Probability and Statistics is 4-6 questions on the FE Mechanical exam. It is one of the smaller areas, but it is also one of the most mechanical: nearly every item reduces to picking the right distribution and reading one handbook table.

It connects to two other parts of the paper. Measurement uncertainty is a statistics question wearing an instrumentation label, and reliability in Mechanical Design is a probability question wearing a machine-element label. Time spent here pays out three times.

Exam weight: NCEES lists Probability and Statistics at 4-6 questions of the 110-question FE Mechanical exam. Work each problem below on paper first, then reveal the worked solution — reading a solution you have not attempted builds recognition, not recall.

What NCEES Tests in Probability and Statistics

The specification covers measures of central tendency and dispersion, probability distributions, expected value, confidence intervals, hypothesis testing, and linear regression. The distributions that actually appear are normal, binomial and exponential; the exam does not ask you to derive them, only to apply them.

Regression items usually give a small data set or a fitted line and ask for a slope, an intercept, or a predicted value. The coefficient of determination shows up as a concept question about how well a fit explains the data, not as a hand calculation.

5 Free Probability and Statistics Practice Problems

Each probability statistics problem below comes from the FEExamClick FE Mechanical question bank, with worked solutions, and matches the style, difficulty and format of the real exam. Attempt each one under a three-minute limit — roughly the pace the exam demands.

Problem 1 — A. Probability distributions

A manufacturing process produces bolts with lengths normally distributed with \(\mu = 50\) mm and \(\sigma = 0.5\) mm. What percentage of bolts are longer than 51 mm? (Use: \(F(2) = 0.9772\))

Figure for FE Mechanical practice problem 1 (Probability and Statistics)

Answer: A) 2.28%

Step 1 – Standardise.

Convert the bolt length to a z-score, which counts how many standard deviations the value sits from the mean:

$$z = \dfrac{x - \mu}{\sigma} = \dfrac{51 - 50}{0.5} = 2.0$$

So 51 mm is exactly two standard deviations above the mean.

Step 2 – Watch which tail the table gives you.

\( F(z) \) in the Unit Normal Distribution Table is the cumulative probability, i.e. the area to the LEFT of \( z \). The question asks for bolts longer than 51 mm, which is the area to the RIGHT, so take the complement:

$$P(X > 51) = P(Z > 2) = 1 - F(2)$$

Step 3 – Evaluate.

$$P(X > 51) = 1 - 0.9772 = 0.0228 = 2.28\%$$

Why the other options fail:
• 15.87% is the \( z = 1 \) tail, \( 1 - 0.8413 \). You get it by dividing by 1 mm instead of \( \sigma = 0.5 \) mm — treating the 1 mm offset as if it were one standard deviation.
• 0.13% is the \( z = 3 \) tail, \( 1 - 0.99865 \).
• 5.0% is the reflexive answer from significance testing and corresponds to no z-value here at all.

The three numeric distractors are the \( z = 1 \), \( 2 \) and \( 3 \) tails, so this question is really testing one thing: whether you formed \( z \) correctly.

Sanity check: the empirical rule says about 95% of a normal population lies within \( \pm 2\sigma \), leaving about 5% split between the two tails — roughly 2.5% in each. 2.28% is right where it should be.

Correct answer: (A)

Problem 2 — A. Probability distributions - Binomial

A manufacturing process produces items with a 10% defect rate. If 5 items are randomly selected, what is the probability that exactly 2 items are defective?

Answer: D) 0.0729

This is a binomial distribution problem.

$$P(x) = C(n,x) \cdot p^x \cdot q^{n-x}$$

where: \(n = 5\), \(x = 2\), \(p = 0.10\), \(q = 0.90\)

$$C(5,2) = \frac{5!}{2!3!} = 10$$

$$P(2) = 10 \times (0.10)^2 \times (0.90)^3$$

$$P(2) = 10 \times 0.01 \times 0.729$$

$$\boxed{P(2) = 0.0729}$$

Problem 3 — B. Measures of central tendencies and dispersions

A data set of 8 values has a sum of 240 and sum of squared values of 7,500. What is the population variance?

