Flashcards · IB Diploma Programme Mathematics: Analysis and Approaches · Topic 4

Statistics and probability flashcards

67 cards on Topic 4, subtopics 4.1 to 4.14. Pick a subtopic or study them all.

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4.1 Concepts of population, sample, sampling techniques and outliers

Population: Every item of interest. Sample: the part you measure. Random sample: every item has an equal chance of selection.
Discrete data: Separate values, usually counts. Continuous data: any value in a range, usually measurements.
Bias: Some members of the population are more likely to be chosen than others, so the sample does not represent it.
Outlier: A data item more than 1.5 × IQR from the nearest quartile.
Quota vs stratified: Both use groups, but only stratified chooses at random within each group.
A gym has 1200 members, 450 of them under 30. How many under-30s go in a stratified sample of 80?: (450/1200) × 80 = 30 For each group: (group size ÷ N) × sample size; check the numbers add up to the sample size.

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

Box and whisker diagram: Minimum, Q₁, median, Q₃, maximum. Outliers are crosses, and the whisker ends at the most extreme value that is not an outlier.
Comparing two box plots: One comment on the median, one on the IQR or range, both in context.
How do you draw and use a cumulative frequency graph?: 1. Add a running-total column. 2. Plot each total at the UPPER class boundary, starting at zero at the lowest boundary. 3. Join with a smooth curve or straight lines. 4. Read across at n/2 (median), n/4 (Q₁), 3n/4 (Q₃).
Frequency histograms in this course: Equal class widths, with bar height equal to frequency. Frequency density is not required. Class intervals have no gaps, such as 20 ≤ t < 30.
Data: 2, 3, 5, 6, 8, 9, 11, 20. Is 20 an outlier?: Q₁ = 4, Q₃ = 10, IQR = 6. Upper fence = 10 + 1.5 × 6 = 19. 20 > 19, so yes, 20 is an outlier.

4.3 Measures of central tendency and measures of dispersion

Mean of a frequency table, and an estimate of the mean for grouped data: x̄ = Σfx/Σf. For grouped data use the mid-interval value of each class as x (not the class boundaries).
Modal class: The class with the highest frequency (equal class intervals only).
Variance: The square of the standard deviation.
Effect on the statistics of adding c to every item, or multiplying every item by k > 0: + c: mean, median and quartiles increase by c; SD, IQR, range and variance unchanged. × k: mean, median, quartiles, SD, IQR and range multiply by k; variance multiplies by k².
Mean 50, SD 8. Every item becomes 3x − 10. Find the new mean, SD and variance.: Mean 3(50) − 10 = 140. SD 3 × 8 = 24. Variance 24² = 576.
σx vs sx on the GDC: Use σx unless the question says otherwise.

4.4 Linear correlation of bivariate data and the regression line of y on x

Pearson’s product-moment correlation coefficient r: range and meaning: −1 ≤ r ≤ 1, found with technology. Its sign gives the direction and its size the strength of LINEAR correlation. A clear curve can give r near 0.
Correlation vs causation: r measures association, not cause.
Mean point: (x̄, ȳ). Lines of best fit by eye and both regression lines pass through it.
In the regression line y = ax + b, what do a and b mean?: a: the change in y for each increase of 1 in x. b: the value of y when x = 0, which may have no sensible meaning in context.
Interpolation vs extrapolation: Inside the data range is reasonable when correlation is strong; outside it is unreliable.
The regression line of y on x is y = 2.5x + 3 and x̄ = 6. Find ȳ.: The line passes through (x̄, ȳ), so ȳ = 2.5(6) + 3 = 18.

4.5 Concepts of trial, outcome, sample space, event and probability

Trial: One run of an experiment. Outcome: one possible result.
Sample space U: All possible outcomes. Event A: a subset of U.
Relative frequency: Number of times A occurred ÷ number of trials (experimental probability).
Complement A′: "Not A". P(A′) = 1 − P(A).
Probability with equally likely outcomes, and expected number of occurrences: P(A) = n(A)/n(U). Expected number = n × P(A). Example: a fair die rolled 150 times gives 150 × 1/6 = 25 expected fives.

4.6 Combined, mutually exclusive, conditional and independent events

Probability of A or B (union): P(A ∪ B) = P(A) + P(B) − P(A ∩ B). For mutually exclusive events P(A ∩ B) = 0.
Mutually exclusive: A and B cannot both happen, so P(A ∩ B) = 0.
Independent: One event does not affect the other, so P(A ∩ B) = P(A)P(B).
Tree diagram without replacement: how do the second branches change?: Reduce the count and the total by one on the branch already taken. Multiply along branches; add the branches that give the event.
Can two events be both mutually exclusive and independent?: Not if P(A) and P(B) are both non-zero: exclusive means P(A ∩ B) = 0, but independent needs P(A ∩ B) = P(A)P(B) > 0.

4.7 Discrete random variables and their probability distributions

Discrete random variable: Takes separate values, each with a probability; Σ P(X = x) = 1.
Expected value of a discrete random variable: E(X) = Σ x P(X = x)
P(X = x) = x/10 for x = 1, 2, 3, 4. Find E(X).: E(X) = (1 + 4 + 9 + 16)/10 = 30/10 = 3
Fair game: E(X) = 0, where X is the player’s gain.
How do you find an unknown constant k in a discrete probability distribution?: Set the sum of all the probabilities equal to 1 and solve for k.

