PE Industrial and Systems Exam Prep
Start with Problem 1 below. It's one of 35 original, unofficial practice problems, 7 for each content area on the current NCEES PE Industrial and Systems specification, each with a full worked solution. A study plan, exam-day rules, and verified exam facts follow the problems.
By Castleport Test Prep Editorial Team · Exam facts last verified October 9, 2026
Practice problems with worked solutions
Original practice written for this page, not NCEES exam questions. Mark each one Correct, Guessed, Slow (more than about 6 minutes), or Wrong. You'll use those marks to build your study plan.
Systems Engineering · Problems 1–7
1. Critical path and float
PEIS-SE-01 · Systems Engineering · Multiple choice
A project has five activities (durations in days). A takes 4 and B takes 6; both can start at once. C takes 3 and follows A. D takes 5 and needs both A and B finished. E takes 2 and follows both C and D. What is the project duration, and how much total float does activity C have?
Show answer and solution
Answer: A. 13 days; C has 4 days of float
List the paths: A–C–E = 4 + 3 + 2 = 9 days. A–D–E = 4 + 5 + 2 = 11 days. B–D–E = 6 + 5 + 2 = 13 days. The longest path, B–D–E, sets the duration: 13 days.
For C's float, run the forward pass: C can finish by day 7 (4 + 3). D can't start until B ends on day 6, so D finishes on day 11. E starts at the later of the two, day 11. C's latest finish is therefore day 11, and its latest start is 11 − 3 = day 8. Float = latest start − earliest start = 8 − 4 = 4 days.
Why the others miss:
- B: 11 days is A–D–E. D also waits on B, which pushes the network to 13 days.
- C: C is not on the longest path, so it has float.
- D: 9 days is the shortest path. Project duration is the longest path.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
2. PERT completion probability
PEIS-SE-02 · Systems Engineering · Multiple choice
One activity on a project's critical path has PERT estimates of 2 days (optimistic), 5 days (most likely), and 14 days (pessimistic). The rest of the critical path has an expected duration of 24 days and a variance of 5 days². Assuming the path duration is approximately normal, what is the probability the path finishes within 33 days?
Show answer and solution
Answer: C. 0.841
PERT expected time = (a + 4m + b) / 6 = (2 + 20 + 14) / 6 = 6 days. Variance = ((b − a) / 6)² = (12 / 6)² = 4 days².
Path mean = 24 + 6 = 30 days. Variances add: 5 + 4 = 9, so σ = 3 days.
z = (33 − 30) / 3 = 1.0, and P(z ≤ 1.0) ≈ 0.841.
Why the others miss:
- A: 0.500 treats 33 days as the mean.
- B: 0.761 adds standard deviations (2 + √5) instead of variances. Variances add; standard deviations do not.
- D: 0.909 uses the most likely time (5 days) instead of the PERT expected time (6 days).
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
3. Single-server queue wait
PEIS-SE-03 · Systems Engineering · Multiple choice
Customers arrive at a single inspection station as a Poisson process at 12 per hour. Inspection times are exponential with a mean rate of 15 per hour. What is the average time a customer waits in line before inspection begins?
Show answer and solution
Answer: C. 16 minutes
This is an M/M/1 queue with λ = 12/h and μ = 15/h. Utilization ρ = λ/μ = 0.8.
Time in queue: Wq = λ / [μ(μ − λ)] = 12 / (15 × 3) = 0.267 h = 16 minutes.
Check: time in system W = 1/(μ − λ) = 1/3 h = 20 min, and W − Wq = 4 min, which is the mean service time (60/15). The numbers agree.
Why the others miss:
- A: 3.2 is Lq, the average number waiting, not a time.
- B: 4 is L, the average number in the system, not a time. It's also equal to the mean service time in minutes, a coincidence that catches people.
- D: 20 minutes is W, total time in system. The question asks for waiting before service starts.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
4. Bottleneck and Little's Law
PEIS-SE-04 · Systems Engineering · Multiple choice
A serial line has three stations. Station 1: one machine, 2.0 min per unit. Station 2: two identical machines in parallel, 5.0 min per unit each. Station 3: one machine, 1.5 min per unit. Average work-in-process is 36 units. What are the line's throughput and average flow time?
Show answer and solution
Answer: A. 24 units/h; 1.5 h
Station capacities: S1 = 60/2.0 = 30/h. S2 = 2 × (60/5.0) = 24/h. S3 = 60/1.5 = 40/h. The slowest station, S2, limits the line: 24 units/h.
Little's Law: WIP = throughput × flow time, so flow time = 36 / 24 = 1.5 h.
Why the others miss:
- B: 12/h counts only one of the two parallel machines at Station 2.
- C: 30/h is Station 1's capacity. The line can't beat its slowest station.
- D: 40/h is the fastest station, the opposite of a bottleneck.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
5. Linear program optimum
PEIS-SE-05 · Systems Engineering · Multiple choice
Maximize Z = 40x + 30y subject to 2x + y ≤ 100, x + y ≤ 80, x ≥ 0, y ≥ 0. What is the optimal value of Z?
Show answer and solution
Answer: C. 2,600
For a two-variable LP, check the corner points of the feasible region.
- (0, 0): Z = 0
- (50, 0): Z = 2,000
- (0, 80): Z = 2,400
- Intersection of 2x + y = 100 and x + y = 80: subtract to get x = 20, so y = 60. Z = 800 + 1,800 = 2,600
The maximum is 2,600 at x = 20, y = 60.
Why the others miss:
- A: 2,000 is the corner (50, 0), which leaves the second constraint slack.
- B: 2,400 is the corner (0, 80). It's feasible but not optimal.
- D: 3,200 comes from (20, 80), which violates x + y ≤ 80.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
6. Two-state Markov chain
PEIS-SE-06 · Systems Engineering · Fill in the blank (numeric)
A machine is checked at the start of each shift. If it is up, it will be up next shift with probability 0.90. If it is down, it will be up next shift with probability 0.60. In the long run, what fraction of shifts start with the machine up? Enter a decimal to three places.
Show answer and solution
Answer: 0.857 (accept 0.856–0.858)
Let π_U and π_D be the long-run probabilities. Balance for the up state: π_U = 0.90 π_U + 0.60 π_D, so 0.10 π_U = 0.60 π_D and π_U = 6 π_D.
