Common mistakes · Maths AI · 10 min read
IB Maths AI mistakes: 14 that cost marks, and the correct numbers
Where AI students lose marks in finance, statistics, networks and modelling, each shown wrong and then right with the numbers worked out.
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Most marks lost in IB Maths AI come from setting up the technology wrongly or interpreting its output carelessly, not from the hard maths. A wrong sign in the finance solver, a cdf where an inverse was needed, a chi-squared test with an expected frequency of 4.5, a regression line pushed far outside its data: below are 14 of the most common mistakes, each with the losing answer next to the correct working.
Every AI paper requires a graphing calculator, so calculator fluency is part of the course. Our GDC mistakes guide covers the button-level traps; this guide covers the maths and the interpretation. It applies to the current AI course, examined until November 2028.
Financial maths mistakes
1. Rounding too early in compound interest
Round the growth factor and the error grows with every period. On calculator papers, answers are accepted as exact or to 3 significant figures unless the question asks otherwise; money is often asked for to two decimal places. Either way, round at the end.
5000 is invested at 4.5% per year, compounded monthly, for 10 years. Find the value to two decimal places.
Loses marks1.00375120 ≈ 1.57, so the value is 5000 × 1.57 = 7850.00.
Earns marks5000 × (1 + 0.045/12)120 = 5000 × 1.566992… = 7834.96.
A 15-unit error from one rounded factor. The finance solver avoids this automatically: N = 120, I% = 4.5, PV = −5000, PMT = 0, P/Y = C/Y = 12, solve for FV.
2. Getting the finance solver's signs wrong
The TVM solver treats money leaving you as negative and money coming to you as positive. If PV and FV have the same sign, there is no solution, and the calculator says so.
How many years will it take for 5000 to grow to more than 8000 at 4% per year, compounded quarterly?
Loses marksPV = 5000, FV = 8000 gives an error; or N = 47.2 quarters, so 11.8 years.
Earns marksPV = −5000, FV = 8000, I% = 4, PMT = 0, P/Y = C/Y = 4 gives N = 47.23 quarters. Interest is only added at the end of each quarter, so it takes 48 quarters: 12 years.
Check: 5000 × 1.0147 ≈ 7981 and 5000 × 1.0148 ≈ 8061. The same sign rule applies to loans: the amount borrowed is positive (you receive it), and each payment is negative.
Regression and correlation mistakes
3. Choosing a model because r is high
A high correlation coefficient says the data is close to some straight line. It does not say a straight line is the right shape. Look at the scatter plot and the residuals before choosing a model.
Loses marksr = 0.96, so the linear model is the most appropriate.
Earns marksr = 0.96, but the points rise steeply and then level off, and the residuals are negative at both ends and positive in the middle. A model that levels off, such as a logistic model, matches that shape better; compare the models' residuals and R2 values before choosing.
Also note that r applies to linear relationships only. For a non-linear regression on your calculator, compare R2 values instead.
4. Extrapolating far beyond the data
A regression model only describes the range it was built on.
For 0 ≤ h ≤ 10 hours of revision, a class's test scores give S = 4.2h + 48.
Loses marksA student revising for 20 hours is predicted to score 4.2 × 20 + 48 = 132%.
Earns marks20 hours is outside the data range of 0–10 hours, so the prediction is unreliable; here it is impossible, since scores cannot exceed 100%.
A related slip: rearranging the y-on-x line to predict x from y. The regression line of x on y is a different line unless the correlation is perfect; check your syllabus and your teacher's notes for which line your course expects you to use.
5. Treating correlation as causation
Loses marksr = 0.78, so using the revision app causes higher grades.
Earns marksr = 0.78 shows a fairly strong positive linear correlation between app use and grade. It does not show causation: students who already work hard may both use the app more and score higher.
Interpret r with both words: strength (weak, moderate, strong) and direction (positive, negative). "Strong correlation" alone usually drops the mark.
Statistical test and distribution mistakes
6. Running a chi-squared test with small expected frequencies, or the wrong degrees of freedom
Expected frequency = (row total × column total) ÷ grand total. If any expected frequency is below 5, combine categories before testing.
In a survey of 80 students, the SL row has 30 students and the "video lessons" column has 12. The expected frequency for that cell is 30 × 12 / 80 = 4.5, which is below 5. Combine "video lessons" with another column so every expected frequency reaches 5.
