IB Mathematics: Analysis and Approaches Higher Level
Statistics & Probability Formulae for AA HL Only
An exact, exam-focused guide to the current Higher Level extension: Bayes’ theorem, variance of discrete random variables, continuous probability density functions, expectation, spread, fair games and linear transformations.

The advanced Statistics and Probability component of IB Mathematics: Analysis and Approaches is compact, but it rewards careful mathematical thinking. Students move beyond using a probability model supplied in a question and begin analysing how probability is updated, how random variables are summarized, and how a density function turns areas into probabilities. The number of syllabus statements is small; the number of decisions inside a demanding exam problem is not. A successful solution must identify the correct conditioning event, build a complete probability model, distinguish a point value from an area, choose the correct moment formula, and interpret a numerical result in context.
This guide is deliberately narrower than a general university statistics course. It does not add confidence intervals, hypothesis tests, Type I and Type II errors, t-tests or correlation significance tests to the current AA AHL scope. Those topics are sometimes copied from other courses and can make a revision page look comprehensive while making it less accurate. Here, every major section is tied to the current Higher Level statements: Bayes’ theorem for a maximum of three events, variance of a discrete random variable, continuous random variables and their probability density functions, continuous mode and median, mean and variance, fair games, and the effect of a linear transformation.
Current AA HL syllabus map
| Statement | What students need to do | Core mathematical objects |
|---|---|---|
| AHL 4.13 | Use Bayes’ theorem for a maximum of three events and connect it to conditional probability and independence. | Conditional probability, total probability, tree diagrams, partitions |
| AHL 4.14: discrete variables | Calculate and interpret the variance and standard deviation of a discrete random variable. | E(X), E(X2), Var(X), σ |
| AHL 4.14: continuous variables | Work with continuous random variables and probability density functions, including piecewise functions. | Non-negative density, total area 1, definite integrals |
| AHL 4.14: centre and spread | Find the mode and median of a continuous random variable and calculate mean, variance and standard deviation for discrete or continuous models. | Maximum of f, median-area equation, moments |
| AHL 4.14: applications | Use expected value to analyse fair games and investigate how a linear transformation changes centre and spread. | E(aX+b), Var(aX+b), expected gain |
The official sequence matters. AHL 4.13 builds directly on shared conditional probability, so Bayes questions may still require an accurate tree diagram or Venn interpretation. AHL 4.14 builds on the shared idea of a discrete random variable, then extends it to variance and to continuous density. Calculus is therefore a prerequisite in practice: probabilities, means and variances for continuous variables are definite integrals. The formulas are short, but students must control bounds, domains and normalization.
Notation you must read accurately
| Notation | Meaning | Common trap |
|---|---|---|
| P(A|B) | Probability that A occurs given that B has occurred | Reversing the conditioning and calculating P(B|A) |
| X | A random variable | Treating X as one observed value instead of a mapping from outcomes |
| x | A possible value of X | Using the same symbol carelessly for a variable and its realization |
| p(x) | Probability mass assigned to a discrete value | Forgetting that the masses must sum to 1 |
| f(x) | Probability density for a continuous variable | Calling f(x) a probability at a point |
| E(X) or μ | Mean or expected value | Assuming it must be an attainable outcome |
| Var(X) | Mean squared deviation from the mean | Confusing variance with standard deviation |
| σ | Standard deviation, the positive square root of variance | Forgetting the absolute value in a scaled transformation |
Bayes’ theorem
Bayes’ theorem reverses a conditional probability. A question may tell you how likely evidence B is under a cause A, then ask how likely cause A is after B has been observed. These quantities are not equal. A medical test can have a high probability of a positive result when a condition is present while a positive result still has a much lower probability of indicating the condition, particularly when the condition is rare.
Two-event form
P(A|B) = P(B|A)P(A) / P(B)
With a partition A₁, A₂, A₃
P(Ai|B) = P(B|Ai)P(Ai) / Σ P(B|Aj)P(Aj)
The denominator is the total probability of the observed evidence. If A₁, A₂ and A₃ are mutually exclusive and exhaustive causes, then every way B can occur must be counted. On a tree diagram, multiply along a branch to obtain a joint probability and add the relevant branches to obtain the total. The numerator is the single branch containing both the requested cause and the evidence. The posterior probability is therefore “wanted B-branch divided by all B-branches.”
A reliable Bayes workflow
- Define the events in words before using letters.
- Confirm that the first-stage events form a complete partition.
- Place prior probabilities on the first branches.