Answer: C) 37.5

Population variance formula:

$$\sigma^2 = \frac{\sum X_i^2}{N} - \left(\frac{\sum X_i}{N}\right)^2$$

Given:
- \(N = 8\)
- \(\sum X_i = 240\)
- \(\sum X_i^2 = 7500\)

Mean:

$$\mu = \frac{240}{8} = 30$$

Variance:

$$\sigma^2 = \frac{7500}{8} - 30^2 = 937.5 - 900$$

$$\boxed{\sigma^2 = 37.5}$$

Note: For sample variance, divide by \((n-1)\), giving 42.86

Problem 4 — A. Probability distributions - Normal

The breaking strength of cables is normally distributed with μ = 2000 N and σ = 150 N. What minimum strength guarantees only 5% of cables will fail? (Z\(_{0}\).\(_{05}\) = -1.645)

Answer: B) 1753 N

We need to find x such that P(X < x) = 0.05

Using the z-transformation:
\(z = \frac{x - \mu}{\sigma}\)

For 5% in lower tail: z = -1.645

$$-1.645 = \frac{x - 2000}{150}$$

$$x = 2000 + (-1.645)(150)$$

\(x = 2000 - 246.75 = 1753.25 \approx 1753\) N

If load is kept below 1753 N, only 5% of cables would be expected to fail.

Problem 5 — C. Expected value in decision making

A company can bid on a project. Preparing the bid costs \(\$5{,}000\). If they win (30% chance), profit is \(\$50{,}000\). If they lose, they lose the bid cost. What is the expected value of bidding?

Answer: A) \(\$11{,}500\)

Expected value formula:

$$E[\text{Value}] = \sum P_i \times \text{Outcome}_i$$

If win (\(P = 0.30\)): Net profit \(= \$50,000 - \$5,000 = \$45,000\)

If lose (\(P = 0.70\)): Net loss \(= -\$5,000\)

\(E[\text{Value}] = (0.30)(\$45,000) + (0.70)(-\$5,000)\)

\(E[\text{Value}] = \$13,500 - \$3,500\)

\(\boxed{E[\text{Value}] = \$11,500}\)

Since \(E[\text{Value}] > 0\), bidding is favorable on average.

Using the FE Reference Handbook for Probability and Statistics

The unit normal distribution table is the single most-used page in this section, and it is the one candidates fumble because the handbook tabulates one specific area convention. Read the column heading once during practice and write down what it gives you, because the difference between the area to the left of z and the area between the mean and z is a wrong answer that is always among the options.

Four Mistakes That Cost Points

Frequently Asked Questions

How many probability and statistics questions are on the FE Mechanical exam?

NCEES specifies 4-6 questions out of 110. Mathematics is counted separately at 6-9 questions.

Which distributions do I actually need?

Normal, binomial and exponential cover almost everything asked. The normal distribution dominates because it supports tolerance, quality and measurement-uncertainty questions elsewhere on the exam. Know how to standardise a value to z and how to read the handbook table in both directions.

Do I need to memorise statistical formulas?

No. Mean, variance, standard deviation, confidence interval and regression formulas are all in the handbook. What is not in the handbook is the judgement of which one applies, so practise reading the question for the words sample, population, with replacement and confidence level.

How is this different from the statistics on the FE Civil exam?

FE Civil folds statistics into a single Mathematics and Statistics area. FE Mechanical splits them, and leans slightly more toward reliability and measurement uncertainty because of the Mechanical Design and Controls areas that depend on them.

Keep Going

These topics feed into each other on the exam:

Done with probability statistics? Browse every knowledge area from the free FE Mechanical practice problem hub, see what the full bank covers on the FE Mechanical exam prep page, or plan your schedule with the FE study timeline. Or test yourself against the clock with the free FE Mechanical practice exam: 20 timed questions in the CBT format.