With π_U + π_D = 1: 7 π_D = 1, π_D = 1/7, and π_U = 6/7 = 0.857.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
7. Verification vs. validation
PEIS-SE-07 · Systems Engineering · Select all that apply
You built a simulation of a distribution center. Which of these are validation activities rather than verification? Select all that apply.
Show answer and solution
Answer: A and C
Verification asks, "Did we build the model right?" It checks that the code does what the design says. Validation asks, "Did we build the right model?" It checks that the model represents the real system.
- A compares the model with real data: validation.
- B checks code against its own design: verification.
- C is face validity, an expert judging realism: validation.
- D checks the logic against a known analytical answer: verification.
Correct selections: A and C. Like the real exam, this item is all-or-nothing; there is no partial credit.
Source: Standard simulation terminology, explained in our own words; topic: NCEES exam specifications
Facilities Engineering and Planning · Problems 8–14
8. Machines required
PEIS-FA-01 · Facilities Engineering and Planning · Multiple choice
A cell must ship 9,000 good units per week. The process scraps 5% of what it starts. Standard time is 4.0 minutes per unit started. The plant runs 2 shifts × 8 hours × 5 days, and machines are available and efficient 85% of scheduled time. How many machines are required?
Show answer and solution
Answer: C. 10
Units to start = 9,000 / 0.95 = 9,473.7.
Machine-hours needed = 9,473.7 × 4.0 / 60 = 631.6 h per week.
Effective hours per machine = 2 × 8 × 5 × 0.85 = 68 h per week.
Machines = 631.6 / 68 = 9.29. You can't buy 0.29 of a machine, so round up: 10 machines.
Why the others miss:
- A: 8 ignores the 85% factor (631.6 / 80 = 7.9).
- B: 9 either rounds 9.29 down, which leaves the cell short, or ignores scrap (600 / 68 = 8.8).
- D: 19 uses only one shift (631.6 / 34 = 18.6).
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
9. Center-of-gravity location
PEIS-FA-02 · Facilities Engineering and Planning · Multiple choice
A new cross-dock will serve three customers. Customer locations (grid miles) and annual loads: (2, 8) with 300 loads, (6, 2) with 200 loads, and (10, 6) with 500 loads. Using the center-of-gravity method, where should the cross-dock go?
Show answer and solution
Answer: C. (6.8, 5.8)
Weight each coordinate by load.
X = (2 × 300 + 6 × 200 + 10 × 500) / 1,000 = 6,800 / 1,000 = 6.8
Y = (8 × 300 + 2 × 200 + 6 × 500) / 1,000 = 5,800 / 1,000 = 5.8
The center of gravity is a good starting point, not a guaranteed optimum. It doesn't exactly minimize rectilinear or straight-line distance, and real sites depend on roads and land.
Why the others miss:
- A: (6.0, 5.3) is the unweighted average of the three locations.
- B: (5.8, 6.8) swaps the X and Y results.
- D: (10, 6) just picks the biggest customer.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
10. Factor-rating site choice
PEIS-FA-03 · Facilities Engineering and Planning · Multiple choice
Two sites are scored 0–100 on three factors. Weights: labor 0.40, transportation 0.35, taxes 0.25. Site X scores 70, 90, 60. Site Y scores 85, 70, 75. Which site wins, and what is its weighted score?
Show answer and solution
Answer: B. Site Y, 77.25
Site X: 0.40(70) + 0.35(90) + 0.25(60) = 28 + 31.5 + 15 = 74.5.
Site Y: 0.40(85) + 0.35(70) + 0.25(75) = 34 + 24.5 + 18.75 = 77.25.
Site Y wins.
Why the others miss:
- A: 74.5 is Site X's correct score, but it is lower than Y's.
- C: 230 is Y's unweighted total. Factor rating multiplies by weights.
- D: The weighted scores differ by 2.75 points, so there is no tie.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
11. Cartons per pallet layer
PEIS-FA-04 · Facilities Engineering and Planning · Fill in the blank (numeric)
A 48 in × 40 in pallet carries cartons that are 16 in × 12 in × 10 in. Cartons must stand with the 10 in dimension vertical, and nothing may overhang the pallet. What is the maximum number of cartons per layer?
Show answer and solution
Answer: 10
Start with the area bound: (48 × 40) / (16 × 12) = 1,920 / 192 = 10. No pattern can beat 10.
Single-orientation patterns fall short. With the 16 in side along 48 in, you get 3 × 3 = 9. With the 12 in side along 48 in, you get 4 × 2 = 8.
A split pattern reaches the bound. Divide the 40 in depth as 16 + 12 + 12 = 40. In the 16 in-deep row, turn cartons so their 12 in face runs along the 48 in edge: 4 cartons. In each 12 in-deep row, run the 16 in face along the 48 in edge: 3 cartons each. Total = 4 + 3 + 3 = 10.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
12. Block-stack storage area
PEIS-FA-05 · Facilities Engineering and Planning · Multiple choice
You must store 600 pallets, each 48 in × 40 in, block-stacked 3 high. Add an aisle and honeycombing allowance equal to 40% of the net pallet footprint. About how much floor area is needed?
Show answer and solution
Answer: B. 3,733 ft²
Floor positions = 600 / 3 = 200.
One pallet footprint = (48 × 40) / 144 = 13.33 ft². Net footprint = 200 × 13.33 = 2,667 ft².
Add 40% of net: 2,667 × 1.40 = ≈3,733 ft².
Why the others miss:
- A: 2,667 ft² is the net footprint with no aisle allowance.
- C: 4,444 ft² divides by 0.60, treating the 40% as a share of gross area. The question says 40% of net.
- D: 11,200 ft² forgets the 3-high stacking.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
13. Conveyor throughput
PEIS-FA-06 · Facilities Engineering and Planning · Multiple choice
A belt conveyor runs at 120 ft/min. Cartons are 24 in long and spaced with a 12 in gap between them. What is the conveyor's throughput in cartons per hour?
Show answer and solution
Answer: B. 2,400
Each carton occupies its length plus one gap: 24 + 12 = 36 in = 3 ft of belt.
Cartons per minute = 120 / 3 = 40. Per hour = 40 × 60 = 2,400.
Why the others miss:
- A: 40 is cartons per minute, not per hour.
