Sixteen lessons on accuracy, finance, modelling, geometry, Voronoi diagrams, statistics, hypothesis tests, calculus and the exploration for the Applications and Interpretation course.
A skills course on common problem structures with technology, at standard-level depth. It does not replace the syllabus. Higher-level-only content (complex numbers, matrices, graph theory beyond what is shown, differential equations and Paper 3 topics) is not covered.
Algebra, coordinate geometry and basic statistics.
Course outline
Accuracy: significant figures and percentage error
Round sensibly, use standard form and calculate percentage error in context.
Compound interest and the finance solver
Use time-value-of-money calculations and know what each solver entry means.
Exponential and sinusoidal models
Build and interpret exponential and sinusoidal models from context.
Geometry and trigonometry in context
Use volume, surface area, bearings and the sine and cosine rules in applied problems.
Voronoi diagrams: boundaries from midpoints
Construct Voronoi diagrams and use them to decide the nearest site.
Descriptive statistics and displaying data
Summarise data with the correct measure and read box plots and histograms.
Correlation and regression: interpolate with confidence
Calculate and interpret correlation coefficients and a regression line, with limits.
Probability: Venn diagrams, conditional and tree diagrams
Combine probabilities with the right rule and interpret conditional statements.
Binomial and normal distributions with technology
Recognise a binomial or normal setting and calculate probabilities and inverse values.
Chi-squared test for independence
Set up hypotheses, expected counts, degrees of freedom and a conclusion in context.
Hypothesis testing with the t-test
Choose between one-sample, two-sample and paired tests and interpret p-values.
Choosing a model: linear, quadratic, power and piecewise
Match a model to the pattern in data and justify it with fit and context.
Differentiation in context: rates and optimisation
Differentiate polynomials and interpret the gradient as a rate of change.
Integration and the trapezoidal rule
Find areas by integration and estimate areas from data with the trapezoidal rule.
Solving with technology: show the setup
Present calculator work so that method marks are awarded and answers can be checked.
The exploration as a modelling cycle
Plan a mathematical exploration around a question you can model and finish.
Sources and curriculum note
Reviewed October 5, 2026 against the IB subject brief for the current course (first assessments 2021). The brief gives SL Paper 1 40 percent, Paper 2 40 percent and the exploration 20 percent. The IB lists a new Mathematics: applications and interpretation course with first teaching August 2027 and first assessment May 2029; this note is carried over from the earlier version and its content is not described here. Check the official IB page for changes.
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. Accuracy: significant figures and percentage error
Learning goal: Round sensibly, use standard form and calculate percentage error in context.
Mathematics in real contexts uses numbers that are measured or rounded, so you must know how accurate an answer should be. In IB examinations an answer is normally given to three significant figures unless the question says otherwise. Significant figures count from the first non-zero digit. The number 0.004560 has four significant figures, since the leading zeros only place the decimal point. Rounding to three significant figures means looking at the fourth digit: if it is 5 or more, round up. Do not round intermediate values; keep the full calculator value and round only the final answer.
Standard form writes a number as a x 10^k, where a is at least 1 and less than 10 and k is an integer. The speed 300 000 000 m/s is 3 x 10^8 and 0.00052 is 5.2 x 10^-4. Standard form makes very large and very small numbers readable and helps compare orders of magnitude. When multiplying, multiply the leading numbers and add the powers; when dividing, divide the leading numbers and subtract the powers. Check that the result is back in standard form, for example 12 x 10^3 becomes 1.2 x 10^4.
The percentage error compares an estimate to the exact value: percentage error = |v_A - v_E| / |v_E| x 100, where v_A is the approximate value and v_E is the exact value. The absolute value means the error is positive, whether the estimate is too high or too low. The denominator is the exact value, not the estimate, a point that is often reversed. If the exact value of a length is 12.5 m and a measurement says 12.0 m, the error is 0.5 / 12.5 x 100 = 4 percent.
