Know Your
Numbers.

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A normal distribution is always symmetric and bell-shaped, centered at its mean (μ). Option C describes the standard normal distribution specifically — not all normal distributions. Option A describes a right-skewed distribution.

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Z = (85 − 75) / 10 = 10 / 10 = 1.0. A Z-score of +1.0 means the student scored exactly one standard deviation above the mean. Always plug in: (your value − mean) ÷ std dev.

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Memorize: 68 → 95 → 99.7 for 1σ → 2σ → 3σ. Within ±2σ: about 95% of all data. This is one of the most tested facts in statistics. Think: "2 standard deviations catches almost everyone."

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Z = (70 − 64) / 3 = 6 / 3 = 2.0. This woman is exactly 2 standard deviations above the mean. Using the Empirical Rule, only about 2.5% of women are taller than her (since 95% fall within ±2σ, and 5% fall outside — split evenly, 2.5% on each side).

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95% of bags fall within Z = ±2, meaning they are accepted. Therefore, 100% − 95% = 5% are rejected. Common mistake: choosing 0.3% (that's the area outside ±3σ) or confusing accepted with rejected.

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68% falls within ±1σ → 32% falls outside ±1σ → 16% falls above Z = +1 (by symmetry). So the top 16% starts at Z = +1, which is: x = μ + 1σ = 500 + 100 = 600. Key insight: "top 16%" means you need to find the score at Z = +1.

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Student A: Z = (78−70)/8 = 1.0. Student B: Z = (82−75)/4 = 7/4 = 1.75. Higher Z-score means better performance relative to the class. This is exactly WHY Z-scores exist — to compare across different tests fairly. Raw score alone is misleading.

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H₀ (null hypothesis) always represents the "boring, default" position: nothing is happening, no effect exists. Think of H₀ as the prosecution saying "innocent until proven guilty." You only reject it if evidence is strong enough.

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p-value (0.03) < α (0.05) → Reject H₀. A small p-value means the data is very unlikely if H₀ were true. Important: we never "accept H₀" — we only fail to reject it. Option C is always wrong wording in statistics.

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H₀ always uses "=" (equality). The teacher's claim goes into Hₐ — and since she claims scores are "above 80," Hₐ: μ > 80. This is a one-tailed (right-tailed) test. Option B would be a two-tailed test (≠ means "different in either direction").

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p-value (0.08) > α (0.05) → Fail to reject H₀. The evidence is not strong enough. Since the engineer suspects a difference in either direction, this is a two-tailed test — do NOT halve the p-value. Option D is wrong here.

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Type I = False Positive = "false alarm" — you cry wolf when there is no wolf. Its probability equals α. Type II = False Negative = "missed it" — the wolf was there but you didn't notice. Its probability is β. Memory trick: Type I → "I cried wolf" (rejected truth).

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Type II Error = Fail to reject a false H₀. Here H₀ says "no effect," but in reality the vaccine works. Failing to reject H₀ means we conclude "no effect" — and a working vaccine gets shelved. Option A describes a Type I Error (approving something ineffective = false positive).

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The p-value is a conditional probability: given H₀ is true, how likely is this data? It is NOT the probability that H₀ is true (Option A), and it does not prove anything (Option D). Option C confuses the p-value with power. Option B is the precise, correct definition.

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The confidence interval is: 47% ± 3% = [44%, 50%]. The margin of error is not about survey mistakes or uncertainty in responses — it quantifies the sampling variability. We are (typically 95%) confident the true population value lies in this range.

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MoE = z* × (σ / √n). A larger n makes √n bigger, making σ/√n smaller → MoE decreases. Options A and C both increase MoE. Option B decreases n → increases MoE. Always remember: more data = more precision = smaller MoE.

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MoE = z* × (σ/√n) = 1.96 × (1.0/√100) = 1.96 × (1.0/10) = 1.96 × 0.1 = 0.196 hours. So the 95% CI is: 3.5 ± 0.196 → [3.304, 3.696]. Common mistake: forgetting to take √n (just dividing by n = 100 instead of 10).

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This is the most commonly misunderstood concept! Once an interval is computed, the true mean either IS or ISN'T in it (probability is 0 or 1). The "95%" refers to the process: if we repeated this study many times, 95% of those intervals would capture the true mean. Option A sounds right but is technically wrong — the true mean is a fixed value, not random.

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For proportions: MoE = z* × √(p(1−p)/n). Solving for n: n = (z*/MoE)² × p(1−p) = (1.96/0.04)² × 0.5 × 0.5 = (49)² × 0.25 = 2401 × 0.25 = 600.25 → round up to 601. Note: n = 385 is the answer for MoE = 5%, a very common mix-up on exams.

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Key insight: A 95% CI is directly equivalent to a two-tailed hypothesis test at α = 0.05. The CI is [2,250; 2,550]. The null value 2,300 falls inside this interval... wait — 2,300 IS inside [2,250, 2,550]. So we fail to reject? Actually: 2,250 < 2,300 < 2,550 ✓ — 2,300 is inside the interval, so Fail to reject H₀. Option A is correct! This is an intentionally tricky "trick question" — always check whether the null value is inside or outside the CI carefully.