Loses marks3 × 4 table, so 12 degrees of freedom (or 11).
Earns marksDegrees of freedom = (rows − 1)(columns − 1) = (3 − 1)(4 − 1) = 6. After combining columns, recount: a 3 × 3 table has 4.
For a goodness-of-fit test with n categories and no parameters estimated from the data, the degrees of freedom are n − 1.
7. Writing the wrong conclusion to a hypothesis test
Compare the p-value with the significance level, state the decision about H0, then say what it means in context. Never "accept H0" or "prove" anything.
Loses marksp = 0.031, so we accept H1. This proves year group affects revision method.
Earns marksp = 0.031 < 0.05, so reject H0. There is evidence at the 5% level that revision method is not independent of year group.
If p were 0.12, the correct conclusion is "there is insufficient evidence at the 5% level to reject H0", not "the variables are independent". For choosing tests in a project, see our guide to statistical tests for the IA.
8. Using the normal cdf when you need the inverse, or the variance instead of the standard deviation
If you know a value and want a probability, use the cdf. If you know a probability and want a value, use the inverse normal. And enter the standard deviation, not the variance: for N(150, 122) the calculator needs 12, not 144 (more on this in the GDC guide).
The battery life of a phone model, X hours, satisfies X ~ N(150, 122). Find the battery life exceeded by the best 10% of phones.
Loses marksInverse normal with area 0.1: m = 134.6 hours.
Earns marksP(X > m) = 0.1, so P(X < m) = 0.9. Inverse normal with area 0.9, mean 150, SD 12: m = 165.4 ≈ 165 hours.
134.6 hours is the value below which the worst 10% fall. Inverse normal works from the area to the left unless you choose a tail setting, so sketch the curve and shade the area first.
Voronoi diagram and Markov chain mistakes
9. Drawing the wrong perpendicular bisector
A Voronoi edge between two sites is the perpendicular bisector of the segment joining them: through the midpoint, with the negative reciprocal gradient.
Find the equation of the boundary between the regions of A(2, 1) and B(6, 5).
Loses marksMidpoint (4, 3), gradient of AB = 1, so y − 3 = 1(x − 4), giving y = x − 1.
Earns marksMidpoint (4, 3). Gradient of AB = (5 − 1)/(6 − 2) = 1, so the perpendicular gradient is −1. y − 3 = −(x − 4), giving y = −x + 7.
The losing line passes through A and B themselves: it is the line AB, not the boundary between them. Check any point on your answer: (0, 7) is √40 from both A and B. In "largest empty circle" questions, check the vertices of the Voronoi diagram and the points where its edges meet the boundary of the region: the centre is at one of those.
10. Setting up a transition matrix the wrong way round
If you multiply a transition matrix by a column state vector, each column describes where one state goes next, so each column must sum to 1. (If your class uses row vectors, the rows sum to 1. Either way, be consistent.)
If today is sunny, tomorrow is sunny with probability 0.8; if today is rainy, tomorrow is sunny with probability 0.4.
Loses marksFirst row (0.8, 0.2), second row (0.4, 0.6), used with a column vector. Its columns sum to 1.2 and 0.8, and solving Tv = v gives s = r: "half the days are sunny".
Earns marksFirst column (0.8, 0.2) "from sunny", second column (0.4, 0.6) "from rainy", each summing to 1. Steady state: s = 0.8s + 0.4r with s + r = 1, so s = 0.4s + 0.4, 0.6s = 0.4 and s = 2/3. In the long run, 2/3 of days are sunny and 1/3 rainy.
The equation from Tv = v on its own only gives a ratio (s = 2r); the extra condition s + r = 1 turns it into probabilities. Check: T × (2/3, 1/3) = (2/3, 1/3). From a sunny day, the probability of sun three days later is the first entry of T3 × (1, 0), which is 0.688.
Graph and model interpretation mistakes
11. Misreading a graph's scale or units
Gradients carry the units of the axes. Read the axis labels before you calculate.
Loses marksThe distance–time graph goes from (0, 0) to (20, 5), so the speed is 0.25 km/h.
Earns marksTime is in minutes and distance in km, so the gradient is 0.25 km per minute, which is 15 km/h.
On a logarithmic axis, equal spacing means equal ratios. The point halfway between 10 and 100 is 101.5 ≈ 31.6, not 55.