In a context question, the units and the reasonableness of the answer matter. A cost of 12.457 dollars should be given as 12.46 dollars because money is rounded to the nearest cent. A number of people must be a whole number, and rounding direction matters: a bus that must hold 130 people with 40 seats each needs 4 buses, not 3.25, so round up. Report the unit and a short sentence that answers the question. Estimation is a useful check: if a calculator answer is a hundred times larger than expected, look for a decimal or unit slip.
Worked example
A length is estimated as 4.8 m. Its exact value is 5.0 m. Find the percentage error.
Error = |4.8 - 5.0| = 0.2.
Divide by the exact value: 0.2 / 5.0 = 0.04.
Multiply by 100.
Percentage error = 4 percent.
Practice problem and solution
Write 0.00072 in standard form as a x 10^k. Enter the value of k.
0.00072 = 7.2 x 10^-4, so k = -4.
Mental model: Round only the final answer.
Common trap: Dividing by the estimate in percentage error.
2. Compound interest and the finance solver
Learning goal: Use time-value-of-money calculations and know what each solver entry means.
Interest is a payment for the use of money. Simple interest is calculated only on the original amount, so it grows linearly: I = P r t / 100, with r in percent per year and t in years. Compound interest is calculated on the amount plus the interest already earned, so it grows exponentially. The future value after n years with annual compounding is FV = PV x (1 + r / 100)^n. For 2000 dollars at 3 percent per year for 5 years, the future value is 2000 x 1.03^5, which is about 2318.55 dollars.
When interest is compounded more often than once a year, the formula is FV = PV x (1 + r / (100k))^(kn), where k is the number of compounding periods per year. Quarterly compounding has k = 4 and monthly k = 12. A higher frequency gives a slightly larger amount for the same nominal annual rate. The nominal rate is the stated yearly rate, while the effective rate is the true annual growth after compounding. Compare investments by their effective rates, not their nominal ones.
The finance solver on a graphing calculator uses the variables N for the total number of payments, I% for the annual interest rate, PV for the present value, PMT for the payment per period, FV for the future value, P/Y for payments per year and C/Y for compounding periods per year. Money paid out is entered as negative and money received as positive. A common error is entering a payment with the wrong sign, which gives a result with the opposite sign, or forgetting to set P/Y and C/Y to match the question.
For loans and savings plans, state the solver entries in your answer. Write the values of N, I%, PV, PMT, FV, P/Y and C/Y that you used, because marks are awarded for a correct setup even if the final number is slightly off. Rounding: money is reported to two decimal places. Depreciation is the opposite of growth: a car that loses 12 percent of its value each year has a multiplier of 0.88. Inflation reduces purchasing power, and the real value of an amount is its nominal value adjusted by the cumulative effect of inflation. Interpret each answer with a sentence that includes the unit.
Worked example
Find the future value of 5000 dollars invested at 4 percent per year, compounded annually, for 3 years.
FV = PV x (1 + r / 100)^n.
FV = 5000 x 1.04^3.
1.04^3 = 1.124864.
FV = 5624.32 dollars.
Practice problem and solution
A car worth 20 000 dollars loses 10 percent of its value each year. Find its value after 2 years in dollars.
Value = 20 000 x 0.9^2 = 20 000 x 0.81 = 16 200.
Mental model: State your solver entries.
Common trap: Entering a payment with the wrong sign.
3. Exponential and sinusoidal models
Learning goal: Build and interpret exponential and sinusoidal models from context.
A real quantity that changes by a fixed percentage in equal time periods follows an exponential model, y = a x b^t. The parameter a is the initial value and b is the multiplier per time unit. If b is greater than 1 the quantity grows, and if b is between 0 and 1 it decays. A population that grows 5 percent a year has b = 1.05. The doubling time is the time at which the quantity is twice as large, and it does not depend on the starting amount. Solve a x b^t = 2a by cancelling a and using technology or logarithms.
When fitting an exponential model to data, use the regression function on the calculator, which gives the values of a and b. Write the equation with the numbers and the meaning of each parameter in context. To predict a future value, substitute the time. To find when the quantity reaches a given value, set up an equation and solve it with the graphing tools. Mention the limits: no real quantity grows exponentially forever, and models apply only within a suitable range of time.
