Probability and Statistics practice questions

Answer: C) 13

• AIncorrect. That adds 3 only.
• BIncorrect. That omits the 3.
• CCorrect. E[2X + 3] = 2E[X] + 3.
• DIncorrect. That isn't linearity of expectation.

Answer: C) 2.92

For a discrete random variable, Var(X) = E[X²] − (E[X])². For a fair die this is 91/6 − 12.25 = 35/12 ≈ 2.92.

• AIncorrect. 3.50 is the expected value E[X] = (1+2+3+4+5+6)/6, not the variance.
• BIncorrect. 1.71 is the standard deviation, the square root of the variance (√2.92 ≈ 1.71).
• CCorrect. Var(X) = E[X²] − (E[X])² = 91/6 − 3.5² = 15.17 − 12.25 = 35/12 ≈ 2.92.
• DIncorrect. 15.17 is E[X²] = 91/6. The square of the mean still has to be subtracted to get the variance.

Answer: D) Reject the null hypothesis

• AIncorrect. α is set in advance.
• BIncorrect. p < α means reject.
• CIncorrect. Tests don't give certainty.
• DCorrect. p is below α.

Answer: C) 0.21

Use the general addition rule: P(A or B) = P(A) + P(B) − P(A and B) = 0.15 + 0.10 − 0.04 = 0.21.

• AIncorrect. This just adds 0.15 + 0.10 and counts the parts with both defects twice. Subtract the joint probability once.
• BIncorrect. This is 1 − (0.85)(0.90), which assumes the two defects are independent. Under independence, the joint probability would be 0.015, but the problem gives 0.04.
• CCorrect. P(M or P) = P(M) + P(P) − P(M and P) = 0.15 + 0.10 − 0.04 = 0.21.
• DIncorrect. This subtracts the joint probability twice (0.25 − 0.08). The overlap is subtracted only once.

Answer: A) The two variables have a strong positive linear association

A correlation coefficient near +1 shows a strong positive linear association, but on its own it does not prove causation. Note also that r² (not r) is the share of variation explained, and r is not the regression slope.

• ACorrect. An r near +1 means a strong positive linear relationship. With observational data, it does not show which variable causes the other or whether a confounder drives both.
• BIncorrect. Correlation measures association, not causation. A confounding factor, such as line speed or the material lot, could drive both variables. Showing cause needs a controlled experiment or other causal evidence.
• CIncorrect. The share of variation explained by a linear fit is the coefficient of determination, r² = 0.92² ≈ 0.85, or about 85%. It is not r itself.
• DIncorrect. r has no units and is not the regression slope. The slope depends on the units and standard deviations of both variables.

Answer: C) Reduces the Type II error probability

• AIncorrect. α is set by the analyst.
• BIncorrect. Larger samples raise power.
• CCorrect. Power rises with sample size.
• DIncorrect. Power depends on n.

Answer: A) 99.7%

• ACorrect. About 0.27% falls outside.
• BIncorrect. That's about ±2σ.
• CIncorrect. That's ±1σ.
• DIncorrect. A small fraction lies outside.

Answer: D) Paired t-test

• AIncorrect. Use the paired t-test.
• BIncorrect. The samples are paired.
• CIncorrect. That's for categorical counts.
• DCorrect. It analyzes the per-worker differences.

Answer: A) 0.81

• ACorrect. R² = r² = 0.9².
• BIncorrect. Square r.
• CIncorrect. That's √0.9.
• DIncorrect. Square, don't double.

Answer: A) 6.93 hours

For an exponential distribution, the median is θ ln 2. With a 10-hour mean, that is about 6.93 hours.

• ACorrect. Setting F(t) = 1 − e^(−t/θ) = 0.5 gives t = θ ln 2 = 10 × 0.693 ≈ 6.93 hours.
• BIncorrect. 10 hours is the mean. The exponential distribution is right-skewed, so its median is below its mean.
• CIncorrect. This just halves the mean, as if the distribution were uniform starting at zero. That is not how an exponential median works.
• DIncorrect. This divides the mean by ln 2 (10 / 0.693) instead of multiplying by it.