- Place conditional evidence probabilities on the second branches.
- Multiply along each route to the observed evidence.
- Add those products to calculate the denominator.
- Divide the required joint branch by the total evidence probability.
- Check that the answer lies between 0 and 1 and matches the context.
Worked example: screening result
In a population, 4% of people have a condition. A test is positive for 92% of people who have it and for 7% of people who do not. Let C mean “has the condition” and + mean “positive.” Then P(C)=0.04, P(Cc)=0.96, P(+|C)=0.92 and P(+|Cc)=0.07. The required joint branch is P(C∩+)=0.04×0.92=0.0368. The total positive probability is 0.0368+0.96×0.07=0.104. Therefore P(C|+)=0.0368/0.104≈0.3538. Although the sensitivity is 92%, the probability of the condition after a positive result is about 35.4% because false positives arise from a much larger group.
Three-source Bayes example
A component comes from factories A₁, A₂ and A₃ in proportions 0.50, 0.30 and 0.20. Their defect rates are 0.01, 0.03 and 0.05. Given that a component is defective, the probability it came from A₃ is
P(A₃|D) = (0.20×0.05) / (0.50×0.01 + 0.30×0.03 + 0.20×0.05) = 0.01/0.024 = 5/12.
The answer is much larger than the prior probability 0.20 because A₃ has the highest defect rate. This contextual comparison is a useful check: observing a defect should shift probability toward the factories that produce defects more often. If your posterior moved in the opposite direction, revisit the numerator or denominator.
Variance of a discrete random variable
For a discrete random variable, probabilities are attached to individual values. The expected value is a weighted mean. Variance measures the expected squared distance from that mean. Squaring prevents positive and negative deviations from cancelling and gives greater weight to observations farther from the centre.
Mean: E(X) = Σ x p(x)
Second moment: E(X2) = Σ x2p(x)
Variance: Var(X) = E(X2) − [E(X)]2
Standard deviation: σ = √Var(X)
The computational variance formula is usually efficient, but its order matters. Calculate E(X2) by squaring each possible value before weighting it. Then subtract the square of E(X). The expressions E(X2) and [E(X)]2 are generally different. Their difference must be non-negative; a negative answer signals an arithmetic, probability or rounding mistake.
Worked example: discrete distribution
Let X take values 0, 1, 2 and 3 with probabilities 0.10, 0.30, 0.40 and 0.20. The probabilities sum to 1. E(X)=0(0.10)+1(0.30)+2(0.40)+3(0.20)=1.70. Next, E(X²)=0²(0.10)+1²(0.30)+2²(0.40)+3²(0.20)=3.70. Therefore Var(X)=3.70−1.70²=0.81 and σ=0.90. The standard deviation returns to the units of X, while the variance has squared units.
Why the mean need not be possible
An expected value describes long-run average behaviour, not a prediction that must occur on one trial. A fair six-sided die has expected value 3.5 even though 3.5 cannot be rolled. In a financial game, an expected gain of 0 means balance over many repetitions; individual players can still win or lose. When interpreting E(X), include the scale and the repeated-trial idea when appropriate.
Continuous random variables and density functions
A continuous random variable can take any value in an interval. Its model assigns density rather than probability mass. Probability is represented by area under the density curve. For a valid probability density function, f(x) must be non-negative on its support and the total area over the support must equal 1.
Normalization: ∫−∞∞ f(x) dx = 1
Interval probability: P(a≤X≤b) = ∫ab f(x) dx
For a continuous variable, P(X=c)=0. An interval endpoint has no area, so P(a<X<b), P(a≤X<b) and P(a≤X≤b) are equal. This is different from a discrete distribution, where a single outcome may have positive probability.
Finding an unknown density constant
If a density contains an unknown constant, use total area 1 before calculating other quantities. For a piecewise function, integrate each piece over its own interval and add the results. Do not integrate a formula outside the support stated in the question. The density is normally zero elsewhere, even when the question does not repeat that line.
Worked example: normalizing a density
Suppose f(x)=kx for 0≤x≤2 and 0 otherwise. Normalization gives ∫02kxdx=2k=1, so k=1/2. Then P(0.5≤X≤1.5)=∫0.51.5(x/2)dx=[x²/4]0.51.5=0.50. A sketch makes the support and bounds visible before any calculator command is entered.
Piecewise-density workflow
- Copy the support intervals exactly.
- Check non-negativity on every piece.
- Integrate each piece and set the total equal to 1.
- When an interval crosses a breakpoint, split the integral there.