Periodic phenomena such as tides, temperatures through a year and the height of a point on a wheel are modelled by sinusoidal functions, y = a sin(b(x - c)) + d. The amplitude is |a|, the period is 2 pi / b (or 360 / b with degrees), the vertical shift d gives the midline and c gives the horizontal shift. The maximum value is d + |a| and the minimum is d - |a|. To find a model from data, use the maximum and minimum to find a and d, and the distance between peaks to find the period.
Interpret the model with the context. For a tidal height with maximum 4.2 m, minimum 0.8 m and a period of 12 hours, the amplitude is (4.2 - 0.8) / 2 = 1.7 m, the midline is 2.5 m and b = 2 pi / 12 = pi / 6. To find when the height exceeds 3 m, solve the inequality by graphing the function and the line y = 3 and reading the intersection points. Check the calculator mode: radians or degrees must match the period you used. Give times with a unit and in a form appropriate to the question, such as hours and minutes.
Worked example
A tide has maximum height 4.2 m and minimum 0.8 m. Find the amplitude and the midline.
Amplitude = (max - min) / 2 = (4.2 - 0.8) / 2.
Amplitude = 1.7 m.
Midline = (max + min) / 2 = 5.0 / 2.
Midline = 2.5 m.
Practice problem and solution
A model for a quantity is y = 3 sin(2x) + 5. Find the maximum value of y.
Maximum = d + |a| = 5 + 3 = 8.
Mental model: Check calculator mode.
Common trap: Using the wrong mode for the period.
4. Geometry and trigonometry in context
Learning goal: Use volume, surface area, bearings and the sine and cosine rules in applied problems.
Applied geometry begins with a clear diagram. For three-dimensional solids, the volume of a prism is the area of the cross-section times the length. For a pyramid or cone, the volume is one third of the base area times the height, and for a sphere it is 4/3 pi r^3. Surface area adds up the faces. Many questions give a composite solid, such as a cylinder with a cone on top. Split it into parts, compute each and add, taking care not to include faces that are joined.
Right-angled trigonometry gives sin theta = opposite / hypotenuse, cos theta = adjacent / hypotenuse and tan theta = opposite / adjacent. Angles of elevation and depression are measured from the horizontal. In three dimensions, find a right triangle that contains the required angle, such as the angle between a line and a plane, and use the lengths in that triangle. Draw it separately, label the sides, and compute step by step, keeping accuracy until the final step.
For triangles without a right angle, use the sine rule a / sin A = b / sin B when you have an opposite pair, and the cosine rule c^2 = a^2 + b^2 - 2ab cos C when you know two sides and the included angle or all three sides. The area of a triangle is (1/2) a b sin C. When the sine rule is used to find an angle, check for the possibility of a second, obtuse angle with sin(180 - A) = sin A. The calculator will give only the acute one, so check the angle sum for validity.
Bearings are measured clockwise from north, with three digits, such as 045 or 270. To solve a bearing problem, draw a north line at each point, mark the angles given, and use alternate angles between parallel north lines to find angles inside the triangle. The return bearing from B to A differs from the bearing from A to B by 180 degrees. Include the unit in the answer: distances in kilometres or metres, angles in degrees. A final sentence stating the result in the language of the question earns the communication mark.
Worked example
A ship sails 10 km on bearing 060, then 8 km on bearing 150. Find the distance from the start.
The turn is 90 degrees, so the path forms a right angle.
Distance^2 = 10^2 + 8^2 = 164.
Distance = sqrt(164).
Distance = 12.8 km.
Practice problem and solution
Find the volume of a cone with radius 3 cm and height 5 cm in cm^3, to three significant figures.
V = (1/3) pi r^2 h = (1/3) x pi x 9 x 5 = 15 pi = 47.1 cm^3.
Mental model: Draw a clear diagram first.
Common trap: Using degrees when the angles are in radians.
5. Voronoi diagrams: boundaries from midpoints
Learning goal: Construct Voronoi diagrams and use them to decide the nearest site.
