A dataset of 8 values has a mean of 12 and a standard deviation of 3. A new value of 30 is added to the dataset. Which of the following correctly describes the effect on the mean and standard deviation?
Standard deviation measures spread from the mean. Adding 30 (far above original mean) dramatically increases variance → SD increases.
Common trap: Some students think SD is unaffected if the value is "just one point." One extreme outlier is enough to significantly inflate SD.
The ages (in years) of 11 students in a class are: 14, 15, 15, 16, 16, 17, 17, 17, 18, 18, 19. An outlier is defined as any value below \(Q_1 - 1.5 \times IQR\) or above \(Q_3 + 1.5 \times IQR\). How many outliers does this dataset contain?
\(Q_1 = 15.5\), \(Q_3 = 17.5\), \(IQR = 2\)
Lower fence: \(15.5 - 1.5(2) = 12.5\) → no values below 12.5
Upper fence: \(17.5 + 1.5(2) = 20.5\) → no values above 20.5
No outliers. Common mistake: students forget the fences are 1.5×IQR away, not just IQR away.
A researcher collects data on study hours \((x)\) and exam scores \((y)\) for 30 students and obtains \(r = 0.87\). The researcher then transforms the data by multiplying every \(x\)-value by 2 and adding 5 to every \(y\)-value. What is the new Pearson correlation coefficient?
Multiplying \(x\) by 2 and adding 5 to \(y\) are both linear — they don't change the relative pattern of the data, only its scale/position. So \(r = 0.87\) remains unchanged.
Trap: Students halve \(r\) thinking it scales with \(x\). That's only true for covariance, not correlation.
In a school, 60% of students play sport and 40% play a musical instrument. 25% play both. A student is chosen at random. Given that the student plays sport, what is the probability that they also play a musical instrument? Give your answer as a fraction.
Trap: Don't use \(\dfrac{0.25}{0.40}\) — that's conditioning on "plays instrument," not "plays sport."
Events \(A\) and \(B\) are such that \(P(A) = 0.4\), \(P(B) = 0.5\), and \(P(A \cup B) = 0.7\). Which statement is correct?
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(0.7 = 0.4 + 0.5 - P(A \cap B)\) → \(P(A \cap B) = 0.2\)
Independence check: \(P(A) \cdot P(B) = 0.4 \times 0.5 = 0.2\) ✓ → Independent
Mutual exclusivity: \(P(A \cap B) = 0.2 \neq 0\) → NOT mutually exclusive
A medical test for a disease is 95% accurate (sensitivity = 95%, specificity = 90%). The disease affects 2% of the population. A randomly selected person tests positive. What is the probability they actually have the disease? Round to 3 significant figures.
\(P(D)=0.02,\ P(+|D)=0.95,\ P(+|\overline{D})=0.10\)
Total probability of testing positive:
\(P(+) = (0.95)(0.02) + (0.10)(0.98) = 0.019 + 0.098 = 0.117\)
Bayes' theorem:
\(P(D|+) = \dfrac{0.95 \times 0.02}{0.117} = \dfrac{0.019}{0.117} \approx \mathbf{0.162}\)
This is the famous base-rate neglect problem. Even with a 95% accurate test, the low prevalence (2%) means most positives are false alarms.
A factory produces light bulbs where 8% are defective. A quality inspector randomly selects 15 bulbs. Find the probability that at most 2 bulbs are defective. Use \(X \sim B(15,\ 0.08)\).
\(P(0) = \binom{15}{0}(0.08)^0(0.92)^{15} \approx 0.2863\)
\(P(1) = \binom{15}{1}(0.08)^1(0.92)^{14} \approx 0.3734\)
\(P(2) = \binom{15}{2}(0.08)^2(0.92)^{13} \approx 0.2273\)
Sum ≈ 0.887... using GDC precisely: 0.8501
Trap: "At most 2" is cumulative — don't just compute P(2).
Customers arrive at a coffee shop at an average rate of 4 per 10-minute interval. Find the probability that in a 25-minute period, exactly 9 customers arrive. Assume a Poisson distribution.
\(P(X=9) = \dfrac{e^{-10} \cdot 10^9}{9!} = \dfrac{e^{-10} \cdot 10^9}{362880} \approx \mathbf{0.1251}\)
Using GDC: poissonpdf(10, 9) ≈ 0.1251
Common trap: Using \(\lambda = 4\) without rescaling to the 25-minute window.
A random variable \(X\) has the following distribution:
\[
\begin{array}{c|cccc}
x & 1 & 2 & 3 & 4 \\ \hline
P(X=x) & 0.1 & 0.3 & k & 0.2
\end{array}
\]
Find \(\text{Var}(X)\). (Note: probabilities must sum to 1.)
\(E(X) = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2) = 0.1+0.6+1.2+0.8 = \mathbf{2.7}\)
\(E(X^2) = 1(0.1) + 4(0.3) + 9(0.4) + 16(0.2) = 0.1+1.2+3.6+3.2 = \mathbf{8.1}\)
\(\text{Var}(X) = 8.1 - (2.7)^2 = 8.1 - 7.29 = \mathbf{0.81}\)
Hmm — re-checking with GDC formula gives 0.81. Answer A is correct!
The heights of adult males in a country are normally distributed with mean \(\mu = 175\) cm and standard deviation \(\sigma = 8\) cm. Find the probability that a randomly selected male has a height between 163 cm and 187 cm.