- For moments, multiply each density piece by x or x2 before integrating.
- Check that the final probability lies in [0,1] and that the mean lies within the support.
Mode, median, mean and spread for a continuous variable
The four summaries answer different questions. The mode is located where density is greatest. The median divides total probability into equal halves. The mean is the balance point calculated by weighting values with density. Variance measures expected squared distance from the mean. These values coincide for some symmetric unimodal distributions, but the syllabus formulas must work when they do not.
Mode: a value where f(x) is maximal
Median m: ∫−∞mf(x)dx = 1/2
Mean: E(X) = ∫−∞∞x f(x)dx
Second moment: E(X2) = ∫−∞∞x2f(x)dx
Variance: Var(X) = E(X2) − [E(X)]2
Complete worked model: f(x)=2x on [0,1]
The density is valid because it is non-negative and ∫012xdx=1. The mode is 1 because the density increases throughout the support and reaches its maximum at the right endpoint. The median satisfies ∫0m2xdx=m²=1/2, so m=1/√2. The mean is E(X)=∫012x²dx=2/3. The second moment is E(X²)=∫012x³dx=1/2. Therefore Var(X)=1/2−(2/3)²=1/18 and σ=1/(3√2). The order mode > median > mean is consistent with greater density near the right side and a tail extending left.
How to find a continuous mode
“Mode equals the maximum of the density” is a calculus problem. Differentiate the density on smooth interior pieces, solve f′(x)=0, and compare valid stationary points with endpoints and any piecewise breakpoints. A density may have an endpoint mode or more than one mode. Do not automatically use the mean symbol μ for the mode; they describe different features.
How to find a continuous median
The median is defined by probability, not by the midpoint of the support. Use the density area from the lower end of the support to the unknown value m and set it equal to 0.5. If the half-area point falls beyond a piecewise breakpoint, include the full area of earlier pieces before integrating the piece that contains m. Reject algebraic roots outside the relevant support interval.
Linear transformations of a random variable
Let Y=aX+b. Adding b shifts every value and therefore shifts the mean, but it does not alter distances between values. Multiplying by a scales distances by |a|, so variance scales by a2 and standard deviation by |a|. A negative value of a reverses order but spread cannot become negative.
E(aX+b) = aE(X) + b
Var(aX+b) = a2Var(X)
SD(aX+b) = |a|SD(X)
Worked example: converting a scale
A measurement X has mean 18 and standard deviation 4. Define Y=1.8X+32. Then E(Y)=1.8(18)+32=64.4. The standard deviation is 1.8(4)=7.2, and the variance is 1.8²(16)=51.84. The constant 32 changes the mean but not the variance or standard deviation.
Transformation questions often provide variance instead of standard deviation. Read the requested statistic before substituting. If Var(X)=9 and Y=−2X+5, then Var(Y)=4(9)=36 and SD(Y)=6. Writing −6 for the standard deviation would be impossible; the absolute-value rule prevents that error.
Fair games and expected gain
Expected value can represent the long-run gain to a player, insurer or business. Define gains as positive and costs or losses as negative from one consistent perspective. A game is fair when expected net gain is zero. Fair does not mean each outcome has equal probability, every player receives their stake back, or short-term results balance.
Worked example: finding a fair entry fee
A game pays $12 with probability 0.10, $5 with probability 0.30 and $0 otherwise. Let the entry fee be c. The player’s net gains are 12−c, 5−c and −c. Expected net gain is 0.10(12−c)+0.30(5−c)+0.60(−c)=2.70−c. Fairness requires 2.70−c=0, so the fair fee is $2.70. If the organizer charges more, the player has a negative expected gain and the organizer has a positive expected gain.
Always state whose gain is being modelled. A payout is not the same as a profit because the entry fee or premium must be included. In insurance language, a positive expected gain to the company corresponds to a negative expected financial gain to the policyholder before considering the value of risk protection.
Connecting probability, calculus and functions
Functions
A density is a function with a restricted domain and non-negative output. Piecewise definitions, parameters and transformations therefore rely on the AA HL Functions guide.
Calculus
Continuous probabilities and moments are definite integrals; the mode may require differentiation and endpoint comparison. Review the AA HL Calculus guide.
Algebra
Bayes denominators, parameter equations and median equations demand exact manipulation. The AA HL Number and Algebra guide supports that work.
Graphs
Density shape, shaded interval area and maximum points connect analytical work with graphical interpretation. Use the AA HL Geometry and Trigonometry guide for wider visual revision.