A Voronoi diagram divides the plane into regions, called cells. Each cell belongs to a site, such as a school or a hospital, and contains all points that are closer to that site than to any other. The boundary between two cells is made of points that are the same distance from both sites. This boundary is part of the perpendicular bisector of the line segment joining the two sites. Voronoi diagrams are used to decide the nearest facility, such as the closest emergency station, and in planning where to place a new one.
To construct a diagram for two sites, find the midpoint of the segment joining them and the gradient of the segment, then write the perpendicular bisector. The perpendicular gradient is the negative reciprocal of the gradient of the segment. For sites A(0, 0) and B(4, 2), the midpoint is (2, 1) and the gradient of AB is 1/2, so the bisector has gradient -2 and equation y - 1 = -2(x - 2), which gives y = -2x + 5. Check that the midpoint lies on the line and that the sites are on opposite sides.
With three sites, construct the three perpendicular bisectors of the segments joining pairs. They meet at a single point, called a vertex, which is the same distance from all three sites. This point is the centre of the circle through the three sites. A vertex can be found by solving two bisector equations simultaneously. The cell boundaries are then the parts of the bisectors from the vertex outward, away from the third site. A sketch with each cell labelled shows which part of each line is used.
To decide which site is closest to a point, locate the point in the diagram and read the cell. To place a new site where it is as far as possible from existing sites, the best candidates are the vertices of the diagram, because each vertex is the largest empty circle around it. Add the new site and construct the changes to the cells. In an exam, include coordinates for each vertex and state which sites it is equidistant from. Mention a limit of the model: straight-line distance may not match travel distance along roads.
Worked example
Sites are A(0, 0) and B(6, 0). Find the equation of the boundary between their cells.
Midpoint = (3, 0).
The segment AB is horizontal.
The perpendicular bisector is vertical.
Boundary: x = 3.
Practice problem and solution
Sites are A(1, 1) and B(5, 1). Find the x-coordinate of the boundary between their cells.
The boundary is the perpendicular bisector through the midpoint (3, 1), so x = 3.
Mental model: Check which cell a test point lies in.
Common trap: Using the segment's gradient instead of its negative reciprocal.
6. Descriptive statistics and displaying data
Learning goal: Summarise data with the correct measure and read box plots and histograms.
A data set is summarised by its centre and spread. The mean is the sum divided by the number of values, and the median is the middle value in order. For grouped data, use the midpoint of each class as an estimate and find an estimated mean. The mode, or modal class for grouped data, is the most frequent. Choose the measure to fit the shape: the median is more resistant to extreme values, so it is preferred for skewed data such as incomes, while the mean uses all the values and is suitable for symmetric data.
Spread is described by the range, the interquartile range (IQR) and the standard deviation. The IQR is the difference between the upper and lower quartiles and shows the spread of the middle half of the data. A larger standard deviation means that the values are more spread out around the mean. On the calculator, enter the data in a list and use the one-variable statistics function. Read the mean, standard deviation, median and quartiles from the output. Be careful to use the population standard deviation as the course requires, and check the label.
A box-and-whisker plot shows the minimum, lower quartile, median, upper quartile and maximum. A value is an outlier if it is more than 1.5 times the IQR above the upper quartile or below the lower quartile. Outliers are plotted separately and the whisker ends at the largest value that is not an outlier. When comparing two box plots, comment on the median, the IQR and the overall range, in context. For example, say that the median score is higher in class A, but the IQR is also larger, so the class is less consistent.
Histograms show the frequency of data grouped in classes of equal width. A cumulative frequency graph plots the running total against the upper class boundary and is used to estimate the median and quartiles by reading across from half and a quarter or three quarters of the total. When estimating, say that the values are estimates because of grouping. Interpret results in the context: units, population and what the numbers mean. Avoid vague words such as 'more spread'; say 'the IQR is larger, so the middle half of the scores covers a wider range'.
Worked example
Q1 = 20 and Q3 = 30. Find the upper fence for outliers.
IQR = 30 - 20 = 10.
1.5 x IQR = 15.
Upper fence = 30 + 15.
Upper fence = 45.
