FE Electrical Probability & Statistics Practice Problems

Answer: C) \(3.71\text{ V}\)

The median is defined on the ordered data, so sort first:

$$3.68,\; 3.69,\; 3.70,\; 3.72,\; 3.72,\; 3.75$$

With \(n = 6\) even, the Handbook takes the average of the \((n/2)\)th and \((n/2 + 1)\)th items, the 3rd and 4th:

$$\text{median} = \dfrac{3.70 + 3.72}{2} = 3.71\text{ V}$$

It happens that the mean of these six readings, \(22.26/6\), is also 3.71 V, so here the two measures agree; that is a coincidence of this data set, not a rule, and the way to tell them apart is that the median never uses the actual values of the readings beyond the two in the middle - the 3.75 V cell could read 4.5 V and the median would not move.

3.70 V and 3.72 V are the 3rd and 4th sorted values taken alone, the average between them skipped. For an even-sized sample the median sits between the two middle readings, not on either of them; 3.72 V is also the mode, which appears twice.

3.68 V is the minimum, the first item of the sorted list rather than the middle one.

Answer: C) \(1.39\)

Handbook: FE Reference Handbook 10.6, Engineering Probability and Statistics → Probability Functions, Distributions, and Expected Values, p. 67 (PDF viewer: page 73). For a discrete random variable with mass function \(f(x_k)\) that page gives both the mean and the variance:

$$\mu = E[X] = \sum_{k=1}^{n} x_k\, f(x_k) \qquad\qquad \sigma^2 = V[X] = \sum_{k=1}^{n} \left(x_k - \mu\right)^2 f(x_k)$$

Step 0 – Check the distribution before using it. \(0.40 + 0.30 + 0.15 + 0.10 + 0.05 = 1.00\). A mass function that does not sum to 1 is not a distribution, and this takes five seconds.

Step 1 – Mean.

$$E[X] = 0(0.40) + 1(0.30) + 2(0.15) + 3(0.10) + 4(0.05) = 0 + 0.30 + 0.30 + 0.30 + 0.20 = 1.10$$

Step 2 – Variance straight from the Handbook definition.

$$\text{Var}(X) = \mathbf{1.39}$$

Step 3 – Cross-check with the shortcut form. The same Handbook page also writes the variance as \(V[X] = E[X^2] - \mu^2\), which is faster on a calculator:

$$E[X^2] = 0^2(0.40) + 1^2(0.30) + 2^2(0.15) + 3^2(0.10) + 4^2(0.05) = 0.30 + 0.60 + 0.90 + 0.80 = 2.60$$

$$\text{Var}(X) = 2.60 - (1.10)^2 = 2.60 - 1.21 = 1.39 \quad \checkmark$$

Step 4 – Answer: (C).

Read the units. A variance carries the square of the units of \(X\), here interruptions\(^2\) per day\(^2\) — which is why you would report the standard deviation, \(\sigma = \sqrt{1.39} = 1.18\) interruptions per day, to anyone operating the network. The question asks for the variance, so the square root is not taken.

Step 5 – Eliminate the distractors:

(A) \(2.60\) is \(E[X^2]\) with nothing subtracted at all. It can never be the variance unless the mean happens to be zero, and here the mean is \(1.10\).

(B) \(1.18 = \sqrt{1.39}\) is the standard deviation. Nothing in the arithmetic is wrong; one extra keystroke was pressed. Re-read the word in the question before you answer — this is the most common way to lose this point.

(D) \(1.50 = 2.60 - 1.10 = E[X^2] - E[X]\) subtracts the mean instead of the mean squared. The Handbook line is \(E[X^2] - \mu^2\), and \(\mu^2 = 1.21\), not \(1.10\).

Answer: A) \(95.45\%\)

Handbook: FE Reference Handbook 10.6, Engineering Probability and Statistics → Normal Distribution (Gaussian Distribution), p. 68 (PDF viewer: page 74), which gives the transformation and defines the table's columns:

$$z = \frac{x - \mu}{\sigma}$$

The table itself is at the end of that section, on p. 77 (PDF viewer: page 83), and is reproduced in the figure. Note the Handbook calls it the unit normal distribution table, and it has no second-decimal columns — the resolution is one decimal in \(x\).

