Probability and Statistics practice questions

Answer: D) 0.19

For independent events, P(at least one) = 1 − P(none) = 1 − (0.9)² = 0.19. Simply adding the probabilities double-counts the case where both valves fail.

• AIncorrect. 0.81 = (0.9)(0.9) is the probability that neither valve fails, which is the complement of the event asked for.
• BIncorrect. Adding 0.1 + 0.1 counts the case where both valves fail twice. You must subtract P(both) = 0.01, which gives 0.19.
• CIncorrect. 0.01 = (0.1)(0.1) is the probability that both valves fail, not that at least one fails.
• DCorrect. P(at least one fails) = 1 − P(neither fails) = 1 − (0.9)(0.9) = 1 − 0.81 = 0.19. Equivalently, 0.1 + 0.1 − 0.01 = 0.19.

Answer: A) 5

• ACorrect. Variances add: \(\sqrt{9 + 16}\).
• BIncorrect. Add variances, not standard deviations.
• CIncorrect. That's the variance.
• DIncorrect. Variances add.

Answer: B) Rejecting a batch that actually meets specification because the test rejected H0

A Type I error means rejecting a true H0 (here, scrapping a good batch), and its probability is α. Releasing a bad batch, which is failing to reject a false H0, is a Type II error.

• AIncorrect. Failing to reject a false null hypothesis is a Type II error, and its probability is β.
• BCorrect. Rejecting a null hypothesis that is true is a Type I error (a false positive), and its probability is the significance level α.
• CIncorrect. Rejecting a false null hypothesis is a correct decision. Its probability is the power of the test, 1 − β.
• DIncorrect. Failing to reject a true null hypothesis is a correct decision, and its probability is 1 − α.

Answer: D) 7

The mode is the most frequently occurring value in a data set. Here 7 occurs twice, so the mode is 7. The median (7.5), mean (8), and range (10) are different measures.

• AIncorrect. 7.5 is the median: the average of the two middle values, (7 + 8)/2.
• BIncorrect. 8 is the arithmetic mean: 48/6 = 8.
• CIncorrect. 10 is the range: 13 − 3 = 10.
• DCorrect. The mode is the value that occurs most often, and 7 appears twice while every other value appears once.

Answer: C) 7

One fair die has an expected value of (1+2+…+6)/6 = 3.5. Because expectation is linear, the expected sum of two rolls is 3.5 + 3.5 = 7.

• AIncorrect. 3.5 is the expected value of a single roll. The sum of two rolls has twice that expectation.
• BIncorrect. 6 is the largest face on one die, not the mean of the sum.
• CCorrect. Expectation is linear: E[X1 + X2] = E[X1] + E[X2] = 3.5 + 3.5 = 7.
• DIncorrect. 12 is the largest possible sum (two sixes), not its expected value.

Answer: D) 1.0

• AIncorrect. That isn't the calculation.
• BIncorrect. The standard error divides by \(\sqrt{n}\).
• CIncorrect. Divide by the square root of n.
• DCorrect. \(s/\sqrt{n} = 6/6\).

Answer: A) Student's t-distribution

• ACorrect. It accounts for estimating σ from the sample.
• BIncorrect. That's used for variances.
• CIncorrect. Use the t-distribution.
• DIncorrect. That's for counts.

Answer: A) 10

When order does not matter, use combinations: C(n,r) = n!/[r!(n − r)!]. For n = 5 and r = 2, this gives 10.

• ACorrect. The number of combinations is C(5,2) = 5!/(2!·3!) = 120/12 = 10.
• BIncorrect. 20 is the number of permutations, P(5,2) = 5!/3!, which counts different orderings of the same pair separately.
• CIncorrect. 25 = 5², which counts ordered picks with replacement. That allows the same catalyst to be picked twice.
• DIncorrect. 120 = 5!, the number of ways to arrange all five catalysts in order. It is not the number of pairs.

Answer: A) 10.1

• ACorrect. The sum is 50.5, divided by 5.
• BIncorrect. Recheck the sum, 50.5.
• CIncorrect. Recheck the sum, 50.5.
• DIncorrect. That's the sum; divide by 5.

Answer: D) 3

• AIncorrect. That inverts it.
• BIncorrect. Divide by σ.
• CIncorrect. That divides 130 by 10.
• DCorrect. (130 − 100)/10.