Practice problem and solution
The values 4, 6, 8, 10 and 12 are given. Find the mean.
Sum = 40; 40 / 5 = 8.
Mental model: Use the one-variable statistics function.
Common trap: Reporting the sample standard deviation without checking the label.
7. Correlation and regression: interpolate with confidence
Learning goal: Calculate and interpret correlation coefficients and a regression line, with limits.
Bivariate data records two variables for each item, and a scatter diagram shows whether they are related. Describe the relationship by direction (positive, negative), strength (strong, moderate, weak) and form (linear or not). Pearson's product-moment correlation coefficient r measures the strength of a linear relationship, from -1 to +1. Values near +1 or -1 indicate a strong linear relationship, and values near 0 indicate little linear relationship. A scatter diagram with a curved pattern may have a low r, although the variables are strongly related in a non-linear way.
Spearman's rank correlation coefficient r_s is used when the relationship is monotonic but not necessarily linear, or when the data are ranks. It is calculated by ranking each variable and finding Pearson's coefficient of the ranks. A value near +1 shows that as one variable increases the other tends to increase. This method is more resistant to outliers than Pearson's. State clearly which coefficient you are using and why, based on the shape of the scatter diagram.
The regression line of y on x, y = a x + b, is found by technology. The gradient a is the predicted change in y per unit increase in x, and the intercept b is the predicted y when x is zero. Interpret both in context with units, noting that the intercept may not make sense if x = 0 is outside the data. Use the line to predict y from x. Prediction within the range of the data is interpolation and is generally reliable when the correlation is strong. Prediction outside the range, extrapolation, is much less reliable.
Correlation does not imply causation. Two variables may both depend on a third, or the relationship may be a coincidence. To argue for causation, you need more evidence, such as a controlled experiment and a plausible mechanism. When a question asks you to comment on the validity of a prediction, mention three things: the strength of the correlation, whether the value is inside the data range and whether the model is reasonable in context. Do not use the regression of y on x to predict x. A residual is the observed value minus the predicted value, and patterns in residuals suggest a poor model.
Worked example
A regression line is y = 2.4x + 15, where x is hours of practice and y is the score. Predict y for x = 10.
Substitute x = 10.
y = 2.4 x 10 + 15.
= 24 + 15.
y = 39.
Practice problem and solution
A regression line is y = 0.5x + 3. Find y when x = 8.
y = 0.5 x 8 + 3 = 7.
Mental model: State the reliability of the prediction.
Common trap: Using the line of y on x to predict x.
8. Probability: Venn diagrams, conditional and tree diagrams
Learning goal: Combine probabilities with the right rule and interpret conditional statements.
Probability measures the chance of an event on a scale from 0 to 1. The probability of an event A is the number of favourable outcomes divided by the number of equally likely outcomes. The complement of A, written A', has probability 1 - P(A). Questions with the words 'at least one' are often easiest through the complement: P(at least one) = 1 - P(none). A Venn diagram shows overlaps clearly. Fill in the overlap first, then the remaining parts of each set, and check that all regions sum to the total.
For two events, P(A or B) = P(A) + P(B) - P(A and B). The subtraction removes the overlap, which was counted twice. Events are mutually exclusive if they cannot occur together, so P(A and B) = 0. Conditional probability is P(A | B) = P(A and B) / P(B), the probability of A given that B has occurred. The condition becomes the new sample space, so the denominator is the group after the word 'given'. In a two-way table, use the row or column total of the condition.
Events A and B are independent if P(A and B) = P(A) x P(B), which is equivalent to P(A | B) = P(A). To test for independence, compute the product of the two probabilities and compare it with the probability of both. Mutually exclusive events with nonzero probabilities are not independent. A common mistake is to call events independent because they seem unrelated. Check with numbers, using the table or the Venn diagram.
A tree diagram shows sequential events. Multiply the probabilities along a path to get the probability of that sequence and add the probabilities of paths that lead to the required outcome. With replacement, the branch probabilities stay the same, and without replacement, they change because the totals change. After building the tree, label each branch and check that the branches from each node sum to 1. For a question that asks for a conditional probability, find the path probabilities needed and divide.