Step 1 – Standardize both specification limits.

$$z_1 = \frac{4.80 - 5.00}{0.10} = -2.00 \qquad\qquad z_2 = \frac{5.20 - 5.00}{0.10} = +2.00$$

The specification band is symmetric about the mean: it is exactly \(\mu \pm 2\sigma\).

Step 2 – Because the limits are \(\pm x\), read the W column. That column exists precisely for a two-sided band, and no subtraction is needed. At the row \(x = 2.0\):

$$P(4.80 \leq X \leq 5.20) = P(-2.00 \leq Z \leq 2.00) = W(2.0) = 0.9545 = \mathbf{95.45\%}$$

Step 3 – Answer: (A).

Cross-check with the other columns. Everything on that row is consistent:

$$W = F(2.0) - F(-2.0) = F(2.0) - \left[1 - F(2.0)\right] = 2(0.9772) - 1 = 0.9544$$
$$W = 1 - 2R(2.0) = 1 - 2(0.0228) = 0.9544$$

Those routes give \(0.9544\) rather than \(0.9545\) only because \(F\) and \(R\) are themselves already rounded to four decimals; the W column is computed before rounding. When the table offers a column that answers the question directly, use it — that is why the question says most nearly.

The fallout is on the same row. Units outside specification are \(1 - 0.9545 = 0.0455\), which the table prints directly as \(2R(2.0) = 0.0455\) — about 4.6%, or roughly 1 unit in 22. That is why these parts are trimmed before they ship.

Step 4 – Eliminate the distractors:

(B) \(97.72\% = F(2.0)\) is the area from \(-\infty\) to \(+2\): everything below the upper limit, which still includes every unit that falls below 4.80 V. The lower tail was never removed. Reaching for the F column on a two-sided problem is the single most common way to lose this question.

(C) \(99.73\% = W(3.0)\) is the \(\pm 3\sigma\) row, one row too far down for the same reason.

(D) \(68.27\% = W(1.0)\) comes from \(z = 1\), which is what you get by dividing the specification half-width by itself — treating \(0.20\text{ V}\) as \(\sigma\). Read the two numbers apart: the half-width is \(0.20\text{ V}\) and \(\sigma\) is \(0.10\text{ V}\), so the band is two standard deviations wide on each side, not one.

Answer: D) \(0.441\)

This is the Handbook's Bayes' theorem: \(P(B_j \mid A) = P(A \mid B_j)P(B_j) / \sum_i P(A \mid B_i)P(B_i)\), with \(A\) = defective and \(B_j\) = made by machine \(j\). The denominator is the overall defect rate, built from the three machines' shares and rates:

$$P(D) = 0.03(0.50) + 0.05(0.30) + 0.02(0.20) = 0.015 + 0.015 + 0.004 = 0.034$$

$$P(M_2 \mid D) = \dfrac{0.05 \times 0.30}{0.034} = \dfrac{0.015}{0.034} = 0.441$$

Machine 2 makes only 30% of the boards but 44% of the defective ones, because its defect rate is the highest. Notice that Machine 1 contributes the same 0.015 to the numerator - a larger share at a lower rate - so a defective board is equally likely to have come from Machine 1; the remaining 12% come from Machine 3.

0.300 is the prior, Machine 2's share of production, with the evidence that the board is defective ignored.

0.050 is \(P(D \mid M_2)\), the defect rate, mistaken for \(P(M_2 \mid D)\). Reversing a conditional probability is the error Bayes' theorem exists to correct.

0.118 is \(0.004/0.034\), the posterior for Machine 3.

Answer: D) \(\$3{,}250\)

Each severity contributes its cost weighted by its probability, and the Handbook's \(E[X] = \sum x_k f(x_k)\) adds them:

\(E[\text{cost}] = 5{,}000(0.15) + 50{,}000(0.03) + 500{,}000(0.002) = 750 + 1{,}500 + 1{,}000 = \$3{,}250\)

The three terms are of the same order even though the costs span two orders of magnitude, because the probabilities fall as steeply as the costs rise. That is typical of risk budgets: the rare catastrophe and the routine nuisance can matter about equally in expectation, and neglecting either one misstates the budget by a third.

$2,250 is \(750 + 1{,}500\), the catastrophic term dropped - the natural slip when a probability of 0.002 looks negligible. Its contribution is a thousand dollars a month.

$1,750 is \(750 + 1{,}000\), the major-outage term dropped.

$2,500 is \(1{,}500 + 1{,}000\), the minor-outage term dropped.