Think in Data.
Reason with Confidence.

Correct: B. Mean = (70+72+75+78+150)/5 = 89. Median = 75 (middle value). The extreme outlier 150 pulls the mean upward. The median is resistant to outliers — it only depends on the middle rank, not the actual value. Always prefer median when data is skewed or has outliers.

Correct: B. Both classes have mean = 80. But Class A's values range from 70 to 90 (wider spread), while Class B clusters between 79–81 (narrow spread). Standard deviation measures how far values deviate from the mean on average. Larger deviations → larger SD. Equal means ≠ equal spread!

Correct: B. \(z = (88 - 80)/4 = 8/4 = 2\). A z-score of +2 means the student scored 2 standard deviations above the mean. Positive z = above mean; negative z = below mean; z = 0 = exactly at the mean. About 95% of data falls within \(|z| \leq 2\) in a normal distribution.

Correct: C. IQR = Q3 − Q1 = 50 − 20 = 30. Wait — let's recalculate properly: Lower fence = 20 − 1.5(30) = 20 − 45 = −25. Upper fence = 50 + 1.5(30) = 50 + 45 = 95. Since 90 < 95, Max = 90 is NOT an outlier. Min = 10 > −25, also not an outlier. So there are no outliers. Answer C correctly identifies IQR = 30 and no outlier at 90. Always check BOTH fences!

Correct: B. Skewness is named after the direction of the tail, NOT the peak. Long tail to the left → left-skewed (negatively skewed). In left-skewed distributions, mean < median < mode. A common trap: students say "peak is on right, so right-skewed" — wrong! Follow the tail!

Correct: A. \(P(S \mid M) = P(S \cap M) / P(M) = 0.15 / 0.40 = \mathbf{0.375}\). Think of it as: "Among the 40% who study math, 15% also study science." So \(15/40 = 37.5\%\). The common mistake is using 0.30 (unconditional P(S)) or 0.15 (the intersection) without dividing.

Correct: C. Mutually exclusive means \(P(A \cap B) = 0\) ✓ (given). For independence, we need \(P(A \cap B) = P(A) \cdot P(B) = 0.4 \times 0.3 = 0.12\). But \(P(A \cap B) = 0 \neq 0.12\), so NOT independent. In fact, mutually exclusive events (when both have positive probabilities) are always dependent — if A occurs, B cannot!

Correct: B. P(no rain on one day) = 1 − 0.35 = 0.65. For 3 independent days: \(0.65^3 = 0.65 \times 0.65 \times 0.65 \approx \mathbf{0.274}\). Common errors: (A) only calculates one day; (C) incorrectly cubes 0.35; (D) multiplies 0.65 × 3 (addition instead of multiplication for independent AND events).

Correct: A. \(E(X) = (10)(0.2) + (5)(0.5) + (-3)(0.3) = 2 + 2.5 - 0.9 = \mathbf{\$3.60}\)... Wait, that gives B! Let me recheck: \(2.0 + 2.5 - 0.9 = 3.6\). Correct answer is B: $3.60. Don't forget: losses are negative values! Multiply carefully and keep track of signs.

Correct: A. \(P(X=4) = \binom{6}{4}(0.5)^4(0.5)^2 = 15 \times 0.0625 \times 0.25 = 15 \times 0.015625 = \mathbf{0.234}\). Note: \(\binom{6}{4} = \binom{6}{2} = 15\). The "choose" (combination) accounts for all the different orderings of 4 heads — without it, you'd only count one specific sequence.

Correct: B. \(\sigma_{\bar{x}} = 20/\sqrt{100} = 20/10 = \mathbf{2}\). This is called the Standard Error (SE). Larger samples reduce variability of the sample mean. SE ≠ SD — SE is the standard deviation of the sampling distribution (how much \(\bar{x}\) varies from sample to sample), not the original population's spread.

Correct: C. "Consistently too high" = high bias (systematic error, hitting the wrong target). "Very close to each other" = low variability (precise, consistent results). Think of it like an archery target: high bias = arrows cluster away from the bullseye; low variability = arrows cluster tightly together (even if in the wrong spot).

Correct: B. This is the most commonly missed concept in AP Stats. The true μ is a fixed (unknown) value — it's either in (42, 58) or it's not. The "95%" refers to the procedure: if we repeat the sampling many times, 95% of the intervals constructed this way will contain μ. Never say "95% probability that μ is in THIS specific interval."

Correct: B. ME = \(z^* \cdot \sigma/\sqrt{n}\). To halve ME: \(\sigma/\sqrt{n_{\text{new}}} = \frac{1}{2} \cdot \sigma/\sqrt{n_{\text{old}}}\), which gives \(\sqrt{n_{\text{new}}} = 2\sqrt{n_{\text{old}}}\), so \(n_{\text{new}} = 4n_{\text{old}}\). The square root makes the relationship non-linear — halving ME requires 4× the sample size, not 2×. This surprises many students!

Correct: B. H₀: person is innocent (true). We "rejected" H₀ (convicted them) even though H₀ was true → Type I error (false positive). Type I error rate = α (significance level). Type II error = acquitting a guilty person (fail to reject false H₀). Reducing α decreases Type I errors but increases Type II errors — there's always a tradeoff.

Correct: C. Since p = 0.03 < α = 0.05, we reject H₀. The p-value is NOT the probability that H₀ is true (option A — very common mistake!). It's the probability of observing results this extreme assuming H₀ is true. Also, we never "prove" anything in statistics (D) — we only find sufficient evidence to reject or insufficient evidence to reject.

Correct: B. The lurking variable is summer heat — it increases both ice cream sales AND swimming (thus drowning risk). High r only tells us there's a linear relationship, not its cause. To establish causation: you need a randomized controlled experiment. Observational studies can NEVER prove causation, regardless of how high r is.

Correct: B. The slope b₁ = 8 means: for each 1-unit increase in x (study hours), the predicted y (test score) increases by 8. The intercept 50 means a student with 0 study hours is predicted to score 50. Always state slope interpretation in context using "for each additional [x-unit], [y] is predicted to [increase/decrease] by [slope]."

Correct: C. r² = 0.64 means 64% of the variability in test scores is explained by the linear relationship with study hours. The remaining 36% is due to other factors. Full template: "[r² × 100]% of the variation in [y] is explained by the linear relationship with [x]." Never skip "variation" or "linear relationship" in your answer!

Correct: C. A curved residual plot means the linear model is systematically missing a curve in the data — it's underfitting a non-linear relationship. A good residual plot should look like random scatter around zero with no pattern. Patterns (U-shape, funnel, wave) all signal that the linear model is inappropriate. Try a transformation (like log or square root) or a polynomial model.