Worked example
A bag has 4 red and 6 blue balls. One is drawn, not replaced, then another. Find P(both red).
P(first red) = 4/10.
P(second red) = 3/9.
Multiply: 4/10 x 3/9 = 12/90.
P = 2/15, about 0.133.
Practice problem and solution
Two dice are rolled. Find the probability of at least one six. Give a decimal to three decimal places.
Common trap: Calling events independent without checking.
9. Binomial and normal distributions with technology
Learning goal: Recognise a binomial or normal setting and calculate probabilities and inverse values.
A binomial random variable counts the number of successes in a fixed number of independent trials, each with the same probability of success. We write X ~ B(n, p). Check four conditions: a fixed number of trials, two outcomes, a constant probability of success and independence. The mean is np and the variance is np(1 - p). On a calculator, use binomial pdf for P(X = k) and binomial cdf for P(X at most k). Write down the probability statement first, then the calculator command with its inputs.
Wording matters. 'At least 4' means P(X at least 4) = 1 - P(X at most 3). 'More than 4' means 1 - P(X at most 4), and 'fewer than 4' means P(X at most 3). The discrete boundary changes the answer, so read the wording carefully and write the statement in symbols. Give the answer to three significant figures. If the question asks for the probability in context, add a short sentence with the context.
A normal random variable X ~ N(mu, sigma^2) is continuous with a bell-shaped curve. The parameter sigma is the standard deviation, so be careful when a calculator asks for sigma and the question gives the variance. P(X = a) is zero, so P(X at most a) and P(X less than a) are the same. Sketch the curve, mark the mean and shade the area. Use the normal cdf with lower and upper bounds, and use a large value such as 10^99 for infinity.
Inverse normal questions give a probability and ask for a value. Use the inverse normal function with the area to the left of the value. For the top 5 percent, the area to the left is 0.95. For z-scores, z = (x - mu) / sigma tells how many standard deviations a value is from the mean, which helps compare values from different distributions. Write the setup, including the distribution and the bounds. Marks are given for a correct statement even if the numerical output is slightly off. Check that answers are reasonable by comparing with the empirical rule.
Worked example
X ~ B(10, 0.3). Find the mean and variance of X.
Mean = np = 10 x 0.3.
Mean = 3.
Variance = np(1 - p) = 10 x 0.3 x 0.7.
Variance = 2.1.
Practice problem and solution
X ~ N(100, 15^2). A value is 130. Find its z-score.
z = (130 - 100) / 15 = 2.
Mental model: Write the statement, then the calculator command.
Common trap: Entering the variance where the standard deviation is needed.
10. Chi-squared test for independence
Learning goal: Set up hypotheses, expected counts, degrees of freedom and a conclusion in context.
The chi-squared test for independence checks whether two categorical variables are associated. Data are put in a contingency table of observed counts. The null hypothesis H0 states that the two variables are independent, and the alternative H1 states that they are not independent, meaning that they are associated. Always write the hypotheses in context, for example 'H0: the choice of sport is independent of gender'. A test uses counts, not percentages or averages.
The expected count for each cell is (row total x column total) / grand total. The test statistic is the sum over all cells of (observed - expected)^2 / expected. For the test to be valid, the expected counts should be at least 5 in each cell; if not, combine categories. On the calculator, enter the observed matrix and run the chi-squared test, which gives the statistic, the p-value, the degrees of freedom and the expected matrix. Quote the numbers you use in the answer.
Degrees of freedom for a table with r rows and c columns are (r - 1)(c - 1). For a 3 by 2 table, this is 2 x 1 = 2. The significance level, commonly 5 percent, is the threshold for the decision. If the p-value is less than the significance level, reject H0; if it is greater, do not reject H0. Alternatively, compare the statistic with the critical value for the given degrees of freedom and significance level. State the decision clearly, with the reason.
