Flashcards · IB Diploma Programme Mathematics: Applications and Interpretation · Topic 4

Statistics and probability flashcards

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4.1 Population, sample, sampling techniques, reliability and bias, interpretation of outliers

Population and sample: Population: the whole group you want to know about. Sample: the part of the population you collect data from.
Random sample: Each member of the population has an equal chance of selection.
Discrete vs continuous data: Discrete: separate values, usually counted (number of pets). Continuous: any value in a range, measured (reaction time, mass).
Bias: The sample systematically over- or under-represents part of the population.
Stratified vs quota sampling: Both take set numbers from each group, but stratified chooses at random within each group; quota does not.
Systematic vs simple random sampling: Systematic picks at a fixed interval (every k-th item, k = population ÷ sample size) after one random start; simple random picks every member at random.
Outlier: A value more than 1.5 × IQR below Q₁ or above Q₃. Whether it is an error depends on the context.

4.2 Presentation of data: frequency distributions, histograms, cumulative frequency graphs, box-and-whisker diagrams

Drawing a cumulative frequency graph: where are the points plotted?: Each running total at the upper boundary of its class (not the mid-interval value), starting from (lower boundary of the first class, 0).
Reading the median, quartiles and percentiles from a cumulative frequency graph: Read across at n/2 (median), n/4 (Q₁), 3n/4 (Q₃), and (p/100) × n for the p-th percentile, then down to the x-axis.
Percentile: The p-th percentile is the value below which p% of the data lie. Q₁ is the 25th, the median the 50th, Q₃ the 75th.
Box and whisker diagram with outliers: Box from Q₁ to Q₃ with a line at the median. Whiskers go to the smallest and largest values that are not outliers. Mark each outlier with a cross.
How do you compare two distributions from box plots?: Compare the medians in context (‘on average, group A took longer’), then the IQRs or ranges in context (‘group B’s times were more consistent’). Numbers alone are not enough.

4.3 Measures of central tendency and dispersion; effect of constant changes on data; quartiles

Mean of a frequency table, and the estimate for grouped data: x̄ = Σfx/Σf. For grouped data use the mid-interval value as x; the result is an estimate because the actual values are unknown.
Mode vs modal class: Mode is a value; modal class is the class with the highest frequency, and the guide uses it only for equal class intervals.
Effect on summary statistics of adding k to every value: Mean, median and quartiles increase by k. Standard deviation, variance, IQR and range do not change.
Effect on summary statistics of multiplying every value by k (k > 0): Mean, median, quartiles, standard deviation, IQR and range are multiplied by k. Variance is multiplied by k². Example: sd 6, ×3 then +2: sd 18, variance 324.
Standard deviation vs variance: Variance is the square of the standard deviation, so its units are squared. Both are found with technology; at SL treat the data as the population and use σx.
Hand quartiles vs technology quartiles: Different methods exist, so a GDC or spreadsheet may give slightly different values from a hand method.

4.4 Linear correlation of bivariate data; Pearson’s product-moment correlation coefficient; regression line of y on x

Bivariate data: Pairs of values (x, y) from the same individual or item.
Pearson's r: Strength and direction of a linear relationship, −1 ≤ r ≤ 1. Find it with technology. Describe it with both direction and strength, and the word ‘linear’.
Regression line of y on x: y = ax + b, found with technology. Use it to predict y from x, never x from y. a is the change in y per unit increase in x; b is y when x = 0.
Mean point: (x̄, ȳ). A line of best fit by eye passes through it, and so does the regression line.
Interpolation vs extrapolation: Inside the data range is reasonable; outside is unreliable.
Correlation vs causation: A strong correlation does not show that one variable causes the other.