- C: 3,600 ignores the gap (2 ft spacing → 60/min).
- D: 7,200 uses only the gap (1 ft) as the spacing.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
14. Equivalent annual cost
PEIS-FA-07 · Facilities Engineering and Planning · Multiple choice
Two conveyor systems each last 10 years with no salvage value. System A costs $50,000 and $6,000 per year to operate. System B costs $30,000 and $10,000 per year. At 8% interest, which has the lower equivalent uniform annual cost (EUAC)?
Show answer and solution
Answer: A. System A, about $13,451 per year
Capital recovery factor (A/P, 8%, 10) = 0.08(1.08)^10 / [(1.08)^10 − 1] = 0.14903.
System A: 50,000 × 0.14903 + 6,000 = 7,451 + 6,000 = $13,451/yr.
System B: 30,000 × 0.14903 + 10,000 = 4,471 + 10,000 = $14,471/yr.
System A is cheaper by about $1,020 per year over its life, despite the higher first cost.
Why the others miss:
- B: $14,471 is B's correct EUAC, but it is the higher one.
- C: First cost alone ignores $4,000 per year in extra operating cost.
- D: $11,000 spreads first cost with no interest (5,000 + 6,000). Money has a time value.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
Operations Engineering · Problems 15–21
15. Economic order quantity
PEIS-OE-01 · Operations Engineering · Multiple choice
Annual demand is 12,000 units. Each order costs $150 to place, and holding costs $4 per unit per year. What are the EOQ and the resulting annual ordering-plus-holding cost?
Show answer and solution
Answer: A. 949 units; $3,795
EOQ = √(2DS / H) = √(2 × 12,000 × 150 / 4) = √900,000 = 949 units.
At the EOQ, ordering and holding costs are equal. Ordering = (12,000 / 949) × 150 = $1,897. Holding = (949 / 2) × 4 = $1,897. Total = $3,795 (equivalently √(2DSH)).
Why the others miss:
- B: $5,692 charges holding on the full order quantity instead of the average inventory, Q/2.
- C: 671 drops the 2 in the EOQ formula, and its cost is evaluated at that wrong quantity.
- D: $1,897 is only the ordering cost (or only the holding cost).
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
16. Exponential smoothing
PEIS-OE-02 · Operations Engineering · Fill in the blank (numeric)
Last period's forecast was 100 units and actual demand was 120 units. Using simple exponential smoothing with α = 0.3, what is the forecast for next period?
Show answer and solution
Answer: 106
F(next) = F + α(A − F) = 100 + 0.3(120 − 100) = 100 + 6 = 106 units.
The same thing written another way: 0.3 × 120 + 0.7 × 100 = 36 + 70 = 106.
Source: NIST/SEMATECH e-Handbook §6.4.3.1, Single exponential smoothing; topic: NCEES exam specifications
17. Reorder point with safety stock
PEIS-OE-03 · Operations Engineering · Multiple choice
Daily demand averages 50 units with a standard deviation of 10 units, and days are independent. Lead time is a constant 9 days. The target cycle service level is 95% (z = 1.645). What is the reorder point?
Show answer and solution
Answer: B. 499 units
Demand during lead time: mean = 50 × 9 = 450. Standard deviation = 10 × √9 = 30.
Safety stock = 1.645 × 30 = 49.35 ≈ 49.
Reorder point = 450 + 49 = ≈499 units.
Why the others miss:
- A: 450 has no safety stock, so it gives only about a 50% chance of avoiding a stockout.
- C: 520 uses z = 2.33, which is a 99% service level.
- D: 598 multiplies σ by 9 instead of √9. Independent daily variances add, so standard deviations scale with the square root.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
18. Present worth decision
PEIS-OE-04 · Operations Engineering · Multiple choice
A machine costs $80,000, saves $20,000 per year for 6 years, and has a $10,000 salvage value at the end of year 6. MARR is 10%. What is the present worth, and should the machine be bought?
Show answer and solution
Answer: A. About +$12,750; buy it
(P/A, 10%, 6) = 4.3553, so savings are worth 20,000 × 4.3553 = $87,106.
(P/F, 10%, 6) = 0.5645, so salvage is worth 10,000 × 0.5645 = $5,645.
PW = −80,000 + 87,106 + 5,645 = ≈ +$12,750. Positive PW at the MARR means the investment earns more than 10%: buy it.
Why the others miss:
- B: +$7,106 leaves out the salvage value.
- C: +$50,000 adds cash flows without discounting.
- D: +$1,461 subtracts the salvage value instead of adding it.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
19. Break-even volume
PEIS-OE-05 · Operations Engineering · Multiple choice
A product line has fixed costs of $240,000 per year. Each unit sells for $50 and has a variable cost of $30. What is the break-even volume?
Show answer and solution
Answer: D. 12,000 units
Contribution margin per unit = 50 − 30 = $20.
Break-even = fixed cost / contribution margin = 240,000 / 20 = 12,000 units per year.
Why the others miss:
- A: 3,000 divides by price plus variable cost ($80).
- B: 4,800 divides by price alone. Each unit only contributes $20 toward fixed costs.
- C: 8,000 divides by variable cost.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
20. Shortest processing time
PEIS-OE-06 · Operations Engineering · Multiple choice
Four jobs wait at one machine, all available now: J1 takes 6 h, J2 takes 2 h, J3 takes 8 h, and J4 takes 4 h. Which sequence minimizes mean flow time, and what is that mean?
Show answer and solution
Answer: A. J2–J4–J1–J3; 10 h
Shortest processing time (SPT) first minimizes mean flow time on a single machine. Order: J2 (2), J4 (4), J1 (6), J3 (8).
Completion times: 2, 6, 12, 20. Mean flow time = (2 + 6 + 12 + 20) / 4 = 10 h.
Why the others miss:
- B: Longest-first gives completions 8, 14, 18, 20 and a mean of 15 h, the worst choice here.
- C: First-come order gives 6, 8, 16, 20 and a mean of 12.5 h.
- D: Right sequence, but 20 h is the makespan (when the last job finishes), not the mean flow time.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
21. Truckload vs. LTL
PEIS-OE-07 · Operations Engineering · Multiple choice
A carrier quotes $1,800 flat for a full truckload or $9.00 per hundredweight (cwt, 100 lb) for less-than-truckload. Your shipment weighs 16,000 lb. Which mode is cheaper, and at what weight would the two cost the same?
