GATE - Statistics (ST) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersEvery standard distribution a statistician must recognise and derive from: discrete and continuous families, their relations and limiting forms, compound and truncated distributions, order statistics, the bivariate normal and the sampling distributions chi-square, t and F.
In the exams: AICE-JRF Ag Stat Unit 1 and Unit 2 'compound and truncated distribution, order statistics' (heavy); GATE ST distributions and order statistics (heavy); IIT JAM MS (heavy); UPSC ISS Statistics I (heavy)
GATE Statistics (ST) and IIT JAM Mathematical Statistics
Numerical and multiple-select items: moments from MGFs, identifying distributions of sums, ratios and transformations, densities and expectations of order statistics, bivariate normal conditionals, chi-square, t and F relations, and exponential-family membership.
How our questions for GATE spread across the units of Probability Distributions & Sampling Distributions. Difficulty: 18 easy, 26 medium, 17 hard.
| Order statistics: distributions of single, joint and range | 10 | |
| Bivariate normal: marginal and conditional distributions | 8 | |
| Sampling distributions: chi-square, t and F and their properties | 8 | |
| Exponential family of distributions | 7 | |
| Continuous distributions: uniform, normal, exponential, gamma, beta, Cauchy, Weibull, lognormal | 6 | |
| Discrete distributions: Bernoulli, binomial, Poisson, geometric, negative binomial, hypergeometric and multinomial | 5 | |
| Relations and limiting forms among distributions | 5 | |
| Compound, mixture and truncated distributions | 5 | |
| Non-central chi-square, t and F | 3 | |
| Distributions of quadratic forms in normal variables (Cochran's theorem) | 3 | |
| Pearsonian system and fitting of distributions to agricultural data | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Wet-day rainfall X is gamma with shape 3 and rate 0.5 per mm, so f(x) ∝ x² e^(−0.5x). What is Var(X)?
C. 12 mm²
With rate β, mean = α/β = 6 mm and variance = α/β² = 3/0.25 = 12 mm². 6 is the mean; 0.75 = αβ² wrongly treats 0.5 as a scale parameter (a parameterisation slip).
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Johnson, Kotz and Balakrishnan, Continuous Univariate Distributions; Johnson, Kemp and Kotz, Univariate Discrete Distributions
X has a Cauchy distribution with location 5 and scale 2. What is its interquartile range?
A. 4
The Cauchy quartiles are θ ± σ = 3 and 7, so the IQR is 2σ = 4. The Cauchy has no mean or variance, but its quantiles exist. 1.349σ is the IQR of a normal.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Johnson, Kotz and Balakrishnan, Continuous Univariate Distributions; Johnson, Kemp and Kotz, Univariate Discrete Distributions
V has a chi-square distribution with 2 degrees of freedom. What is P(V > 6)?
B. 0.0498
χ²₂ is exponential with mean 2: P(V > 6) = e^(−6/2) = e^(−3) = 0.0498. e^(−6) treats the mean as 1.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Johnson, Kotz and Balakrishnan, Continuous Univariate Distributions; Johnson, Kemp and Kotz, Univariate Discrete Distributions
X₁, ..., X₄ are independent exponential variables with rate 2. What is the distribution of their sum?
A. Gamma with shape 4 and rate 2, mean 2
The MGF of the sum is [2/(2 − t)]⁴, the gamma (Erlang) MGF with shape 4 and rate 2: mean 4/2 = 2. Rate 8 is the distribution of the minimum of the four, not the sum.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Johnson, Kotz and Balakrishnan, Continuous Univariate Distributions; Johnson, Kemp and Kotz, Univariate Discrete Distributions
Insects per plant are Poisson with mean λ, and λ varies from plant to plant according to a gamma distribution. What is the unconditional distribution of insects per plant?
D. Negative binomial
Integrating e^(−λ)λˣ/x! against a gamma(k, β) density gives C(x + k − 1, x)(β/(1 + β))ᵏ(1/(1 + β))ˣ: negative binomial with r = k. The variance exceeds the mean by the variance of the mixing distribution, so it cannot be Poisson.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Johnson, Kotz and Balakrishnan, Continuous Univariate Distributions; Johnson, Kemp and Kotz, Univariate Discrete Distributions
A large sample of n = 100 plot yields is drawn from a normal population with σ = 10. What is the approximate standard error of the sample median?
D. 1.2533
Var(median) ≈ 1/[4nf(ξ)²] with f(ξ) = 1/(σ√(2π)), giving πσ²/(2n) = 1.5708; SE = 1.2533. 1.0000 is the standard error of the mean, σ/√n; 1.5708 is the variance, not the standard error.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Johnson, Kotz and Balakrishnan, Continuous Univariate Distributions; Johnson, Kemp and Kotz, Univariate Discrete Distributions
Plant height X (cm) and grain yield Y (g) are bivariate normal with μ_X = 100, σ_X = 10, μ_Y = 50, σ_Y = 8 and ρ = 0.6. What is the mean yield of plants 110 cm tall?
D. 54.8 g
E(Y | X = x) = μ_Y + ρ(σ_Y/σ_X)(x − μ_X) = 50 + 0.6 × 0.8 × 10 = 54.8 g. Using σ_X/σ_Y inverts the slope and gives 57.5 g; 58.0 g assumes perfect correlation.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Johnson, Kotz and Balakrishnan, Continuous Univariate Distributions; Johnson, Kemp and Kotz, Univariate Discrete Distributions
A random sample of 16 is drawn from N(μ, 4). What is the variance of the sample variance s² (divisor n − 1)?
B. 2.1333
(n − 1)s²/σ² ~ χ²₁₅ with variance 30, so Var(s²) = σ⁴ × 2(n − 1)/(n − 1)² = 2σ⁴/(n − 1) = 32/15 = 2.1333. 30 is the variance of the chi-square itself; dividing by n gives 2.0000.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Johnson, Kotz and Balakrishnan, Continuous Univariate Distributions; Johnson, Kemp and Kotz, Univariate Discrete Distributions
X has the F distribution with (5, 12) degrees of freedom. What is E(X)?
B. 1.2
E(F) = n/(n − 2) for denominator df n > 2: 12/10 = 1.2, regardless of the numerator df 5. 1.67 = 5/3 uses the numerator df.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Johnson, Kotz and Balakrishnan, Continuous Univariate Distributions; Johnson, Kemp and Kotz, Univariate Discrete Distributions
For a random sample from a beta(a, b) distribution with both parameters unknown, what is the natural sufficient statistic?
A. (Σ log Xᵢ, Σ log(1 − Xᵢ))
log f = (a − 1) log x + (b − 1) log(1 − x) − log B(a, b), so T = (log x, log(1 − x)). Note ΣXᵢ and Σ(1 − Xᵢ) carry the same information and are not sufficient here. ΠXᵢ alone covers only a.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Johnson, Kotz and Balakrishnan, Continuous Univariate Distributions; Johnson, Kemp and Kotz, Univariate Discrete Distributions
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