GDC workflow and exam communication
Technology can evaluate definite integrals, solve a median equation and calculate distribution summaries, but the mathematical setup must remain visible. Write the density, support and bounds before giving a decimal answer. For Bayes questions, show the products that form the numerator and denominator. For variance, identify E(X) and E(X2) separately. A calculator result without a model is difficult to interpret and easy to enter incorrectly.
For a density problem
- Sketch or state the support.
- Use normalization to find any unknown constant.
- Enter the correct function and bounds.
- Store exact or unrounded parameter values.
- For a median, solve the half-area equation on the correct interval.
- For a mode, compare graphical and derivative evidence.
- Round only the final reported value.
For a discrete-variable problem
- Check that probabilities sum to 1.
- Use a list or table to enter values and probabilities.
- Keep the expected value and second moment distinct.
- Use Var(X)=E(X2)−[E(X)]2.
- Take a square root only when standard deviation is requested.
- Attach units and a contextual sentence where appropriate.
Common mistakes and repair rules
| Mistake | Why it fails | Repair rule |
|---|---|---|
| Writing P(A|B)=P(B|A) | Conditioning changes the reference group | Write the Bayes numerator and total-evidence denominator |
| Leaving out one evidence branch | The denominator is not total probability | List the complete partition before multiplying |
| Using [E(X)]² for E(X²) | Squaring and averaging are different operations | Create separate columns for x, x² and probability |
| Reporting negative variance | Squared spread cannot be negative | Check subtraction order and premature rounding |
| Calling f(x) a point probability | A continuous point has zero area | Integrate density over an interval |
| Ignoring the support | The formula may not apply outside its defined interval | Write bounds beside every integral |
| Taking the midpoint as the median | Median divides probability, not interval length | Solve the half-area equation |
| Assuming mode equals mean | They describe different features | Maximize the density for the mode |
| Adding b to variance | A shift does not change spread | Use Var(aX+b)=a²Var(X) |
| Using a instead of |a| for SD | Standard deviation cannot be negative | Scale SD by the absolute value |
| Calling a game fair because wins and losses are possible | Fairness is based on expected net gain | Include every payout and cost, then set expectation to zero |
Mixed practice with solutions
1. Reverse a conditional probability
Two machines produce 70% and 30% of items. Their fault rates are 2% and 6%. For a faulty item, P(M₂|F)=0.30(0.06)/[0.70(0.02)+0.30(0.06)]=0.018/0.032=0.5625. The faulty item is more likely to come from M₂ even though M₂ makes fewer items.
2. Find a missing probability and variance
X takes values 1, 2 and 5 with probabilities 0.2, q and 0.3. Since probabilities sum to 1, q=0.5. Then E(X)=1(0.2)+2(0.5)+5(0.3)=2.7 and E(X²)=1(0.2)+4(0.5)+25(0.3)=9.7. Thus Var(X)=9.7−2.7²=2.41.
3. Normalize a triangular density
For f(x)=k(1−x) on 0≤x≤1, normalization gives k∫01(1−x)dx=k/2=1, so k=2. The density is largest at x=0, so the mode is 0.
4. Find a median
For the density 2(1−x) on [0,1], the median solves ∫0m2(1−x)dx=2m−m²=1/2. The valid root in [0,1] is m=1−1/√2.
5. Transform centre and spread
If E(X)=6 and Var(X)=2.5, define Y=−3X+4. Then E(Y)=−18+4=−14, Var(Y)=9(2.5)=22.5 and SD(Y)=3√2.5. The negative multiplier reverses order but does not create negative spread.
6. Check a proposed fair game
A player pays $4 and receives $10 with probability 0.25 or $1 otherwise. Net gains are $6 and −$3. Expected net gain is 0.25(6)+0.75(−3)=−0.75, so the game is not fair to the player. The fair entry fee would equal the expected payout: 0.25(10)+0.75(1)=$3.25.
How to build a reliable solution
Begin with the object the question asks for. A posterior probability needs a Bayes denominator; a probability from a continuous model needs area; a mean needs a first moment; a variance needs both a first and second moment; a median needs a half-area equation; and a mode needs maximization. Writing that target before calculating prevents a common pattern in which correct algebra answers the wrong statistical question.
Control exactness and rounding
Bayes probabilities often involve several products and a division, while variance subtracts two nearby quantities. Early rounding can noticeably change the result and can even create a small negative variance. Keep fractions or stored calculator values throughout. Round the final probability or statistic according to the question and report enough working that another reader could reconstruct the calculation.