Show answer and solution
Answer: A. LTL at $1,440; break-even at 20,000 lb
LTL cost = (16,000 / 100) × $9 = 160 cwt × $9 = $1,440, less than $1,800. Ship LTL.
Break-even weight: 1,800 / 9 = 200 cwt = 20,000 lb. Above that, the flat truckload rate wins.
Why the others miss:
- B: $14,400 divides by 10 instead of 100. A hundredweight is 100 lb.
- C: 2,000 lb confuses cwt with lb; the break-even is 200 cwt.
- D: At 16,000 lb LTL is still cheaper; the costs only meet at 20,000 lb.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
Work Design · Problems 22–28
22. Standard time
PEIS-WD-01 · Work Design · Multiple choice
A time study gives a mean observed time of 1.20 min. The operator was rated at 110%. Personal, fatigue, and delay (PF&D) allowance is 15%, applied as a percentage of normal time. What is the standard time?
Show answer and solution
Answer: C. 1.518 min
Normal time = observed × rating = 1.20 × 1.10 = 1.32 min.
Standard time = normal time × (1 + allowance) = 1.32 × 1.15 = 1.518 min.
Watch the convention. If allowance were stated as a fraction of total time, you'd use 1.32 / (1 − 0.15) = 1.553 min. The stem decides which one you use.
Why the others miss:
- A: 1.320 is normal time, with no allowance.
- B: 1.380 applies the allowance but skips the 110% rating.
- D: 1.553 treats the 15% as a share of total time. This question says it applies to normal time.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
23. Minimum stations and efficiency
PEIS-WD-02 · Work Design · Multiple choice
An assembly line must make 400 units in a shift with 440 productive minutes. Task times total 290 seconds. What is the theoretical minimum number of stations, and what is line efficiency if you achieve it?
Show answer and solution
Answer: B. 5 stations; 87.9%
Cycle time = available time / demand = (440 × 60) / 400 = 66 s per unit.
Minimum stations = ⌈290 / 66⌉ = ⌈4.39⌉ = 5.
Efficiency = total task time / (stations × cycle time) = 290 / (5 × 66) = 87.9%.
The theoretical minimum is a lower bound. Precedence and task sizes may force a sixth station in practice.
Why the others miss:
- A: Rounding 4.39 down gives efficiency above 100%, which is impossible. Stations always round up.
- C: 80.6% uses a 480-minute shift (72 s cycle time) and ignores the stated 440 productive minutes.
- D: 6 stations is not the theoretical minimum.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
24. Work-sampling observations
PEIS-WD-03 · Work Design · Multiple choice
A pilot work-sampling study suggests an operator is idle 20% of the time. You want the idle proportion within ±0.04 (absolute) at 95% confidence (z = 1.96). How many observations are needed?
Show answer and solution
Answer: C. 385
n = z² p(1 − p) / e² = (1.96)² (0.20)(0.80) / (0.04)² = 3.8416 × 0.16 / 0.0016 = 384.2.
Always round sample sizes up: 385.
Why the others miss:
- A: 271 uses z = 1.645, a 90% confidence level.
- B: 384 rounds down, which falls just short of the required precision.
- D: 9,604 treats ±4% as relative to p (e = 0.04 × 0.20 = 0.008). The question says absolute.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
25. DART rate
PEIS-WD-04 · Work Design · Fill in the blank (numeric)
A plant recorded 6 cases involving days away from work, restricted work, or job transfer (DART) last year. Employees worked 760,000 hours. What is the DART rate per 100 full-time workers? Round to two decimals.
Show answer and solution
Answer: 1.58 (accept 1.57–1.58)
DART rate = (DART cases × 200,000) / hours worked = (6 × 200,000) / 760,000 = 1.58.
The 200,000 is 100 full-time workers × 40 hours × 50 weeks. It puts every employer on the same per-100-workers basis.
Source: BLS, How to compute nonfatal incidence rates; topic: NCEES exam specifications
26. NIOSH lifting index
PEIS-WD-05 · Work Design · Multiple choice
A worker lifts a 30 lb tote. At the origin of the lift, the hands are 15 in in front of the midpoint between the ankles (H) and 20 in above the floor (V). The tote travels 30 in vertically (D), and the lift involves 30° of trunk twisting (A). For this task, use a frequency multiplier (FM) of 0.88 and a coupling multiplier (CM) of 0.95. What are the recommended weight limit (RWL) and lifting index (LI)?
Show answer and solution
Answer: A. RWL ≈ 20.9 lb; LI ≈ 1.43
U.S. customary multipliers from the NIOSH Applications Manual:
- HM = 10/H = 10/15 = 0.667
- VM = 1 − 0.0075|V − 30| = 1 − 0.0075(10) = 0.925
- DM = 0.82 + 1.8/D = 0.82 + 0.06 = 0.880
- AM = 1 − 0.0032A = 1 − 0.096 = 0.904
- FM = 0.88 and CM = 0.95 (given)
RWL = 51 × 0.667 × 0.925 × 0.880 × 0.904 × 0.88 × 0.95 = ≈20.9 lb.
LI = load / RWL = 30 / 20.9 = ≈1.43. An LI above 1.0 means the load exceeds the NIOSH recommended weight limit for the stated lift conditions, so this lift is a candidate for redesign, such as bringing the tote closer or raising the origin.
Why the others miss:
- B: 23.8 lb leaves out the frequency multiplier.
- C: 0.70 divides RWL by the load. LI is load ÷ RWL.
- D: 51 lb is the load constant, the limit only under ideal conditions.
Source: NIOSH Applications Manual for the Revised NIOSH Lifting Equation, Pub. 94-110 (rev. Sept. 2021), §1.3 p. 7 and §1.4; topic: NCEES exam specifications
Lifting-index interpretation: An LI above 1.0 signals elevated risk for some workers, so this lift is a candidate for redesign, such as bringing the tote closer or raising the origin.
27. Mixed noise exposure
PEIS-WD-06 · Work Design · Multiple choice
An operator's 8-hour shift includes 4 h at 95 dBA, 2 h at 90 dBA, and 2 h at 100 dBA. Under OSHA's Table G-16 limits for general industry, what is the daily noise dose, and is the exposure within the permissible limit?
