NCEES PE Agricultural and Biological Engineering · USA · NCEES (ASABE is the content partner)
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
NCEES PE Agricultural and Biological Engineering
Statistics is listed among the common system applications. It appears as applied analysis: a confidence interval on test data, a regression of machine or process performance, frequency analysis for a hydrological design (with the national guideline method), and quality or uniformity measures in processing and irrigation.
How our questions for NCEES PE Agricultural and Biological Engineering spread across the units of Probability, Statistics & Data Analysis. Difficulty: 0 easy, 18 medium, 11 hard.
| Frequency analysis of rainfall and floods | 6 | |
| Tests of hypothesis (z, t, chi-square, F) | 4 | |
| Statistical quality control charts | 4 | |
| Data analysis with spreadsheets and scripts | 4 | |
| Probability and random variables | 3 | |
| Sampling and sampling distributions | 3 | |
| Binomial, Poisson and normal distributions | 2 | |
| Correlation and linear regression | 2 | |
| Measures of central tendency and dispersion | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Two grain lots are blended: 10 t at 14 per cent moisture (wet basis) and 30 t at 18 per cent moisture (wet basis). What is the moisture content of the 40 t blend, wet basis?
C. 17.0 per cent
Wet-basis moisture is water mass per unit wet mass, so weight by wet mass: water = 0.14 × 10 + 0.18 × 30 = 1.4 + 5.4 = 6.8 t; 6.8/40 = 17.0 per cent. The simple average (14 + 18)/2 = 16 per cent ignores the lot sizes; 15 per cent weights the wrong way round (0.14 × 30 + 0.18 × 10 = 6.0 t, /40). 20.5 per cent is the blend's dry-basis moisture, 6.8/(40 − 6.8) = 6.8/33.2.
Further reading: Walpole et al., Probability and Statistics for Engineers and Scientists
A combine harvester averages 0.5 breakdowns per 100 hours of operation, and breakdowns follow a Poisson process. What is the probability of no breakdown in a 300-hour season?
C. 0.2231
Scale the rate to the interval: λ = 0.5 × 3 = 1.5. P(0) = e^(−1.5) = 0.2231. 0.6065 forgets to scale λ; 0.3347 is P(exactly one); 0.7769 is P(at least one).
Further reading: Montgomery and Runger, Applied Statistics and Probability for Engineers
Bags of fertiliser from a filler have normally distributed net mass with standard deviation 0.5 kg. The mean is set at 25.0 kg. Below what mass do the lightest 5 per cent of bags fall? (One-tailed 5 per cent point of Z = 1.645.)
D. 24.18 kg
x = μ − 1.645σ = 25.0 − 1.645 × 0.5 = 24.18 kg. Using 1.96 (the two-tailed 5 per cent point) gives 24.02 kg, which cuts off only 2.5 per cent; 25.82 kg is the upper 5 per cent point; 24.59 kg uses the variance (0.25) as if it were σ.
Further reading: Montgomery and Runger, Applied Statistics and Probability for Engineers
A survey is planned to estimate the proportion of farms using drip irrigation to within ±5 percentage points at 95 per cent confidence (z = 1.96). Nothing is known in advance about the proportion. What sample size is needed?
B. 385
With no prior estimate use p = 0.5, which maximises p(1 − p): n = z²p(1 − p)/E² = 1.96² × 0.25/0.05² = 384.16, rounded up to 385. 271 uses 90 per cent confidence; 100 is 1/E² × 0.25; 20 forgets to square.
Further reading: Walpole et al., Probability and Statistics for Engineers and Scientists
Two moisture meters are compared for precision. Meter A gives a sample variance of 0.64 from 11 readings; meter B gives 0.25 from 9 readings of the same sample. Testing equal variances at 5 per cent with the larger variance on top (upper 5 per cent F for 10 and 8 df = 3.35), what is the result?
A. F = 2.56; the variances are not significantly different
F = s1²/s2² = 0.64/0.25 = 2.56 with (11 − 1, 9 − 1) = (10, 8) df. Since 2.56 < 3.35 (tabled 3.35), equality of variances is not rejected. 1.60 is the ratio of SDs, not variances; 0.39 puts the smaller variance on top.
Further reading: Montgomery and Runger, Applied Statistics and Probability for Engineers
A fertiliser spreader specification requires the standard deviation of spread rate between trays to be no more than 0.40 kg/ha. Fifteen trays give s = 0.50 kg/ha. Testing H0: σ = 0.40 against H1: σ > 0.40 at 5 per cent (upper 5 per cent χ² for 14 df = 23.685), what is the conclusion?
B. χ² = 21.875; do not reject H0
χ² = (n − 1)s²/σ0² = 14 × 0.25/0.16 = 21.875 on 14 df. Since 21.875 < 23.685 (tabled 23.685), the evidence does not show σ exceeds 0.40. Using n = 15 gives 23.438, which crosses the critical value and wrongly rejects; using SDs instead of variances gives 17.500.
Further reading: Montgomery and Runger, Applied Statistics and Probability for Engineers
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
Annual rainfall at a station is approximately normal with mean 850 mm and standard deviation 170 mm. What annual rainfall is equalled or exceeded in 90 per cent of years? (z for 10 per cent in one tail = 1.282.)
C. 632.1 mm
Rainfall exceeded in 90 per cent of years is the 10th percentile: x = μ − 1.282σ = 850 − 1.282 × 170 = 632.1 mm. 1067.9 mm is exceeded in only 10 per cent of years; 570.4 mm is the 95 per cent dependable value; 765 mm is simply 90 per cent of the mean.
Further reading: Chow, Maidment and Mays, Applied Hydrology; Subramanya, Engineering Hydrology
An engineer writes a script that simulates 10,000 seasons of random rainfall to estimate the probability that a tank runs dry. Colleagues get slightly different answers each time they run it. What is the simplest step that makes the result exactly reproducible?
C. Fix the random number generator's seed in the script and record it
A fixed seed makes the pseudo-random sequence, and therefore the result, identical on every run. More simulations reduce the variation between runs but do not remove it; rounding hides it; computer speed is irrelevant.
Further reading: NIST/SEMATECH e-Handbook of Statistical Methods
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