NCEES FE Other Disciplines · USA · NCEES
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 FE Other Disciplines
Probability and Statistics is its own area with 6–9 questions, answered with the FE Reference Handbook on screen, which supplies the formulas and the distribution tables. Expect descriptive statistics, probability distributions, expected value, confidence intervals, hypothesis tests, regression and goodness of fit. Speed in reading the right table and choosing sample versus population formulas matters more than memorising.
How our questions for NCEES FE Other Disciplines spread across the units of Probability, Statistics & Data Analysis. Difficulty: 22 easy, 25 medium, 13 hard.
| Tests of hypothesis (z, t, chi-square, F) | 11 | |
| Probability and random variables | 9 | |
| Binomial, Poisson and normal distributions | 9 | |
| Statistical quality control charts | 9 | |
| Sampling and sampling distributions | 7 | |
| Correlation and linear regression | 6 | |
| Data analysis with spreadsheets and scripts | 5 | |
| Measures of central tendency and dispersion | 4 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Events A and B are independent with P(A) = 0.4 and P(B) = 0.3. What is P(A ∪ B)?
B. 0.58
P(A ∩ B) = 0.4 × 0.3 = 0.12 for independent events, so P(A ∪ B) = 0.4 + 0.3 − 0.12 = 0.58. 0.70 treats them as mutually exclusive; 0.12 is the intersection; 0.46 subtracts the intersection twice.
Further reading: Montgomery and Runger, Applied Statistics and Probability for Engineers
The operating life of a filter element is exponentially distributed with mean 500 hours. A filter has already run 300 hours without failing. What is the probability that it lasts at least another 200 hours?
B. 0.6703
The exponential distribution is memoryless: P(T > 500 | T > 300) = P(T > 200) = e^(−200/500) = 0.6703. 0.3679 is P(T > 500) from new, ignoring the condition; 1.2214 divides the wrong way and exceeds 1; 0.3297 is the probability of failing within the next 200 h.
Further reading: Montgomery and Runger, Applied Statistics and Probability for Engineers
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
Sugarcane stalk mass has mean 12 N and standard deviation 3 N. For a random sample of 36 stalks, what is the probability that the sample mean exceeds 12.8 N? (Φ(1.6) = 0.9452.)
A. 0.0548
SE = σ/√n = 3/6 = 0.5 N. z = (12.8 − 12)/0.5 = 1.6. P(x̄ > 12.8) = 1 − 0.9452 = 0.0548. Using σ instead of the standard error gives z = 0.267 and 0.3949; 0.9452 is the lower area; 0.1096 doubles the tail.
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
A seed company claims 90 per cent germination. In a test of 200 seeds, 170 germinate. Testing H0: p = 0.90 against H1: p < 0.90 at the 5 per cent level (critical z = −1.645), what is the result?
C. z = -2.36; reject the claim
p̂ = 170/200 = 0.85. Under H0 the SE uses p0: √(0.9 × 0.1/200) = 0.02121. z = (0.85 − 0.90)/0.02121 = -2.36 < −1.645, so the claim is rejected. The 2.576 value belongs to a two-tailed 1 per cent test; a negative z is exactly what H1: p < 0.90 predicts, so it is compared with −1.645; omitting n from the SE gives -0.167; the test of a hypothesised p uses p0, not p̂, in the SE.
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
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
A spreadsheet t-test function returns p = 0.03, and its help text says the value is one-tailed. The engineer needs a two-sided test at the 5 per cent level. What is the correct conclusion?
D. The two-tailed p-value is 0.06, so H0 is not rejected at 5 per cent
For a symmetric t distribution the two-tailed p-value is twice the one-tailed value: 2 × 0.03 = 0.06 > 0.05, so H0 is not rejected. Halving instead of doubling, or using the complement 0.97, are common slips; check which convention a function uses before reading its output.
Further reading: NIST/SEMATECH e-Handbook of Statistical Methods
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