12. Using a model outside its sensible domain
Questions that ask you to "comment" or to state a limitation want the model's assumptions in context.
A ball's height is modelled by h(t) = −4.9t2 + 12t + 1.5 metres.
Loses marksAfter 3 seconds, the height is h(3) = −6.6 m.
Earns marksh(t) = 0 when t ≈ 2.57 s, when the ball lands, so the model only applies for 0 ≤ t ≤ 2.57. A negative height has no meaning here. The model also ignores air resistance.
13. Miscounting the strips in the trapezoidal rule
Five readings make four strips. The width is h = (b − a)/n, where n is the number of strips, not the number of points.
Readings at x = 0, 2, 4, 6, 8 give y = 3, 5, 6, 4, 1. Estimate the area under the curve.
Loses marksh = 8/5 = 1.6, area ≈ 0.8 × (3 + 1 + 2(5 + 6 + 4)) = 27.2.
Earns marksh = 8/4 = 2, area ≈ (2/2) × (3 + 1 + 2(5 + 6 + 4)) = 34.
Accuracy, units and sequence mistakes
14. Dividing by the wrong value in a percentage error, and other small losses
Percentage error = |approximate − exact| ÷ |exact| × 100%. Divide by the exact value.
Loses marksEstimate 48, exact 52.5: error = 4.5/48 × 100 = 9.375
Earns marksError = |48 − 52.5| / 52.5 × 100% = 8.57%
Two other small losses add up across a paper. First, missing units or the wrong accuracy: give the unit (%, cm2, km/h) and 3 significant figures unless told otherwise. Second, the n − 1 in arithmetic sequences: if you save 50 in week 1 and 5 more each week after that, week 20 is 50 + 19 × 5 = 145 (not 150), and the total over 20 weeks is (20/2)(50 + 145) = 1950.
How do you stop making the same AI mistakes?
- Write the setup before you press anything. For TVM, list N, I%, PV, PMT, FV with signs. For the normal distribution, sketch the curve and shade the area. Both earn method marks, and both catch errors. Our guide to how mark schemes award marks explains why the setup counts.
- Interpret every number in context. A probability, a correlation or a regression prediction is only finished when you have said what it means and whether it is reasonable.
- Keep a mistake log by type: "used cdf for inverse", "same-sign TVM", "expected < 5". Your three most common types are your revision plan.
- Know which topics are HL only. Markov chains, the Poisson distribution and the network algorithms (minimum spanning trees, Chinese postman, travelling salesman) are AI HL content; see how the AA and AI papers compare for the paper structure.
Questions students ask
How many significant figures should I give in IB Maths AI?
Give the exact value or 3 significant figures unless the question asks for something else, such as money to two decimal places or a whole number of people. Keep full values in your calculator until the final answer.
Why does my TVM solver give an error?
When you solve for N or I%, the usual cause is a present value and future value with the same sign. Money you pay out and money you receive must have opposite signs, so an investment has a negative PV and a positive FV.
When can I not use a chi-squared test?
When any expected frequency is below 5. Combine adjacent categories until every expected frequency is at least 5, and recalculate the degrees of freedom for the smaller table.
Does a high correlation coefficient mean one variable causes the other?
No. A value of r close to 1 or −1 shows a strong linear relationship, not a cause. A third variable, or reverse causation, can produce the same pattern.
Try one: Maths AI
The time T minutes that students at a school take to travel to school is normally distributed with mean 32 minutes and standard deviation 6 minutes.
(a) Find the probability that a randomly chosen student takes more than 40 minutes.
(b) 15% of students take less than k minutes. Find k.
[4 marks]Show the answer and why
- (a) Uses the normal cdf with lower bound 40, mean 32 and standard deviation 6
- (a) P(T > 40) = 0.0912 (3 s.f.)
- (b) Uses the inverse normal with area 0.15 to the left, mean 32 and standard deviation 6
- (b) k = 25.8 minutes (3 s.f.)
Part (a) finds a probability from a value, so it needs the cdf. Part (b) finds a value from a probability, so it needs the inverse. Using 0.85 in part (b) gives 38.2 minutes, which is the time exceeded by 15% of students, not the one 15% fall below. Entering the variance (36) instead of the standard deviation (6) also loses both marks.