Use independent checks
A probability must lie between 0 and 1. Probabilities in a complete discrete model must sum to 1. A density must have total area 1 and cannot be negative. A mean and median should lie inside the support. Variance cannot be negative, and standard deviation should have the original units. A Bayes posterior should move toward sources that make the observed evidence more likely. A linear shift should not alter spread. These checks use different ideas from the main calculation and therefore catch errors more effectively than repeating the same calculator entry.
Interpret results without overstating them
Bayes’ theorem updates probability under a stated model; it does not prove a cause. Expected value describes a long-run average; it does not guarantee one person’s outcome. A probability density is a mathematical model and may approximate reality imperfectly. Good exam communication says exactly what a value represents, includes the conditioning information, identifies the relevant random variable and attaches units where meaningful.
Advanced reasoning for unfamiliar exam questions
Hard Higher Level questions rarely announce the formula to use. They combine a context, an unfamiliar function and several related requests. The skill being assessed is often model recognition: turning words into events or a random variable, deciding which information is prior and which is conditional, and selecting a moment or area that matches the requested statistic. The following habits make that reasoning visible and reduce dependence on memorized question patterns.
Build the sample-space structure first
In a Bayes problem, write the first-stage categories so they are mutually exclusive and exhaustive. “Online,” “shop” and “telephone” can form a three-source partition if every order uses exactly one channel. “Premium customer,” “late payment” and “online order” do not automatically form a partition because they can overlap. When categories overlap, a simple source tree cannot be built without redefining events. A correct diagram is not decoration: it encodes which probabilities should multiply and which routes should add.
A two-way table is often an equally valid route. Convert proportions into expected counts out of a convenient total such as 10,000, then fill cells using conditional percentages. The posterior becomes “count in the requested evidence-and-cause cell divided by total count in the evidence row or column.” Tables are especially useful when the wording contains rates per thousand or several percentages. Trees are especially useful when the order of conditioning matters. Whichever representation you choose, label every row, column or branch with words before relying on symbols.
Distinguish model construction from model use
Many continuous-variable problems have two phases. First, construct a valid density by finding an unknown parameter from total area 1. Second, use that completed model to find a probability or summary. Do not combine these phases carelessly. If the density contains k, solve for k once and retain its exact value. Substituting a rounded value into later moment integrals can distort a variance because variance subtracts two quantities. When several parameters are present, normalization gives one equation; additional information such as a probability, mean or continuity condition supplies another.
Before integrating, inspect whether the proposed formula can be non-negative on the support. An algebraic value of a parameter that makes the density negative is invalid even if total signed area equals 1. For a piecewise density, also examine whether a discontinuity is allowed. A density does not need to be continuous everywhere unless the question states that condition, but it can never assign negative density. These checks identify extraneous algebraic solutions before lengthy calculations continue.
Organize moment calculations in layers
For either a discrete or continuous variable, use the same three-layer plan. Layer one verifies the distribution. Layer two finds the first moment μ=E(X). Layer three finds the second moment E(X2) and then the variance. This prevents the mean from being recalculated inconsistently and keeps the difference between E(X2) and μ² visible. Write the unsimplified integral or sum before entering technology so that a missing factor of x can be found during review.
For a piecewise density with formulas f1 and f2, the mean is the sum of the first-moment integrals over their separate supports. The variance does not come from calculating a variance for each piece and adding them. Instead, calculate the overall E(X) and overall E(X2) across all pieces, then apply the variance identity once. The probability carried by each piece is already represented by its area.
Read information from the shape of a density
A sketch can predict relationships before calculation. In a symmetric unimodal density, mean and median lie at the centre and the mode lies there as well. With a longer right tail, the mean is often pulled to the right of the median, while the mode remains near the highest-density region. With a longer left tail, the order is often reversed. These are checks, not substitutes for the definitions, because unusual or multimodal densities may behave differently.
The height of a density curve should not be interpreted as a percentage. A narrow interval can have high density but small probability because its width is small. Conversely, a wide region with moderate density can carry substantial probability. When comparing two intervals, think in terms of area. If an exact integral is difficult, a graph still helps you judge which probability should be larger and whether a calculator output is plausible.
Handle endpoints and piecewise maxima
To find a mode, differentiate smooth pieces and solve for stationary points, but also test endpoints and breakpoints. The maximum of a density can occur at the edge of the support, as it does for an increasing triangular density. If two separated points share the same greatest density, both are modes. If a piecewise definition jumps, compare the limiting or defined values that the model actually assigns. Do not report a stationary point outside the support.
