2.3 Probability & Statistics

What the exam asks

Probability and Statistics questions are formulaic and fast points if you know the right expression. The Handbook supplies the formulas; you supply recognition and clean arithmetic.

Central tendency and dispersion

• Mean (arithmetic average): μ = (sum of x_i) / n.
• Median: the middle value when sorted; robust to outliers.
• Mode: the most frequent value.
• Variance: spread about the mean.
• Standard deviation: σ = sqrt(variance), in the same units as the data.

Watch the denominator. Population variance divides by N; sample variance divides by (n - 1) (Bessel's correction). The FE often supplies a small sample and expects (n - 1).

Worked example (mean, median, std dev): For the data set {2, 4, 4, 6, 9}, the mean is μ = (2+4+4+6+9)/5 = 25/5 = 5. The median (middle of the sorted list) is 4 and the mode is 4. The squared deviations from the mean are 9, 1, 1, 1, 16, summing to 28. The population standard deviation is σ = sqrt(28/5) = sqrt(5.6) = 2.37; the sample standard deviation is s = sqrt(28/4) = sqrt(7) = 2.65. The denominator choice changes the answer, which is exactly the distractor the FE exploits.

Permutations and combinations

When order matters, use permutations P(n,r) = n!/(n-r)!. When order does not, use combinations C(n,r) = n!/[r!(n-r)!]. For example, the number of 3-member committees from 5 people is C(5,3) = 5!/(3!2!) = 10, while the number of ordered finishes for 3 of 5 racers is P(5,3) = 5!/2! = 60. The single most common FE counting error is choosing permutations when the problem does not care about order (or vice versa) — read for the words "arrangement/order" versus "group/committee/selection."

Basic probability rules

For mutually exclusive events, P(A or B) = P(A) + P(B); in general P(A or B) = P(A) + P(B) - P(A and B). For independent events, P(A and B) = P(A)·P(B). Conditional probability is P(A|B) = P(A and B)/P(B). These rules combine with counting: the probability of drawing 2 defective parts from a bin is the count of favorable combinations over the count of total combinations. Keep probabilities between 0 and 1 — a result above 1 means you added when you should have multiplied.

Expected value, distributions, regression, and reliability

Expected value and variance of a random variable

For a discrete random variable, the expected value is the probability-weighted sum E[X] = sum(x_i · p_i), and the variance is Var(X) = E[X²] - (E[X])². This shortcut (mean of the squares minus the square of the mean) is faster than the deviation form under exam timing.

Common distributions

The binomial probability of exactly k successes is P(k) = C(n,k) p^k (1-p)^(n-k). For the normal distribution, standardize with the z-score z = (x - μ)/σ, then read the standard normal table in the Handbook. The empirical rule says about 68% of values lie within ±1σ, 95% within ±2σ, and 99.7% within ±3σ.

Worked example (binomial): A circuit has a 0.1 chance of a defective resistor. In a batch of 5, what is the probability of exactly 1 defect? P(1) = C(5,1)(0.1)¹(0.9)⁴ = 5(0.1)(0.6561) = 0.328, about 33%.

Linear regression and hypothesis testing

Least-squares fits y = a + bx with slope b = S_xy/S_xx; the line passes through the centroid (x̄, ȳ). The correlation coefficient r ranges from -1 to +1; r² is the fraction of variance explained. In hypothesis testing you state a null H0 and alternative H1, choose significance level α (often 0.05), and reject H0 if the test statistic falls in the rejection region. A Type I error rejects a true null (probability α); a Type II error fails to reject a false null (probability β).

Reliability

For n components in series, all must work: R_sys = R1 × R2 × ... × Rn. For parallel redundancy, the system fails only if all fail: R_sys = 1 - (1-R1)(1-R2)...(1-Rn). With a constant failure rate λ, reliability R(t) = e^(-λt) and the mean time to failure (MTTF) = 1/λ.

Worked example (reliability and MTTF): A power-supply component has a constant failure rate λ = 0.001 per hour. Its MTTF is 1/λ = 1,000 hours, and the probability it survives 500 hours is R(500) = e^(-0.001·500) = e^(-0.5) = 0.607, about 61%. Adding a second identical unit in parallel raises survival to 1 - (1 - 0.607)² = 1 - 0.154 = 0.846, about 85% — redundancy buys reliability at the cost of a duplicated part. Worked example (z-score): If exam scores are normal with μ = 70 and σ = 8, a score of 86 has z = (86 - 70)/8 = 2.0, so by the empirical rule only about 2.5% of scores exceed it (half of the 5% lying beyond ±2σ).