ASRB NET - Agricultural Statistics (and Computer Application / Bioinformatics) · India · Agricultural Scientists Recruitment Board (ICAR)
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)
ASRB NET / ARS Agricultural Statistics
Applied judgement: which distribution fits a given agricultural count or measurement, testing for over-dispersion, and planning replication through non-central F.
How our questions for ASRB NET spread across the units of Probability Distributions & Sampling Distributions. Difficulty: 13 easy, 15 medium, 6 hard.
| Compound, mixture and truncated distributions | 8 | |
| Pearsonian system and fitting of distributions to agricultural data | 4 | |
| Continuous distributions: uniform, normal, exponential, gamma, beta, Cauchy, Weibull, lognormal | 3 | |
| Order statistics: distributions of single, joint and range | 3 | |
| Non-central chi-square, t and F | 3 | |
| Exponential family of distributions | 3 | |
| Distributions of quadratic forms in normal variables (Cochran's theorem) | 3 | |
| Discrete distributions: Bernoulli, binomial, Poisson, geometric, negative binomial, hypergeometric and multinomial | 2 | |
| Relations and limiting forms among distributions | 2 | |
| Bivariate normal: marginal and conditional distributions | 2 | |
| Sampling distributions: chi-square, t and F and their properties | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Counts of aphids on 200 leaves have mean 1.5 and variance 4.2. Which conclusion and moment estimate are correct?
A. The counts are aggregated; negative binomial k ≈ 0.83
Index of dispersion 4.2/1.5 = 2.8 > 1 means clumping. Moment estimate k̂ = x̄²/(s² − x̄) = 2.25/2.7 = 0.83. 2.80 is the dispersion index itself, not k. A binomial would need variance below the mean.
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
Plant height in a population is normal with mean 100 cm and standard deviation 10 cm. The taller half is selected (truncation at the mean). What is the mean height of the selected plants?
A. 107.98 cm
Truncated below at a = μ, α = 0: mean = μ + σφ(0)/[1 − Φ(0)] = 100 + 10 × 0.3989/0.5 = 107.98 cm (selection intensity i = 0.798). Forgetting to divide by the proportion selected gives 103.99 cm.
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 of size n from a continuous distribution with CDF F and density f, which is the density of the r-th order statistic X₍ᵣ₎?
D. n!/[(r − 1)!(n − r)!] F(x)^(r−1) [1 − F(x)]^(n−r) f(x)
r − 1 observations fall below x, one at x and n − r above: the multinomial count n!/[(r − 1)! 1! (n − r)!] times F^(r−1) f (1 − F)^(n−r). The C(n, r)Fʳ(1 − F)ⁿ⁻ʳ form is one term of the CDF sum, not a density.
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
Four drip emitters clog independently, each after an exponential time with mean 10 days. What is the mean time until the first emitter clogs?
B. 2.5 days
The minimum of n exponentials with rate λ is exponential with rate nλ: mean 10/4 = 2.5 days. 40 days multiplies instead of divides; 20.8 days = 10(1 + 1/2 + 1/3 + 1/4) is the mean time until the last one clogs.
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 one-way trial has 3 varieties with true effects −2, 0 and 2 kg, 4 replicates each and error variance 4 kg². With λ = rΣτᵢ²/σ², what is the non-centrality parameter of the treatment F-statistic?
C. 8
λ = 4 × (4 + 0 + 4)/4 = 8. Omitting r gives 2; dividing by σ = 2 instead of σ² gives 16; using |τ| instead of τ² gives 4.
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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