repeats regularly within a year = seasonal variation.
Q58. The trend equation is Y = 50 + 4X (origin 2020, X in years). The trend value for 2026 is: Practice
(A) 54
(B) 74
(C) 70
(D) 78
Correct answer: (B)
Solution: 2026 → X = 6; Y = 50 + 4 × 6 = 74.
Q59. The seasonal index of the second quarter is 125 and the actual sales in that quarter are ₹500 lakh. The deseasonalised sales are: Practice
(A) ₹625 lakh
(B) ₹375 lakh
(C) ₹500 lakh
(D) ₹400 lakh
Correct answer: (D)
Solution: Deseasonalised = actual ÷ index × 100 = 500 ÷ 125 × 100 = 400.
Q60. The (adjusted) seasonal indices of the four quarters add up to: Practice
(A) 400
(B) 100
(C) 1200
(D) 4
Correct answer: (A)
Solution: Each quarter averages 100, so the total is 4 × 100 = 400; monthly indices add up to 1200.
Q61. Which index number satisfies both the Time Reversal and Factor Reversal tests? Practice
(A) Fisher's ideal index
(B) Laspeyres index
(C) Paasche index
(D) Simple aggregate index
Correct answer: (A)
Solution: Fisher = √(L × P) satisfies both tests, hence 'ideal'. Laspeyres and Paasche satisfy neither.
Q62. With base 2015 = 100, the index is 150 in 2020 and 240 in 2024. Shifting the base to 2020, the 2024 index becomes: Practice
(A) 90
(B) 140
(C) 160
(D) 62.5
Correct answer: (C)
Solution: 240/150 × 100 = 160.
Q63. A worker’s money wage is ₹18,000 and the consumer price index is 150. The real wage is: Practice
(A) ₹12,000
(B) ₹27,000
(C) ₹16,500
(D) ₹13,500
Correct answer: (A)
Solution: 18,000 ÷ 150 × 100 = ₹12,000.
Q64. In a year the Laspeyres index is 144 and the Paasche index is 121. Fisher’s ideal index is: Practice
(A) 132.5
(B) 144
(C) 132
(D) 121
Correct answer: (C)
Solution: Fisher = √(L × P) = √(144 × 121) = 12 × 11 = 132. (132.5 is the arithmetic mean — Dorbish–Bowley.)

📝 Practice questions

Quantitative Aptitude — Practice MCQs (Bilingual) Ch 25: Statistics (Advanced)

Practice MCQs SF Prep द्वारा तैयार (PYQ नहीं)। सभी उत्तर गणना द्वारा जाँचे गए।