ICAR AIEEA-PG - Group 07 Statistical Sciences (Master's admission and PG scholarship) · India · National Testing Agency for 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)
ICAR AIEEA-PG Group 07, CUET-PG Statistics, SSC JSO and state statistical services
Binomial, Poisson and normal: mean, variance, mode, fitting to data, normal-area calculations and the chi-square, t and F tests that follow.
How our questions for ICAR AIEEA-PG spread across the units of Probability Distributions & Sampling Distributions. Difficulty: 19 easy, 9 medium, 0 hard.
| Discrete distributions: Bernoulli, binomial, Poisson, geometric, negative binomial, hypergeometric and multinomial | 6 | |
| Continuous distributions: uniform, normal, exponential, gamma, beta, Cauchy, Weibull, lognormal | 5 | |
| Pearsonian system and fitting of distributions to agricultural data | 5 | |
| Relations and limiting forms among distributions | 4 | |
| Sampling distributions: chi-square, t and F and their properties | 4 | |
| Bivariate normal: marginal and conditional distributions | 2 | |
| Compound, mixture and truncated distributions | 1 | |
| Distributions of quadratic forms in normal variables (Cochran's theorem) | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A binomial variable has mean 6 and variance 2.4. What is n?
B. 10
q = variance/mean = 2.4/6 = 0.4, so p = 0.6 and n = 6/0.6 = 10. 15 comes from taking p = 0.4 (confusing p with q); 6 confuses n with 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
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 is uniform on (2, 8). What is Var(X)?
A. 3
Var = (b − a)²/12 = 36/12 = 3. 5 is the mean; 36 forgets to divide by 12.
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
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
(X, Y) is bivariate normal with σ_X = 2, σ_Y = 3 and ρ = 0.5. What is the distribution of X − Y?
D. Normal with mean μ_X − μ_Y and variance 7
Linear combinations of a bivariate normal are normal. Var(X − Y) = 4 + 9 − 2(0.5)(2)(3) = 7. 13 ignores the covariance; 19 adds it with the wrong sign.
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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