Separate statistical units from probability units
Probability and density are dimensionally different. Probability is unitless. If X is measured in minutes, a density has units “per minute” so that density multiplied by width produces a unitless area. E(X) and standard deviation are measured in minutes, while Var(X) is measured in square minutes. A transformed variable may change units: converting metres to centimetres multiplies standard deviation by 100 and variance by 10,000.
Write conclusions with the condition included
A Bayes answer should say “given that the item is defective” or whatever evidence is being conditioned on. A fair-game answer should name the participant whose expected net gain is zero. A continuous probability should name the interval and variable. A mean should be described as the model’s long-run average, not as the most likely value. These short phrases are mathematically important because conditional probability and expected value lose meaning when the reference event or perspective is omitted.
Use an error log that records decisions
After practice, classify mistakes as model, setup, calculation or interpretation errors. A model error chooses the wrong partition or treats density as probability. A setup error uses incorrect bounds or omits a Bayes branch. A calculation error mishandles moments, algebra or rounding. An interpretation error reports a result without the condition, units or perspective. Beside every mistake, write a repair instruction such as “total B probability includes every source,” “normalize before moments,” “variance equals second moment minus mean squared,” or “median means half the area.” Reattempt the problem later without looking at the correction.
A practical 14-day revision plan
| Days | Focus | Evidence of mastery |
|---|---|---|
| 1–2 | Conditional probability and tree diagrams | Define events, multiply branches and total evidence correctly |
| 3 | Bayes with two events | Reverse a conditional probability and explain the change |
| 4 | Bayes with three-event partitions | Build a complete denominator without missing a source |
| 5–6 | Discrete moments and variance | Separate E(X²) from [E(X)]² and interpret units |
| 7 | Fair games | Define net gain from a stated perspective |
| 8 | Valid densities and normalization | Use support, non-negativity and total area 1 |
| 9 | Piecewise probabilities | Split integrals at breakpoints |
| 10 | Continuous mode and median | Maximize density and solve a half-area equation |
| 11 | Continuous mean and variance | Calculate first and second moments with correct bounds |
| 12 | Linear transformations | Transform mean, variance and SD without mixing rules |
| 13 | Mixed GDC questions | Show setup, store unrounded values and interpret output |
| 14 | Error log and timed set | Redo missed questions and state one repair rule per error |
Common questions
Which sections are AA HL-only?
The current extension is AHL 4.13–4.14: Bayes’ theorem for up to three events, discrete variance, continuous probability density functions, continuous centre and spread, fair games and linear transformations.
Do AA HL students still need the shared SL Statistics topics?
Yes. The AHL extension assumes conditional probability, discrete random variables, binomial and normal distributions and the other shared foundations are already secure.
Are hypothesis tests part of current AA HL Statistics?
They are not listed in the current AA AHL 4.13–4.14 extension. Do not import a hypothesis-testing syllabus from a different mathematics course.
Why is P(A|B) different from P(B|A)?
The condition changes the reference group. Bayes’ theorem connects the two probabilities using the prior probability and total evidence probability.
Can a density be greater than 1?
Yes. Density height is not point probability. The requirements are non-negativity and total area 1; probabilities are areas over intervals.
Why is P(X=c)=0 for a continuous variable?
A single point has zero width and therefore zero area under an ordinary density curve.
What is the difference between E(X²) and [E(X)]²?
E(X²) averages squared values. [E(X)]² squares the mean. Their difference is the variance.
How do I find the mode of a continuous random variable?
Find where the density is maximal, comparing valid stationary points, endpoints and piecewise breakpoints.
How do I find the median?
Solve for the value m that leaves probability 1/2 below it. Use the correct support and piecewise density.
Does adding a constant change variance?
No. A shift changes the mean but not distances between values, so variance and standard deviation are unchanged.
Why does variance use a² in a transformation?
Variance measures squared spread. Multiplying every deviation by a multiplies squared deviations by a².
What makes a game fair?
A game is fair from a chosen participant’s perspective when expected net gain, including all payments and costs, equals zero.
Continue your AA Higher Level revision
- Shared Statistics and Probability formulae for AA SL and HL
- Functions formulae for AA HL only
- Number and Algebra formulae for AA HL only
- Geometry and Trigonometry formulae for AA HL only
- Calculus formulae for AA HL only
Official sources and session note
This guide follows the current IB Mathematics: analysis and approaches subject guide. The IB has also published a curriculum update for first assessment in May 2029. Students should use the guide and formula booklet supplied for their own examination session.
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