4.5 Concepts of trial, outcome, sample space, event, probability and expected number of occurrences

Trial and outcome: Trial: one repetition of an experiment. Outcome: one possible result.
Sample space U and event A: Sample space: the set of all outcomes. Event: a set of outcomes. For equally likely outcomes, P(A) = n(A)/n(U).
Relative frequency: (Times A occurred) ÷ (number of trials). An experimental estimate of P(A); more trials bring it closer to the theoretical probability.
Complementary event A′: A does not occur. P(A′) = 1 − P(A).
Expected number of occurrences: n × P(A). Example: a 0.04 chance of a false alarm per run gives 350 × 0.04 = 14 expected false alarms in 350 runs. It need not be a whole number.

4.6 Venn diagrams, tree diagrams; combined, mutually exclusive, conditional and independent events

Probability of A or B (combined events): P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
Mutually exclusive events: A and B cannot both occur, so P(A ∩ B) = 0 and P(A ∪ B) = P(A) + P(B).
Independent events, and how to test for independence: The occurrence of one does not change the probability of the other: P(A ∩ B) = P(A)P(B). To test, write P(A ∩ B) and P(A) × P(B) side by side: equal means independent.
Conditional probability: P(A | B) = P(A ∩ B)/P(B). The ‘given’ event B is the denominator. Example: P(A ∩ B) = 0.12, P(B) = 0.4 gives P(A | B) = 0.3.
Rain on 30% of days; a bus is late with probability 0.4 if it rains, 0.1 if dry. Find P(rain | late): P(late) = 0.3 × 0.4 + 0.7 × 0.1 = 0.19. P(rain | late) = 0.12/0.19 = 0.632 (3 s.f.).
Tree diagrams with vs without replacement: With replacement: second-stage probabilities are unchanged (independent draws). Without: the denominator drops by 1, and the numerator may too.

4.7 Discrete random variables and their probability distributions; expected value

Discrete random variable: Takes separate values, each with a probability; the probabilities sum to 1.
Expected value of a discrete random variable: E(X) = Σ x P(X = x). Example: values 1, 2, 3 with probabilities 0.5, 0.3, 0.2 give E(X) = 0.5 + 0.6 + 0.6 = 1.7.
Fair game: E(X) = 0, where X is the player's gain.
P(X = 1, 2, 3, 4) = 0.2, 0.35, k, 0.15. Find k: The probabilities sum to 1: 0.2 + 0.35 + k + 0.15 = 1, so k = 0.3.
Expected gain for a game with a cost to play: the trap: Subtract the cost from each payout first, so X is the gain, then find E(X). Example: pay 2, win 10 (p 0.1) or 5 (p 0.2): gains 8, 3, −2 give E(X) = 0.8 + 0.6 − 1.4 = 0, a fair game.

4.8 The binomial distribution; its mean and variance

Conditions for a binomial model X ~ B(n, p): A fixed number n of trials, two outcomes, constant probability p of success, independent trials.
Mean and variance of X ~ B(n, p): E(X) = np, Var(X) = np(1 − p). Example: B(12, 0.4) has E(X) = 4.8 and Var(X) = 2.88.
Binomial on the GDC: pdf or cdf?: P(X = r): binomial pdf. P(X ≤ r): binomial cdf. Rewrite others first, for example P(X ≥ 3) = 1 − P(X ≤ 2) and P(X < 5) = P(X ≤ 4).
Is the number of successes when drawing without replacement from a small group binomial?: No: p changes after each draw, so the trials are not independent. Use a tree diagram.

4.9 The normal distribution and curve; normal probability and inverse normal calculations

Normal distribution: the 68–95–99.7 facts: For X ~ N(μ, σ²): about 68% of values lie in μ ± σ, 95% in μ ± 2σ and 99.7% in μ ± 3σ.
X ~ N(μ, σ²): which parameter goes into the GDC?: σ, the standard deviation, not σ². N(60, 25) has σ = 5.
Normal probability vs inverse normal on the GDC: Probability P(a < X < b): normal cdf(a, b, μ, σ). Value x with P(X < x) = q: inverse normal(q, μ, σ). For P(X > x) = q, enter 1 − q (the area to the left).
X ~ N(60, 5²). Find P(X < 52) and a with P(X < a) = 0.8: P(X < 52) = 0.0548. a = inverse normal(0.8, 60, 5) = 64.2 (3 s.f.).
P(X < 5) vs P(X ≤ 5): binomial and normal: Binomial (discrete): P(X < 5) = P(X ≤ 4). Normal (continuous): P(X < 5) = P(X ≤ 5).