Show answer and solution
Answer: A. 2.25 (225%); exceeds the limit
Table G-16 permitted times: 95 dBA → 4 h, 90 dBA → 8 h, 100 dBA → 2 h.
Dose = C₁/T₁ + C₂/T₂ + C₃/T₃ = 4/4 + 2/8 + 2/2 = 1.00 + 0.25 + 1.00 = 2.25.
OSHA treats a sum above 1 as exceeding the limit, so this exposure is over it.
Why the others miss:
- B: 1.25 leaves out the 2 hours at 100 dBA.
- C: 1.00 uses 8 h as the permitted time at every level.
- D: Each period alone fits its own permitted time, but exposures add. OSHA evaluates the combined sum.
Source: 29 CFR 1910.95(b), Table G-16 and footnote; topic: NCEES exam specifications
28. Learning curve unit time
PEIS-WD-07 · Work Design · Multiple choice
The first unit of a new assembly takes 100 labor-hours. The process follows an 80% unit learning curve. How long should the 8th unit take?
Show answer and solution
Answer: B. 51.2 h
On an 80% curve, each doubling of cumulative output multiplies unit time by 0.80. Unit 8 is three doublings from unit 1 (1 → 2 → 4 → 8).
T₈ = 100 × 0.80³ = 51.2 h.
General form: Tₙ = T₁ × n^b, where b = log(0.80)/log(2) = −0.322. Here 8^−0.322 = 0.512.
Why the others miss:
- A: 40 h subtracts 20 hours per doubling. The reduction is multiplicative.
- C: 64 h stops after two doublings (unit 4).
- D: 80 h is unit 2.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
Quality Engineering · Problems 29–35
29. X-bar chart control limit
PEIS-QE-01 · Quality Engineering · Fill in the blank (numeric)
Twenty-five subgroups of size n = 5 give a grand mean of 50.00 mm and an average range of 2.00 mm. What is the upper control limit of the X-bar chart? Enter to three decimals. (For n = 5, A₂ = 0.577.)
Show answer and solution
Answer: 51.154 (accept 51.15–51.155)
UCL = grand mean + A₂ × R̄ = 50.00 + 0.577 × 2.00 = 51.154 mm.
For completeness: LCL = 48.846 mm. On the R chart, UCL = D₄ × R̄ = 2.115 × 2.00 = 4.23 mm, and LCL = 0 because D₃ = 0 for n = 5.
Source: NIST/SEMATECH e-Handbook §6.3.2.1, Shewhart X-bar and R charts; topic: NCEES exam specifications
30. Cp and Cpk
PEIS-QE-02 · Quality Engineering · Multiple choice
A shaft diameter has specification limits of 9.70 mm and 10.30 mm. The process is stable with mean 10.06 mm and standard deviation 0.08 mm. What are Cp and Cpk?
Show answer and solution
Answer: A. Cp = 1.25; Cpk = 1.00
Cp = (USL − LSL) / 6σ = 0.60 / 0.48 = 1.25.
Cpk = min[(USL − μ)/3σ, (μ − LSL)/3σ] = min[0.24/0.24, 0.36/0.24] = min[1.00, 1.50] = 1.00.
Cp says the spread would fit with room to spare. Cpk is lower because the mean sits 0.06 mm above center, closer to the upper limit. Centering the process would raise Cpk toward 1.25.
Why the others miss:
- B: 1.50 is the distance to the lower limit. Cpk takes the smaller side.
- C: This swaps the two indices. Cpk can never exceed Cp.
- D: 2.50 divides the tolerance by 3σ instead of 6σ.
Source: NIST/SEMATECH e-Handbook §6.1.6, What is process capability?; topic: NCEES exam specifications
31. Series system with redundancy
PEIS-QE-03 · Quality Engineering · Multiple choice
Two components operate in series: component 1 has reliability 0.95 and component 2 has reliability 0.90. To improve the system, an identical copy of component 2 is added in active parallel with it. Assuming independent failures, what is the new system reliability?
Show answer and solution
Answer: B. 0.9405
The parallel pair works unless both copies fail: R = 1 − (1 − 0.90)² = 1 − 0.01 = 0.99.
That pair is in series with component 1: R_system = 0.95 × 0.99 = 0.9405.
The original system was 0.95 × 0.90 = 0.855, so the redundancy cut the failure probability from 14.5% to about 6%.
Why the others miss:
- A: 0.855 is the system before the redundancy.
- C: 0.99 is only the parallel pair; it still has to work with component 1 in series.
- D: 0.9995 puts all three units in parallel, which is not the configuration described.
Source: NIST/SEMATECH e-Handbook §8.1.8.2, Series model; NIST/SEMATECH e-Handbook §8.1.8.3, Parallel or redundant model; topic: NCEES exam specifications
32. Exponential reliability
PEIS-QE-04 · Quality Engineering · Multiple choice
A sensor has a constant failure rate of 0.002 failures per hour. What is the probability it survives a 100-hour mission?
Show answer and solution
Answer: C. 0.819
With a constant failure rate, reliability is exponential: R(t) = e^(−λt) = e^(−0.002 × 100) = e^(−0.2) = 0.819.
Why the others miss:
- A: 0.181 is the probability of failure, 1 − R.
- B: 0.800 uses the linear approximation 1 − λt, which is only close when λt is very small.
- D: 0.980 uses t = 10 h.
Source: NIST/SEMATECH e-Handbook §8.1.6.1, Exponential distribution; topic: NCEES exam specifications
33. Probability of acceptance
PEIS-QE-05 · Quality Engineering · Multiple choice
A single sampling plan inspects n = 20 items and accepts the lot if 1 or fewer are defective (acceptance number c = 1). If the lot is 5% defective, what is the probability of accepting it? Use the binomial distribution.
Show answer and solution
Answer: D. 0.736
P(accept) = P(0) + P(1).
P(0) = 0.95²⁰ = 0.358.
P(1) = 20 × 0.05 × 0.95¹⁹ = 0.377.
P(accept) = 0.358 + 0.377 = 0.736. This is one point on the plan's operating characteristic (OC) curve.
Why the others miss:
- A: 0.264 is the probability of rejecting the lot.
