FE Electrical Probability & Statistics Practice Problems

Answer: B) \(47.1\,\Omega\)

The Handbook defines the arithmetic mean of a sample as \(\bar{X} = \frac{1}{n}\sum X_i\). Add the five readings and divide by five:

$$\bar{X} = \dfrac{47.2 + 46.8 + 47.5 + 46.5 + 47.5}{5} = \dfrac{235.5}{5} = 47.1\,\Omega$$

A quick check: the readings run from 46.5 to 47.5 \(\Omega\), so the mean has to land inside that range, and it sits a little above the midpoint because three of the five readings are at the high end. The three wrong answers are all legitimate descriptions of this same sample - each is defined on the same Handbook page - which is exactly what the item is testing: knowing which measure is which.

47.2 \(\Omega\) is the median. Sorted, the readings are 46.5, 46.8, 47.2, 47.5, 47.5, and with \(n = 5\) odd the median is the \((n+1)/2 = 3\)rd item.

47.5 \(\Omega\) is the mode, the value that occurs with greatest frequency; it appears twice.

47.0 \(\Omega\) is the midrange, \((46.5 + 47.5)/2\). It uses only the two extreme readings and throws the other three away, so it equals the mean only when the data happen to be symmetric.

Answer: C) \(0.92\)

Eight of the hundred are defective, so the probability of drawing one of them is \(8/100 = 0.08\). Drawing a good LED is the complementary event - the two outcomes are mutually exclusive and one of them must happen - and the Handbook's first law of probability, that a certain event has probability 1, gives

$$P(\text{not defective}) = 1 - 0.08 = 0.92$$

The same answer comes from counting directly, \(92/100\). Reaching for the complement is the habit worth building, because in the harder versions of this question - at least one defective in a sample of ten, say - counting directly is far more work than subtracting the one case that is easy.

0.08 is the probability of the event the question is not asking about.

0.98 subtracts 2 instead of 8, or reads the 8 as a percentage of something other than the batch.

0.80 subtracts 0.2, the wrong power of ten.

Answer: D) Generator B, at \(\$13{,}000\)

The repair cost of each generator is a random variable that is either zero or the full repair bill, so its expected value by the Handbook's \(E[X] = \sum x_k f(x_k)\) is the bill times the probability of incurring it:

\(E[\text{A}] = 15{,}000 + 0.10 \times 8{,}000 = 15{,}000 + 800 = \$15{,}800\)

\(E[\text{B}] = 12{,}000 + 0.20 \times 5{,}000 = 12{,}000 + 1{,}000 = \$13{,}000\)

Generator B is cheaper in expectation by $2,800. Its higher repair probability costs it $200 a year more than A in expected repairs, but that is far less than the $3,000 saved on the purchase. Expected cost is the right basis when the decision is repeated many times or the risk is affordable; a buyer who could not survive an $8,000 bill might weigh the worst case instead, but that is a different criterion from the one the question sets.

Generator B at $17,000 and Generator A at $23,000 add the full repair bill to the purchase price, as if the repair were certain. That is the worst case, not the expected case, and it also picks the wrong generator by a smaller margin.

Generator A at $15,800 has A's expected cost right but names it the cheaper one; $15,800 is more than $13,000.

Answer: D) \(5.1\text{ V}\) and \(5.05\text{ V}\)

Both definitions are printed on the Handbook page, and the second one decides the item. The mode is the value that occurs with greatest frequency: 4.8 and 4.9 appear once each, 5.0 twice, 5.1 three times, 5.3 once, so the mode is 5.1 V.

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

$$4.8,\; 4.9,\; 5.0,\; 5.0,\; 5.1,\; 5.1,\; 5.1,\; 5.3$$

With \(n = 8\) even, the Handbook takes the average of the \((n/2)\)th and \((n/2 + 1)\)th items, the 4th and 5th:

$$\text{median} = \dfrac{5.0 + 5.1}{2} = 5.05\text{ V}$$

Note that the median need not be one of the readings. No sensor reading of 5.05 V was ever taken, and that is normal for an even-sized sample; rejecting 5.05 V because it is not in the list is the instinct the most common wrong answer is built on.

5.1 V and 5.0 V has the mode right and takes the 4th value alone as the median, skipping the average the even case requires.

5.0 V and 5.0 V takes 5.0 as the mode - it does appear twice, but 5.1 appears three times - and repeats the un-averaged median.

5.0 V and 5.05 V gets the median right but reports 5.0 V as the mode - the runner-up, with two occurrences against three.

Answer: C) \(0.955\)

The conditioning event is 'at least one detected', \(A \cup B\). Because the systems are independent, the Handbook's compound-probability law gives \(P(A, B) = P(A)P(B) = 0.855\), and the law of total probability gives

$$P(A \cup B) = 0.95 + 0.90 - 0.855 = 0.995$$

The event 'A detected it and at least one detected it' is just 'A detected it', since A detecting already guarantees at least one. So the conditional probability is

$$P(A \mid A \cup B) = \dfrac{P(A)}{P(A \cup B)} = \dfrac{0.95}{0.995} = 0.955$$

The conditioning raises the probability only slightly, from 0.950 to 0.955, because the information 'at least one fired' rules out very little - only the 0.5% of faults that both systems missed.

0.950 is \(P(A)\) with the conditioning ignored. It is close because the condition is weak, but it is not the conditional probability.

0.995 is \(P(A \cup B)\), the denominator, reported as the answer.

0.859 is \(P(A \cap B)/P(A \cup B) = 0.855/0.995\), the probability that both detected it given that at least one did. It answers a different question; A detecting does not require B to.