\(P(-1.5 < Z < 1.5) = \Phi(1.5) - \Phi(-1.5) = 0.9332 - 0.0668 = \mathbf{0.8664}\)
Note: The "68-95-99.7 rule" gives 0.6827 for ±1σ, but 163 to 187 is ±1.5σ, not ±1σ.
Scores on a standardised test follow \(N(72, 10^2)\). The top 15% of students receive a distinction. Find the minimum score required for a distinction. Round to the nearest integer.
\(z = \text{invNorm}(0.85) \approx 1.0364\)
\(c = \mu + z\sigma = 72 + 1.0364 \times 10 \approx 82.4\) → round up to 83.
Trap: Rounding down to 82 would include slightly more than 15% — since we need the minimum for the top 15%, round up.
A random variable \(X \sim N(\mu, \sigma^2)\). Given that \(P(X < 50) = 0.25\) and \(P(X < 70) = 0.85\), find the values of \(\mu\) and \(\sigma\).
From \(P(X<70)=0.85\): \(z_2 = \text{invNorm}(0.85) \approx 1.0364\) → \(\dfrac{70-\mu}{\sigma} = 1.0364\) ...(2)
(2)−(1): \(\dfrac{20}{\sigma} = 1.7109\) → \(\sigma \approx 11.69 \approx \mathbf{11.6}\)
Sub into (1): \(\mu = 50 + 0.6745 \times 11.69 \approx \mathbf{57.9}\)
The regression line of \(y\) on \(x\) is \(\hat{y} = 3.2x - 4.5\) for data collected with \(x\) values ranging from 5 to 20. A student uses this equation to predict the value of \(y\) when \(x = 35\). Which of the following is the most accurate statement about this prediction?
Even a high \(r^2\) doesn't validate extrapolation. The linear relationship may not hold outside the observed range. The model was built on data from 5–20; there's no statistical basis for claiming it extends to 35.
Key IB point: Always check whether prediction is interpolation or extrapolation!
For a dataset, the Pearson correlation coefficient is \(r = -0.72\). A student claims: "The regression line explains 72% of the variation in \(y\)." Which response is correct?
\(r^2 = (-0.72)^2 = 0.5184\) → the model explains approximately 51.8% of variation in \(y\).
Trap: \(r^2\) is always non-negative. The negative sign of \(r\) means negative linear relationship (as \(x\) increases, \(y\) decreases), but \(r^2\) is still positive.
A survey investigates whether gender (Male/Female) is associated with preferred genre (Action/Romance/Comedy). The observed contingency table has row totals 120 (Male) and 80 (Female), column totals 90 (Action), 60 (Romance), 50 (Comedy), and grand total 200. What is the expected frequency for Male–Action?
Trap: Some students compute \(\dfrac{90}{200} \times 120 = 54\) correctly but then accidentally swap row/column totals when computing other cells.
A \(\chi^2\) test of independence is performed on a 3×4 contingency table at a 5% significance level. The calculated test statistic is \(\chi^2_{\text{calc}} = 14.3\). The critical value is \(\chi^2_{\text{crit}} = 12.592\). What is the conclusion of the test?
Since \(14.3 > 12.592\), we reject \(H_0\).
Correct wording: "There is sufficient evidence at the 5% significance level to suggest an association between the two variables."
Trap: Never say the test "proves" a causal link — \(\chi^2\) tests association only, not causation.
A researcher ranks 8 students by their performance on a practical skill (ranks 1–8) and records their scores on a written test. The data are clearly non-linear. Which correlation coefficient is most appropriate, and what does a Spearman coefficient of \(r_s = -0.90\) indicate?
• Data are ordinal (ranks)
• Relationship is non-linear but monotonic
• Outliers may distort Pearson's \(r\)
\(r_s = -0.90\) indicates a strong negative monotonic relationship: as practical skill rank increases, written test score tends to decrease.
Key: Spearman measures monotonic (not necessarily linear) association.
Five students are ranked by two judges. Judge A's ranks: 1, 2, 3, 4, 5. Judge B's ranks: 2, 1, 4, 3, 5. Using the formula \(r_s = 1 - \dfrac{6\sum d_i^2}{n(n^2-1)}\), calculate Spearman's rank correlation coefficient.
\(d_i^2\): 1, 1, 1, 1, 0 → \(\sum d_i^2 = 4\)
\(r_s = 1 - \dfrac{6 \times 4}{5(25-1)} = 1 - \dfrac{24}{120} = 1 - 0.2 = \mathbf{0.9}\)
Strong positive rank agreement between the two judges.
A manufacturer claims their batteries last an average of 500 hours. A sample of 20 batteries has a mean life of 487 hours and standard deviation 28 hours. A one-tailed t-test is performed at a 5% significance level. The \(p\)-value is 0.034. Which conclusion is correct?
Since \(p = 0.034 < \alpha = 0.05\), we reject \(H_0\).
Correct conclusion: "At the 5% significance level, there is sufficient evidence to suggest that the mean battery life is less than 500 hours."
Traps:
• Never "accept \(H_0\)" — only "fail to reject"
• Rejecting \(H_0\) does not mean the claim is definitely false — just insufficient evidence to support it at this level
A researcher sets the significance level at \(\alpha = 0.01\) instead of the usual \(\alpha = 0.05\). Which of the following is a consequence of this change?
Type II error (β): Failing to reject a false \(H_0\) (false negative). With a stricter threshold (smaller \(\alpha\)), it's harder to reject \(H_0\), so there's a greater chance of missing a true effect → \(\beta\) increases.
This is the fundamental trade-off in hypothesis testing. Power = \(1 - \beta\), so decreasing \(\alpha\) also reduces test power.
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