4.10 Spearman’s rank correlation coefficient; limitations of Pearson’s and Spearman’s coefficients

Spearman's rₛ: Pearson's r applied to ranks. Measures a monotonic relationship. Find it with technology; average tied ranks.
Pearson vs Spearman: Pearson tests only for linearity; Spearman tests for any monotonic relationship and is less sensitive to outliers.
Rank 12, 15, 15, 18, 20 for Spearman's rₛ: 1, 2.5, 2.5, 4, 5. The two 15s share positions 2 and 3, so each gets the average rank 2.5.
x = 1 to 6, y = 2, 3, 5, 9, 17, 33. Compare rₛ and r: rₛ = 1, because y always increases with x. r = 0.906, because the relationship is curved, not linear. Spearman describes it better.
A scatter diagram shows a clear U-shape and r = 0.12. What can you conclude?: Only that there is no linear correlation. The variables are clearly related, but not linearly (nor monotonically), so neither r nor rₛ describes the relationship.

4.11 Hypothesis testing: null/alternative hypotheses, significance levels, p-values, χ² tests, the t-test

H₀ and H₁: H₀: the null hypothesis (‘independent’, ‘fits the distribution’, ‘means equal’). H₁: the alternative hypothesis (what you want evidence for).
p-value: The probability of a result at least as extreme as the one observed, assuming H₀ is true. Reject H₀ if the p-value is less than the significance level (1%, 5% or 10% in exams).
χ² statistic and expected frequency in a contingency table: χ²calc = Σ (f_o − f_e)²/f_e. Expected frequency = (row total × column total)/grand total. In exams every expected frequency is greater than 5.
Independence vs goodness of fit: degrees of freedom: Independence uses a contingency table and df = (r − 1)(c − 1); goodness of fit compares one row of data with a claimed distribution and at SL uses df = n − 1.
p-value vs critical value: Small p-value means reject; large χ²calc means reject. Compare the p-value with the significance level, or χ²calc with the critical value, never across.
One-tailed vs two-tailed: ‘Greater’, ‘higher’, ‘less’ → one-tailed; ‘different’, ‘differ’ → two-tailed.
The t-test at SL: conditions: Two unpaired samples, population variance unknown, populations normal; assume equal variances and use the pooled two-sample t-test on the GDC. H₀: μ₁ = μ₂.
‘Do not reject H₀’ vs ‘accept H₀’: A test never proves H₀. Say there is insufficient evidence, and finish with a sentence in context.

4.12 Design of valid data collection methods; reliability and validity tests (AHL only)

Biased questionHL: Wording leads towards an answer. Unbiased: neutral wording.
Personal questionHL: Sensitive (income, health). Risk of refusal or dishonest answers.
Unstructured vs structured questionHL: Unstructured: open answer. Structured question: fixed, consistent answer choices.
Precise questionHL: Clear time frame, units and categories that do not overlap. Example: ‘How many hours did you spend on homework last week?’
Reliability, and the two reliability testsHL: Reliability: the method gives consistent results when repeated. Test-retest: same test, same group, two different times. Parallel forms: two equivalent versions, same group.
Validity, and the two kinds of validityHL: Validity: the method measures what it claims to. Content validity: covers the whole topic. Criterion-related validity: agrees with an established measure.
χ² goodness of fit with estimated parameters: degrees of freedomHL: df = number of categories − 1 − number of parameters estimated from the data. Merge categories first until every expected frequency is greater than 5. Example: 7 categories, p estimated: df = 5.

