GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
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
GATE — Agricultural Engineering (AG)
Probability and statistics sit inside the Engineering Mathematics section and appeared in every paper Laisha counted from 2019 to 2026, usually as one or two short computational items, often numerical-answer type: a conditional or Bayes probability, a binomial or Poisson probability, a normal-table calculation, the mean or variance of a random variable from its density, a correlation coefficient or regression slope from a small table. Hydrology questions in the soil and water sections reuse return period, risk and Gumbel frequency factors; processing and machinery questions reuse CV and uniformity.
How our questions for GATE spread across the units of Probability, Statistics & Data Analysis. Difficulty: 18 easy, 27 medium, 17 hard.
| Design of experiments: CRD and RBD | 10 | |
| Frequency analysis of rainfall and floods | 9 | |
| Tests of hypothesis (z, t, chi-square, F) | 8 | |
| Binomial, Poisson and normal distributions | 7 | |
| Correlation and linear regression | 7 | |
| Probability and random variables | 6 | |
| Sampling and sampling distributions | 6 | |
| Measures of central tendency and dispersion | 4 | |
| Statistical quality control charts | 4 | |
| 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
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
Two judges rank 8 samples of dried mango slices for colour. The differences between the two ranks for the samples are 1, −1, 2, 0, −2, 1, −1 and 0. What is Spearman's rank correlation coefficient?
A. 0.8571
Σd² = 1 + 1 + 4 + 0 + 4 + 1 + 1 + 0 = 12. rs = 1 − 6Σd²/(n(n² − 1)) = 1 − 6 × 12/(8 × 63) = 1 − 72/504 = 0.8571. Using n² + 1 gives 0.8615; omitting the 6 gives 0.9762; 0.1429 forgets to subtract from 1.
Further reading: Montgomery and Runger, Applied Statistics and Probability for Engineers
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
What does bootstrap resampling do?
B. It repeatedly draws samples of the same size, with replacement, from the observed data to estimate the sampling variability of a statistic without assuming its distribution
Bootstrap resamples the data already collected, with replacement, many times; the spread of the statistic across resamples estimates its standard error and gives confidence intervals. It does not clean data, fit a normal curve or require new fieldwork.
Further reading: NIST/SEMATECH e-Handbook of Statistical Methods
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