Indian Statistical Service Examination (with Indian Economic Service) · India · Union Public Service Commission
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)
How our questions for Indian Statistical Service Examination spread across the units of Probability Distributions & Sampling Distributions. Difficulty: 8 easy, 19 medium, 5 hard.
| Pearsonian system and fitting of distributions to agricultural data | 6 | |
| Discrete distributions: Bernoulli, binomial, Poisson, geometric, negative binomial, hypergeometric and multinomial | 4 | |
| Continuous distributions: uniform, normal, exponential, gamma, beta, Cauchy, Weibull, lognormal | 3 | |
| Relations and limiting forms among distributions | 3 | |
| Bivariate normal: marginal and conditional distributions | 3 | |
| Sampling distributions: chi-square, t and F and their properties | 3 | |
| Distributions of quadratic forms in normal variables (Cochran's theorem) | 3 | |
| Compound, mixture and truncated distributions | 2 | |
| Non-central chi-square, t and F | 2 | |
| Exponential family of distributions | 2 | |
| Order statistics: distributions of single, joint and range | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
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
Calls to a farmers' helpline in a day are Poisson with mean 100. Using the normal approximation with continuity correction, what is P(at least 120 calls)?
B. 0.0256
Mean = variance = 100, SD 10. P(X ≥ 120) ≈ P(Z > (119.5 − 100)/10) = P(Z > 1.95) = 0.0256. Without correction z = 2 gives 0.0228; dividing by the variance gives z = 0.2.
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 binomial distribution is fitted to germination in 100 trays of 5 seeds, with an average of 4 seeds germinating per tray. What is the expected number of trays in which all 5 seeds germinate?
A. 32.77
p̂ = x̄/n = 4/5 = 0.8; expected trays = 100 × 0.8⁵ = 32.77. 0.8⁴ uses the wrong power; 100 × 0.2⁵ = 0.03 is the expected number with none germinating.
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 σ_X = 10, σ_Y = 8 and ρ = 0.6. What is the standard deviation of yield among plants of any fixed height?
A. 6.4 g
Var(Y | X) = σ_Y²(1 − ρ²) = 64 × 0.64, so SD = 8√0.64 = 6.4 g, the same at every height. 5.12 g multiplies σ_Y (not σ_Y²) by 1 − ρ²; 8.0 g ignores the information in X.
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
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