4.13 Non-linear regression; sum of square residuals; coefficient of determination R² (AHL only)

Residual and SSresHL: Residual = observed y − predicted y. SSres = Σ(yᵢ − ŷᵢ)², the sum of the squared residuals. Smaller means a closer fit.
R², the coefficient of determinationHL: Proportion of the variability in y accounted for by the model. 0 ≤ R² ≤ 1.
R² vs rHL: R² = r² only for a linear model. r has a sign; R² does not.
SSres vs R²HL: SSres depends on the units of y. R² is a proportion, so it has no units.
Which regression models are used in AI HL examinations?HL: Linear, quadratic, cubic, exponential, power and sine, fitted with technology. Choose using R² together with the context, not R² alone.
A model's residuals are 0.3, −0.5, 0.1 and 0.4. Find SSresHL: 0.09 + 0.25 + 0.01 + 0.16 = 0.51.

4.14 Linear transformation and combination of random variables; unbiased estimators of μ and σ² (AHL only)

E(aX + b) and Var(aX + b)HL: E(aX + b) = aE(X) + b. Var(aX + b) = a²Var(X). Example: E(X) = 12, Var(X) = 4, Y = 3X − 5 gives E(Y) = 31, Var(Y) = 36.
Mean and variance of a linear combination a₁X₁ + … + aₙXₙHL: Mean: a₁E(X₁) + … + aₙE(Xₙ) (any variables). Variance: a₁²Var(X₁) + … + aₙ²Var(Xₙ), only if the variables are independent.
Var(X − Y) for independent X and YHL: Var(X) + Var(Y). Variances always add, even when the variables are subtracted.
X₁ + X₂ vs 2XHL: Two independent observations: Var(X₁ + X₂) = 2Var(X). One observation doubled: Var(2X) = 4Var(X). The total of n items is X₁ + … + Xₙ, not nX.
Unbiased estimateHL: A sample statistic whose expected value equals the population parameter. x̄ estimates μ; s²ₙ₋₁ estimates σ².
s²ₙ vs s²ₙ₋₁HL: s²ₙ: the sample variance (divide by n). s²ₙ₋₁: the unbiased estimate of the population variance (divide by n − 1). s²ₙ₋₁ = (n/(n − 1))s²ₙ; on many GDCs σx is sₙ and Sx is sₙ₋₁.

4.15 Linear combinations of normal random variables; the central limit theorem (AHL only)

Distribution of the sample mean from a normal populationHL: X ~ N(μ, σ²) ⇒ X̄ ~ N(μ, σ²/n), exactly, for any n. Enter σ/√n as the standard deviation on the GDC.
Central limit theorem (CLT)HL: For large n, X̄ is approximately normal whatever the distribution of X. In examinations, n > 30 counts as large.
How do you find a probability for a combination of normal variables, such as P(A₁ + A₂ > B)?HL: 1. Define T = A₁ + A₂ − B. 2. Find E(T). 3. Find Var(T): square each coefficient and add every term. 4. T is normal, so find P(T > 0) on the GDC with SD √Var(T).
Saying why X̄ is normal: the two reasonsHL: ‘The population is normal’ (exact, any n), or ‘n > 30, so by the central limit theorem’ (approximate). State which one applies.

4.16 Confidence intervals for the mean of a normal population (AHL only)

Confidence intervalHL: A range, built from one sample, that would contain μ in the stated percentage of repeated samples.
Z-interval or T-interval?HL: σ known: Z-interval. σ unknown: T-interval using sₙ₋₁, whatever the sample size. A large sample does not let you switch to z.
σ = 3 is known, n = 25 and x̄ = 48.6. Find the 95% confidence interval for μHL: x̄ ± 1.96 × σ/√n = 48.6 ± 1.96 × 3/5 = 48.6 ± 1.176, giving [47.4, 49.8].
How do you interpret a 95% confidence interval?HL: ‘We are 95% confident that the mean [quantity] lies between … and ….’ About 95% of intervals built this way would contain μ. It is not a 95% probability that μ is in this one interval.
Width of a confidence interval, and what its midpoint isHL: A higher confidence level widens it; a larger sample narrows it. The midpoint is x̄: [12.1, 13.5] has x̄ = 12.8.

