ICAR AIEEA-PG — Agricultural Engineering & Technology (Major Subject Group 10) · India · National Testing Agency for ICAR
Original practice questions written to the published syllabus · not past papersDescribing data, probability distributions, regression and designed experiments, with the frequency analysis that hydrology and quality control rely on.
In the exams: GATE AG Section 1 (probability and statistics asked every year 2019–26); ICAR AIEEA-PG Unit I; NCEES FE 6–9 questions; NCEES PE statistics
ICAR AIEEA-PG — Agricultural Engineering & Technology (Unit I)
Unit I combines mathematics and statistics. Question papers are not published, so emphasis cannot be counted; the syllabus supports single-answer MCQs on measures of central tendency and dispersion, probability rules, standard distributions, correlation and regression, tests of significance and basic experimental design (CRD, RBD, degrees of freedom in the ANOVA).
How our questions for ICAR AIEEA-PG spread across the units of Probability, Statistics & Data Analysis. Difficulty: 32 easy, 29 medium, 17 hard.
| Design of experiments: CRD and RBD | 12 | |
| Measures of central tendency and dispersion | 9 | |
| Binomial, Poisson and normal distributions | 9 | |
| Correlation and linear regression | 9 | |
| Frequency analysis of rainfall and floods | 9 | |
| Probability and random variables | 8 | |
| Sampling and sampling distributions | 8 | |
| Tests of hypothesis (z, t, chi-square, F) | 8 | |
| Statistical quality control charts | 5 | |
| Data analysis with spreadsheets and scripts | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Eight nozzles on a sprayer boom deliver 0.98, 1.02, 1.05, 0.95, 1.00, 1.04, 0.97 and 0.99 L/min. What is the coefficient of variation of the discharge (using the sample standard deviation)?
D. 3.46 per cent
Mean = 1.0000 L/min; sample SD = 0.03464 L/min; CV = s/x̄ × 100 = 3.46 per cent. Using the population SD (divide by n) gives 3.24 per cent; using the variance instead of the SD gives 0.120; forgetting to multiply by 100 but still writing 'per cent' gives 0.0346.
Further reading: Walpole et al., Probability and Statistics for Engineers and Scientists
A moderately skewed distribution of daily evaporation has mean 42 mm and median 40 mm. Using Karl Pearson's empirical relation, what is the approximate mode?
A. 36 mm
Mode ≈ 3 × median − 2 × mean = 3(40) − 2(42) = 120 − 84 = 36 mm. Writing 3 × mean − 2 × median gives 46 mm; the mode lies on the far side of the median from the mean in a skewed distribution, so a value above 40 cannot be right for mean > median. 38 mm places the mode as far below the median as the mean is above it (40 − 2); 44 mm adds the mean–median gap to the mean (42 + 2).
Further reading: Walpole et al., Probability and Statistics for Engineers and Scientists
Filled bags of wheat have weights that are normally distributed with mean 50.2 kg and standard deviation 0.4 kg. What proportion of bags weigh less than 49.5 kg? (Take Φ(1.75) = 0.9599.)
B. 0.0401
z = (49.5 − 50.2)/0.4 = -1.75. P(Z < −1.75) = 1 − Φ(1.75) = 1 − 0.9599 = 0.0401. 0.9599 is the proportion above 49.5 kg; 0.0802 doubles the tail (a two-tailed slip); 0.4599 is the area between the mean and z.
Further reading: Montgomery and Runger, Applied Statistics and Probability for Engineers
Drawbar pull x (kN) and wheel slip y (per cent) for a tractor are: (2, 5.1), (4, 7.9), (6, 10.8), (8, 14.2), (10, 16.5). What is Pearson's correlation coefficient r?
D. 0.9987
Sxx = 40.0, Syy = 84.900, Sxy = 58.20. r = Sxy/√(Sxx·Syy) = 58.20/√(40.0 × 84.900) = 0.9987. 0.9974 is R²; 0.7990 mixes divisors (covariance with n in the numerator, SDs with n − 1); 1.4550 is the slope, not r.
Further reading: Montgomery and Runger, Applied Statistics and Probability for Engineers
A farm has three distinct soil types, and soil nitrate differs sharply between them. To estimate the farm's mean nitrate most precisely for a fixed number of samples, which method is best?
C. Stratified random sampling with the soil types as strata
When the population splits into internally similar groups that differ from one another, sampling each stratum removes the between-strata variation from the sampling error, so the estimate is more precise than a simple random sample of the same size. Cluster sampling is usually less precise per sample; convenience sampling is biased.
Further reading: Walpole et al., Probability and Statistics for Engineers and Scientists
A contingency table classifies tractors into 3 age groups and 4 fault types. How many degrees of freedom does the chi-square test of independence have?
D. 6
df = (r − 1)(c − 1) = (3 − 1)(4 − 1) = 6. 12 is r × c; 11 is rc − 1 (a goodness-of-fit count); 7 is r + c.
Further reading: Montgomery and Runger, Applied Statistics and Probability for Engineers
In a CRD with 4 replications the error mean square is 2.5 on 15 degrees of freedom. What is the least significant difference between two treatment means at 5 per cent (t for 15 df, two-tailed = 2.131)?
C. 2.383
LSD = t × √(2 MSE/r) = 2.131 × √(2 × 2.5/4) = 2.131 × 1.1180 = 2.383. Omitting the 2 gives t times the SE of one mean (1.685); forgetting to divide by r gives 4.765; using z = 1.96 ignores the small error df.
Further reading: Gomez and Gomez, Statistical Procedures for Agricultural Research
A farm-pond spillway is designed for the 50-year flood and has a design life of 25 years. What is the risk that the design flood is equalled or exceeded at least once during its life?
B. 0.3965
R = 1 − (1 − 1/T)ⁿ = 1 − 0.98²⁵ = 1 − 0.6035 = 0.3965. 0.5 comes from n/T = 25/50; 0.6035 is the probability of no exceedance; 0.02 is the annual probability.
Further reading: Chow, Maidment and Mays, Applied Hydrology; Subramanya, Engineering Hydrology
What is the probability that a 10-year storm occurs exactly once in a 10-year period, assuming independent years?
C. 0.3874
Binomial with n = 10, P = 0.1: P(exactly 1) = 10 × 0.1 × 0.9⁹ = 0.3874. 1.0 is the common misreading of 'once in 10 years'; 0.6513 is the probability of at least one; 0.3487 is the probability of none.
Further reading: Chow, Maidment and Mays, Applied Hydrology; Subramanya, Engineering Hydrology
What is the probability that a 20-year flood is not exceeded in any of the next 5 years, assuming independent years?
D. 0.7738
Annual non-exceedance = 1 − 1/20 = 0.95; for 5 independent years 0.95⁵ = 0.7738. 0.2262 is the risk of at least one exceedance; 0.75 = 1 − 5/20 and 0.25 = 5/20 treat probabilities as additive.
Further reading: Chow, Maidment and Mays, Applied Hydrology; Subramanya, Engineering Hydrology
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