- B: 0.358 counts only zero defectives; the plan accepts one.
- C: 0.377 counts only exactly one defective.
Source: NIST/SEMATECH e-Handbook §1.3.6.6.18, Binomial distribution; NIST/SEMATECH e-Handbook §6.2.3, Choosing a single sampling plan; topic: NCEES exam specifications
34. Main effect in a 2² factorial
PEIS-QE-06 · Quality Engineering · Multiple choice
A 2² factorial experiment gives these responses: (1) = 20, a = 30, b = 24, ab = 38, where a letter means that factor is at its high level. What is the main effect of factor A?
Show answer and solution
Answer: C. 12
Main effect of A = average response with A high − average with A low.
A high: (30 + 38) / 2 = 34. A low: (20 + 24) / 2 = 22.
Effect of A = 34 − 22 = 12.
For reference, the effect of B is (24 + 38)/2 − (20 + 30)/2 = 31 − 25 = 6, and the AB interaction is (20 + 38)/2 − (30 + 24)/2 = 29 − 27 = 2.
Why the others miss:
- A: 2 is the AB interaction.
- B: 6 is the main effect of B.
- D: 24 is the contrast (68 − 44) before dividing by 2.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
35. Inherent availability
PEIS-QE-07 · Quality Engineering · Fill in the blank (numeric)
A packaging machine has a mean time between failures (MTBF) of 190 hours and a mean time to repair (MTTR) of 10 hours. What is its inherent availability? Enter as a decimal.
Show answer and solution
Answer: 0.95
Inherent availability = MTBF / (MTBF + MTTR) = 190 / (190 + 10) = 190 / 200 = 0.95.
Inherent availability counts only corrective repair time. Operational availability would also include waiting for parts, logistics, and preventive maintenance downtime, so it is usually lower.
Source: Derived calculation — method and arithmetic shown in the solution; topic: NCEES exam specifications
How did you do? Turn your marks into a study order
Count your marks by area. A guessed or slow correct answer still counts as a review flag; on exam day it costs you time even when it lands.
| Content area | Problems | Correct, clean | Guessed or slow | Wrong | Review flags (guessed + slow + wrong) |
|---|---|---|---|---|---|
| Systems Engineering | 1–7 | ||||
| Facilities Engineering and Planning | 8–14 | ||||
| Operations Engineering | 15–21 | ||||
| Work Design | 22–28 | ||||
| Quality Engineering | 29–35 |
Rank the five areas by review flags. The two with the most flags are your first two study blocks in either plan below. If two areas tie, start with the one your daily job touches least.
Keep this in proportion. Seven problems per area is a starting signal, not a measurement. It says where to look first, not how close you are to passing, and it isn't an NCEES score.
These 35 problems also don't cover every listed subtopic. Not practiced here: quality function deployment, value analysis, systematic layout planning, activity-based costing, distribution networks and cross-docking, methods analysis (therbligs, man-machine charts), predetermined time systems, RULA, anthropometry, link analysis, MIL-STD-105E table lookups, Taguchi methods, FMEA, and root cause analysis. Put them on your list.
What the exam covers: five areas, 14 to 21 questions each
NCEES gives every area the same published range of 14 to 21 questions out of 85 (specification). No area is light enough to skip. Because the ranges are identical, your own weak spots, not the blueprint, should set your study order.
| Content area | Questions | What NCEES lists (condensed) | Problems on this page |
|---|---|---|---|
| Systems Engineering | 14–21 | Analysis and design tools (flowcharts, Pareto, value stream mapping), requirements analysis and QFD, performance measures, modeling (simulation, queuing, linear programming, networks, Markov chains), process types, model verification and validation, bottleneck analysis and theory of constraints, value engineering, project management (PERT/CPM/CCPM, risk, Gantt charts) | 1–7 |
| Facilities Engineering and Planning | 14–21 | Process flow, layout design (SLP, relationship diagrams, center of gravity), space analysis, capacity analysis (people and machines), cost-benefit analysis, site selection, unit loads, facility life-cycle cost, material handling | 8–14 |
| Operations Engineering | 14–21 | Forecasting, production planning (aggregate planning, JIT, lot sizing), engineering economics, costing systems (activity-based costing), production scheduling, inventory (deterministic and stochastic), distribution models, storage and warehousing, transportation modes | 15–21 |
| Work Design | 14–21 | Methods analysis, line balancing, work measurement and predetermined time systems, learning curves, sample size, work sampling, safety codes (ANSI, OSHA, MIL-STD, NIOSH), risk quantification (NIOSH lifting equation, OSHA noise, RULA), human capacity limits, lifting aids, link analysis, workplace design and anthropometry, DART rates, process planning and mistake-proofing | 22–28 |
| Quality Engineering | 14–21 | SPC and control charts, process capability, acceptance sampling (MIL-STD-105E, Dodge-Romig, OC curves), quality systems (Deming, TQM, ISO 9000), DOE, Taguchi, FMEA, reliability, maintenance strategies, root cause analysis | 29–35 |
Adding the five low ends gives 70 and the five high ends gives 105. Neither is a real exam; they're just the outer bounds of the published ranges. The specification also says questions may draw on engineering economics, probability and statistics, operations research, management systems, project management, and codes and standards across all five areas, and that the listed topics are examples rather than a complete list.
Worked example: one work cell, three checks
Real exam questions often cross areas. This one touches Work Design, Facilities capacity, and Systems Engineering takt time in a single setup.
A packing cell must produce 1,500 units per week with 40 productive hours. A time study shows a mean observed time of 2.40 minutes per unit, a 95% performance rating, and a 12% allowance applied to normal time.
Check 1: standard time. Normal time = 2.40 × 0.95 = 2.28 min. Standard time = 2.28 × 1.12 = 2.554 min per unit.
Check 2: staffing. Weekly workload = 1,500 × 2.554 / 60 = 63.8 hours. Each operator supplies 40 hours, so 63.8 / 40 = 1.60 operators. Round up: 2 operators.
Check 3: takt time. Takt = available time / demand = (40 × 60) / 1,500 = 1.6 min per unit. One operator needs 2.554 minutes per unit, which is slower than takt. Two operators working in parallel produce a unit about every 1.28 minutes, which meets it.