4.17 The Poisson distribution, its mean and variance; sum of independent Poisson distributions (AHL only)

Po(m)HL: The Poisson distribution with mean m. X ~ Po(m) means X has this distribution. P(X = x) = e⁻ᵐmˣ/x!, x = 0, 1, 2, …
Mean and variance of X ~ Po(m)HL: E(X) = m and Var(X) = m.
Conditions for a Poisson model, and changing the intervalHL: Events occur independently at a uniform average rate. Scale m to the interval asked about: 2.4 calls per 10 minutes gives Po(4.8) for 20 minutes.
Sum of independent Poisson variablesHL: X ~ Po(m₁) and Y ~ Po(m₂) independent ⇒ X + Y ~ Po(m₁ + m₂).
Emails arrive at 0.6 per minute. Find P(at least one in a given minute)HL: X ~ Po(0.6). P(X ≥ 1) = 1 − P(X = 0) = 1 − e^(−0.6) = 0.451 (3 s.f.).
Binomial vs PoissonHL: Binomial: a fixed number of trials and a maximum value n. Poisson: counts events in an interval, with no fixed maximum.

4.18 Critical values/regions; hypothesis tests for mean, proportion and correlation; Type I and II errors (AHL only)

Critical region and critical valueHL: Critical region: values of the test statistic that lead to rejecting H₀. Critical value: its boundary.
Type I and Type II errorsHL: Type I error: reject H₀ when H₀ is true. Type II error: accept H₀ when H₀ is false.
P(Type I error): continuous vs discrete testsHL: For the normal test P(Type I) equals the significance level. For binomial and Poisson tests it is at most the significance level.
z vs t test for a meanHL: σ known vs σ unknown. Use the t-test whenever σ is unknown, whatever the sample size. z = (x̄ − μ)/(σ/√n).
Binomial critical region: H₀: p = 0.3, H₁: p < 0.3, n = 20, 5%HL: P(X ≤ 2) = 0.0355 and P(X ≤ 3) = 0.107, so the critical region is X ≤ 2.
How do you find P(Type II error) for a discrete test?HL: P(X not in the critical region), using the stated true value of the parameter, not the H₀ value.
Matched pairs testHL: Find the differences for each pair, then do a one-sample test (usually t) on the differences. H₀: μ_d = 0.
Testing for correlation in bivariate normal dataHL: H₀: ρ = 0 against H₁: ρ > 0, ρ < 0 or ρ ≠ 0. The GDC gives r and the p-value to compare with the significance level.

4.19 Transition matrices, regular Markov chains, steady-state and long-term probabilities (AHL only)

Transition matrix THL: Tᵢⱼ = P(moving from state j to state i). Columns sum to 1.
Initial state matrix and the state after n stepsHL: s₀ is a column of starting probabilities, and sₙ = Tⁿs₀ (not s₀T).
Regular Markov chainHL: Some power of T has all entries positive.
Steady state sHL: Ts = s, entries sum to 1. It is the eigenvector for eigenvalue 1, scaled so its entries add to 1.
How do you find the exact steady state by equations?HL: 1. Write Ts = s as equations and drop one (they are dependent). 2. Add ‘entries sum to 1’. 3. Solve, giving fractions if exact answers are asked for.
T has columns (0.8, 0.2) and (0.3, 0.7), s₀ = (0.5, 0.5)ᵀ. Find s₁ and the steady stateHL: s₁ = (0.55, 0.45)ᵀ. Steady state: 0.2a = 0.3b with a + b = 1, so (3/5, 2/5)ᵀ.