4.8 The binomial distribution

Conditions for a binomial model X ~ B(n, p): A fixed number of trials n; two outcomes per trial; constant probability p of success; independent trials.
Mean and variance of X ~ B(n, p), and their values for B(10, 0.3): E(X) = np and Var(X) = np(1 − p). B(10, 0.3): E(X) = 3, Var(X) = 2.1.
How do you rewrite P(X ≥ 4) and P(X < 4) for a binomial cdf?: P(X ≥ 4) = 1 − P(X ≤ 3) P(X < 4) = P(X ≤ 3) Use binomial pdf for "=" and binomial cdf for "≤".
Why does sampling without replacement not give a binomial model?: The probability changes from one trial to the next and the trials are not independent.

4.9 The normal distribution

Normal distribution N(μ, σ²): Symmetric about μ; mean = median = mode. About 68% of values lie within μ ± σ, 95% within μ ± 2σ and 99.7% within μ ± 3σ. The z-value gives the number of standard deviations from the mean.
Normal cdf vs inverse normal on the GDC: Given a value, find a probability with normal cdf. Given a probability, find a value with inverse normal.
X ~ N(70, 5²). What do you enter for the standard deviation?: 5. N(μ, σ²) gives the variance, but the GDC asks for σ. Entering 25 is a common error.
X ~ N(50, 4²) and P(X < a) = 0.9. Find a.: Inverse normal with area 0.9 to the left: a = 55.1 (3 s.f.).
For a continuous variable such as a normal one, how do P(X < a) and P(X ≤ a) compare?: They are equal, because P(X = a) = 0.

4.10 The regression line of x on y

y on x vs x on y: Predict y from x with the y on x line; predict x from y with the x on y line. Never rearrange one to get the other. The lines are different unless r = ±1.
You are given a value of y and need to estimate x. Which regression line do you use?: The regression line of x on y, x = cy + d.
How do you find the regression line of x on y with a GDC?: Run linear regression with the lists swapped (y as the independent list), and write it as x = cy + d. Predict with unrounded coefficients.

4.11 Formal definition of conditional probability and independence

Formal definition of conditional probability: P(A | B) = P(A ∩ B)/P(B), for P(B) > 0.
Formal test for independence: A and B are independent if P(A | B) = P(A) = P(A | B′), equivalently P(A ∩ B) = P(A)P(B).
P(A) = 0.45, P(B) = 0.3, P(A ∩ B) = 0.12. Find P(A | B). Are A and B independent?: P(A | B) = 0.12/0.3 = 0.4 ≠ 0.45 = P(A), so not independent (also 0.45 × 0.3 = 0.135 ≠ 0.12).
Is P(A | B) the same as P(B | A)?: No, in general they are different. P(A | B) = P(A ∩ B)/P(B), while P(B | A) = P(A ∩ B)/P(A).

4.12 Standardization of normal variables (z-values)

Standardized value z: z = (x − μ)/σ: the number of standard deviations x is from the mean. Z ~ N(0, 1).
X ~ N(70, 5²). Find the z-value of x = 62.: z = (62 − 70)/5 = −1.6
How do you find an unknown μ or σ of a normal distribution?: 1. Use inverse normal on N(0, 1) with the area to the LEFT to get z. 2. Solve x = μ + zσ. Two unknowns need two equations.
X ~ N(120, σ²) and P(X > 130) = 0.2. Find σ.: Area to the left of 130 is 0.8, so z = 0.8416. σ = (130 − 120)/0.8416 = 11.9 (3 s.f.)

4.13 Bayes’ theorem (AHL only)

Bayes’ theorem for two eventsHL: P(B | A) = P(B)P(A | B) / [P(B)P(A | B) + P(B′)P(A | B′)]
How do you set up a Bayes’ theorem problem?HL: 1. Tree diagram: the "cause" events first, the observed event second. 2. P(observed) = sum of all branches ending in it. 3. P(cause | observed) = that cause’s branch ÷ P(observed).
P(B) = 0.4, P(A | B) = 0.5, P(A | B′) = 0.2. Find P(B | A).HL: P(A) = 0.4 × 0.5 + 0.6 × 0.2 = 0.2 + 0.12 = 0.32 P(B | A) = 0.2/0.32 = 0.625
Bayes’ theorem with three mutually exclusive causes B₁, B₂, B₃HL: P(Bᵢ | A) = P(Bᵢ)P(A | Bᵢ) / [P(B₁)P(A | B₁) + P(B₂)P(A | B₂) + P(B₃)P(A | B₃)]

4.14 Variance of discrete and continuous random variables (AHL only)

Variance of a random variable in terms of expectationsHL: Var(X) = E(X²) − [E(X)]²
E(aX + b) and Var(aX + b)HL: E(aX + b) = aE(X) + b Var(aX + b) = a²Var(X) Example: E(X) = 4, Var(X) = 3 gives E(2X + 1) = 9 and Var(2X + 1) = 12.
Continuous random variable: conditions on the pdf f(x), and the meanHL: f(x) ≥ 0 and the integral of f(x) over its whole domain is 1. E(X) = ∫ x f(x) dx over the domain.
Median and mode of a continuous random variableHL: Median m: the integral of f(x) from the lower end of the domain to m equals 1/2. Mode: where f(x) is largest. Check the end points as well as stationary points.
f(x) = x²/9 for 0 ≤ x ≤ 3. Find E(X), Var(X) and the median.HL: E(X) = ∫₀³ x³/9 dx = 9/4. E(X²) = ∫₀³ x⁴/9 dx = 27/5. Var(X) = 27/5 − 81/16 = 27/80. Median: m³/27 = 1/2, so m = 3/∛2 = 2.38 (3 s.f.).