Both checks point to 2 operators, which is how you know the setup is consistent. When two methods disagree on an exam problem, go back to the units first. Minutes versus hours is the usual culprit.
Your study plan
Both plans are editorial. NCEES doesn't publish a required number of study hours, and following a plan doesn't guarantee a result. Each week mixes four kinds of work: learning and practicing a weak area, mixed timed problems, a handbook lookup drill, and repairing your error log (both explained in the next section).
If your exam is 14 days away or less
The PE Industrial and Systems exam is given on one day a year; the next date is October 28, 2026 (NCEES). Plan on 2 to 3 hours a day. If you have more than 14 days, add the extra days to Days 3 through 8.
| Day | What to do |
|---|---|
| 1 | Work Problems 1–18 untimed. Mark each Correct, Guessed, Slow, or Wrong. |
| 2 | Work Problems 19–35. Fill in the results table and name your two weakest areas. |
| 3–5 | Weakest area #1. Rework every miss from this page until you can do it cold. Then solve 15 to 20 fresh problems in that area from another source, such as the NCEES practice exam, a review book, or course notes. Do the 20-minute lookup drill each day. |
| 6–8 | Weakest area #2, same method. |
| 9 | One mixed set of about 20 problems from your other three areas. Focus on finding each formula fast. |
| 10 | References and safety day: recheck the current NCEES specification and MyNCEES reference materials, then drill the NIOSH lifting equation, OSHA noise dose, DART rate, and acceptance-sampling concepts. Do not assume an older supplied-standards list still applies. |
| 11 | Timed block: 42 fresh mixed problems in 4 hours 15 minutes, about half the exam at exam pace. |
| 12 | Review only the misses and slow solves from Day 11. Update your error log. |
| 13 | Read your error log. Confirm your Pearson VUE appointment, ID, and route to the test center. |
| 14 | Light review only. No new topics. Sleep. |
12-week plan for the next administration
About 10 hours a week, roughly 120 hours total. NCEES ran a survey in 2026 to update this exam's specifications, so check for a new specification before you start (see What's changing).
| Week | Focus | Finish line |
|---|---|---|
| 1 | Work all 35 problems untimed; fill in the results table. Read the full NCEES specification. Download the reference handbook and find where each area's formulas live. | Ranked list of your five areas |
| 2–3 | Weakest area (two weeks) | You can solve each core problem type without notes |
| 4 | Second-weakest area | Same |
| 5 | Third area | Same |
| 6 | Fourth area | Same |
| 7 | Strongest area: a quick pass to confirm, not relearn | Same |
| 8 | Back to the two weakest areas with fresh problems | Fewer repeat errors in your log |
| 9 | Timed half-exam: 42 problems in 4 hours 15 minutes, then repair | Pace near 6 minutes per problem |
| 10 | Current-reference check, safety calculations, and lookup speed; close remaining gaps | Current NCEES reference set confirmed; lookup terms written down for every formula you missed |
| 11 | Full-length simulation: 85 problems with 8.5 hours of working time plus a 50-minute scheduled break, then repair | You finish with time to review |
| 12 | Error log, light mixed review, exam logistics | Stop adding new sources |
In weeks 2 through 10, split each 10 hours about like this: 6 hours learning and practicing the week's area, 2 hours of mixed timed problems from all areas, 1 hour of lookup drill, and 1 hour of error-log repair.
Adjusting the plan
- Fewer hours a week: stretch weeks 2 through 8 across more calendar time. Don't cut weeks 9 and 11; timed practice is where pacing problems show up.
- Retaking after a fail: NCEES sends failing examinees a diagnostic report showing performance by major topic (NCEES scoring). Use it instead of the results table above to rank your areas, give the two lowest areas weeks 2 through 5, and practice with fresh problems rather than redoing ones you've memorized. The report doesn't show a passing score or exactly how far you were from one.
Practice with the references you'll actually have
The exam is closed book. NCEES supplies the current electronic reference handbook, plus any additional codes or standards listed on the last page of the current exam specification (NCEES Examinee Guide, May 2026, p. 10). Download the current handbook from your MyNCEES account and recheck it before exam day in case NCEES updates the file.
Do not rely on an old standards list. An earlier NCEES version of the Industrial and Systems specification listed MIL-STD-105E and MTM-1 as supplied standards. The current specification linked by NCEES, updated in 2025 and checked October 9, 2026, is three pages long and does not contain a separate standards page. The current NCEES Examinee Guide says supplied codes and standards will be listed on the last page of the exam specification. For the 2026 administration, use the current Industrial and Systems specification and your MyNCEES exam materials as the controlling sources rather than assuming the older MIL-STD-105E/MTM-1 list still applies.
ISO, OSHA, MIL-STD, and NIOSH concepts still appear in the current specification as examples of knowledge that may be tested. That is different from saying the underlying documents will be supplied during the exam.
The 20-minute lookup drill
Do this most days. It turns "I know this" into "I can find and use this in under a minute."
- Pick a problem. Before searching, name the concept in a few words (for example, "safety stock" or "X-bar chart limits").
- Search the handbook PDF for the shortest term that works. Write that term down.
- Note the variables and conditions the formula needs, such as known σ, constant lead time, or allowance as a share of normal time.
- Solve the problem.
- Log the search term that worked. Next time, start there.
Error log
Keep one row per miss, guess, or slow solve.
| Problem | Area | Error type | Correct setup | Lookup term | Next drill |
|---|---|---|---|---|---|
| e.g., 24 (Work sampling) | Work Design | Model choice: used relative error | n = z²p(1 − p)/e², e absolute | "work sampling" | Two fresh sample-size problems |
Error types worth tracking: concept, model choice, setup, units, lookup, calculator, misread, time. After a week, the most common type tells you more than any single score.
PE Industrial and Systems exam at a glance
| Row label | Current rule | Source |
|---|---|---|
| Questions | 85, multiple choice plus alternative item types | NCEES exam page |
| Appointment | 9.5 hours: 2-min nondisclosure agreement, 8-min tutorial, 8.5-hour exam, 50-min scheduled break | NCEES exam page |
| Format | Computer-based at Pearson VUE test centers; closed book with electronic reference | NCEES exam page, specification |
| Availability | One day a year; next date October 28, 2026 | NCEES exam page |
| Units | SI and U.S. Customary | Specification |
| Specification | Effective beginning with the October 2020 exams | Specification |
| Fee | $400 to NCEES; your licensing board may charge a separate application fee | NCEES exam page |
| Results | Typically 7–10 days after the exam; reported pass or fail | NCEES scoring |
| Retakes | One attempt per testing window, no more than three in 12 months; some boards are stricter | NCEES PE exam page |
NCEES encourages reserving a seat for the single annual date as early as possible. You need to be registered with NCEES and approved by your board before you can schedule. Rescheduling or canceling an appointment must be done at least 48 hours ahead and carries a $50 Pearson fee (NCEES Examinee Guide, May 2026, pp. 6–7). Our NCEES exam registration guide walks through the steps.
What's changing
NCEES surveyed licensed industrial and systems engineers from January 20 to June 12, 2026 to update the specifications for this exam (NCEES Licensure Exchange, April 2026, p. 11). As of October 9, 2026, the exam page still links the October 2020 specification and lists no new effective date. If you're preparing for a later administration, check the exam page for a new specification before you build your topic list.
The October 2020 specification file now on the NCEES site is marked "updated 2025." It keeps the same five areas and the same 14–21 ranges. The earlier version described "85 multiple-choice questions"; the updated version says "85 questions," and the NCEES exam page now notes the exam includes alternative item types.
Exam-day rules that change how you practice
- Pace is about 6 minutes per question. 8.5 hours × 60 ÷ 85 = 6.0 minutes. That's why this page uses 6 minutes as the line for marking a problem Slow.
- Not every question is multiple choice. Alternative item types can include select-all-that-apply, point-and-click, drag-and-drop, and fill-in-the-blank. Every item is scored right or wrong with no partial credit (NCEES computer-based testing). Problems 6, 7, 11, 16, 25, 29, and 35 on this page use select-all or fill-in formats so you get used to answers without choices to check against.
- Everyone gets the same questions. Single-day exams like this one give all examinees the same questions (NCEES computer-based testing).
- Use an approved calculator. The 2026 NCEES list is HP 33s and 35s, Casio fx-115 and fx-991 models, and TI-30X and TI-36X models (NCEES 2026 approved calculators memo). The Examinee Guide also notes an on-screen TI-30XS is available during the exam (p. 9). Practice with the one you'll bring.
- Scratch work goes in a booklet. The proctor gives you 2 reusable booklets and 3 markers (Examinee Guide, p. 10). Practice writing setups neatly enough to recheck them.
- Both unit systems appear. Keep conversions in your error log when they trip you up.
What the pass rates tell you
On the NCEES table last updated in January 2026, 69% of 80 first-time examinees and 29% of 21 repeat examinees passed PE Industrial and Systems (NCEES exam page).
That describes one small group of people, not your personal odds. With about 100 examinees a year, a handful of results can move the percentage noticeably. The gap between first-time and repeat rates is a good reason to treat the first attempt seriously, given that a retake means waiting for the next annual date. NCEES doesn't publish a passing score (NCEES scoring).
Is there an official practice exam?
Yes. NCEES sells a PE Industrial and Systems practice exam as a PDF: 85 questions with solutions, based on the October 2020 specification, including alternative item types such as fill-in-the-blank and matching. It was listed at $59.95 when we checked on October 9, 2026, and the PDF is nonrefundable (NCEES exam prep). Because it's NCEES's own full-length exam for this specification, it fits well as your week-11 simulation or Day 11 timed block. If you use it earlier, save at least half of it for timed work.
Quick answers
Is this the same as the FE Industrial and Systems exam?
No. The FE Industrial and Systems exam is a separate NCEES exam with its own specification and format. If that's the exam you need, see our FE Industrial and Systems exam prep.
Does passing the PE exam make me a licensed PE?
Not by itself. Your state or territory licensing board grants the license and sets requirements such as education and experience; the exam is one part. NCEES's page describes the PE as designed for engineers with at least four years of post-college work experience in their discipline. Check your board's rules through the NCEES PE exam page or its licensing board directory.
How long should I study?
There's no official figure. Our 12-week plan assumes about 120 hours. If your work covers only one or two of the five areas, plan for more; if you already work across most of them, you may need less.
Where do I check other PE disciplines?
Our PE exam prep hub covers the other PE exams. For results, retakes, or testing accommodations, see our guides to NCEES exam results, the NCEES retake policy, and NCEES exam accommodations.
Sources
Exam facts were checked against these sources on October 9, 2026.
- NCEES, PE Industrial and Systems exam page: questions, appointment, availability and next date, fee, references, pass rates
- NCEES, PE Industrial and Systems exam specifications, effective October 2020 (updated 2025): content areas, ranges, units, closed-book rule
- NCEES, Examinee Guide, May 2026: rescheduling/cancellation, calculator and test-center rules, current reference-material policy, exam format, and results
- NCEES, PE exam page: retake limits and board-specific registration; Exam scoring and computer-based testing
- NCEES, 2026 approved calculators memo (October 20, 2025)
- NCEES, Licensure Exchange, April 2026, p. 11: PE Industrial and Systems specification survey
- NCEES, PE Industrial and Systems practice exam listing
- NIOSH, Applications Manual for the Revised NIOSH Lifting Equation, Pub. 94-110 (rev. September 2021)
- OSHA, 29 CFR 1910.95, Occupational noise exposure
- BLS, How to compute nonfatal incidence rates
- NIST/SEMATECH, e-Handbook of Statistical Methods: process capability, control charts, exponential smoothing, reliability, binomial distribution (sections cited beside each problem)
How we made this page. The Castleport Test Prep Editorial Team wrote the problems and plans with AI-assisted research and drafting. We checked exam facts against the NCEES documents above, checked safety formulas against the NIOSH, OSHA, and BLS sources, and recomputed every answer. Source checking is not professional engineering review, and the problems have not been reviewed by NCEES or a licensed engineer. More on our process: methodology.
Castleport Test Prep is an independent exam prep publisher. We are not affiliated with, endorsed by, or approved by NCEES, Pearson VUE, or any licensing board. Exam and credential names are used to identify their subjects, and trademarks belong to their respective owners. The practice problems here are original and unofficial, not actual or recalled exam questions. Nothing on this page guarantees a passing result or licensure; confirm your own requirements with your licensing board.