The conclusion must be in context and must not overclaim. If H0 is rejected, say that there is sufficient evidence at the 5 percent level to suggest that the variables are associated. If H0 is not rejected, say there is insufficient evidence of an association, not that independence is proved. A chi-squared test also does not say how strong or in what direction the association is. To describe it, compare observed and expected counts in the cells with the largest contributions. A goodness-of-fit test, which compares counts with an expected distribution, uses degrees of freedom of the number of categories minus one minus any estimated parameters.
Worked example
A 2 by 3 table has row totals 40 and 60, column totals 30, 30 and 40, and grand total 100. Find the expected count for row 1, column 3.
Expected = row total x column total / grand total.
= 40 x 40 / 100.
= 1600 / 100.
Expected count = 16.
Practice problem and solution
A contingency table has 4 rows and 3 columns. Find the degrees of freedom for the chi-squared test of independence.
(4 - 1)(3 - 1) = 3 x 2 = 6.
Mental model: Counts, not percentages.
Common trap: Concluding that the variables are independent when H0 is not rejected.
11. Hypothesis testing with the t-test
Learning goal: Choose between one-sample, two-sample and paired tests and interpret p-values.
A hypothesis test asks whether sample data give enough evidence against a claim about a population. The null hypothesis H0 is the claim, often that there is no difference or that a mean equals a stated value. The alternative H1 states the effect. A test can be one-tailed, for 'greater than' or 'less than', or two-tailed, for 'different from'. Decide the form of H1 from the wording of the question, before looking at the data. The significance level, commonly 5 percent, is the chance of rejecting a true H0 that you are prepared to accept.
A two-sample t-test compares the means of two independent groups, such as the test scores of two classes, assuming the data are roughly normally distributed with similar variances. Enter the two lists into the calculator, select the pooled option as the course states, and read the t-statistic and the p-value. For a paired test, the same subjects are measured twice, such as before and after training. In that case, calculate the differences, and the test is applied to the differences. Choosing the wrong type of test is a common error.
The p-value is the probability of obtaining a result at least as extreme as the one observed, if H0 were true. A small p-value, less than the significance level, is evidence against H0. It does not give the probability that H0 is true. Compare the p-value with the significance level, and for a two-tailed test, make sure the p-value corresponds to two tails. State the conclusion: reject H0 or do not reject H0. Then interpret it in context.
Interpretation matters more than the arithmetic. If H0 is rejected at the 5 percent level, say there is sufficient evidence to suggest that the means differ. If it is not rejected, say there is insufficient evidence. State the assumptions: random samples, approximately normal distributions and independence. A small p-value from a very large sample may show a difference that is statistically significant but too small to matter in practice, so comment on the size of the difference as well. Use words from the situation, such as 'the new teaching method' and 'the mean score'.
Worked example
A paired t-test on the same students before and after training gives p = 0.02. State the conclusion at the 5 percent level.
p = 0.02 is less than 0.05.
Reject H0.
The mean scores differ.
There is sufficient evidence of a change after training.
Practice problem and solution
A test has a p-value of 0.04 at the 5 percent level. Enter 1 if H0 is rejected, 0 if not.
0.04 is less than 0.05, so H0 is rejected.
Mental model: State the assumptions behind the test.
Common trap: Saying that a large p-value proves H0.
12. Choosing a model: linear, quadratic, power and piecewise
Learning goal: Match a model to the pattern in data and justify it with fit and context.
A model is a mathematical description of a real situation. Choosing one starts with a graph of the data. A straight-line pattern suggests a linear model y = mx + c, where m is the rate of change per unit of x and c is the starting value. A curve with one turning point suggests a quadratic model y = ax^2 + bx + c, as for the height of a thrown ball. A curve that rises or falls steeply and passes through the origin may fit a power model y = a x^b. A pattern with a fixed percentage change fits an exponential model, and a repeating wave fits a sinusoidal one.
Fitting a model uses technology. Enter the data into a list, choose the regression type and read the parameters. Then check the fit in three ways. First, look at the graph of the model with the data points. Second, look at the coefficient of determination R^2, which is close to 1 for a good fit. Third, look at the residuals, which are the data values minus the model values; they should look random, with no pattern. A curve in the residuals shows that the model is missing something, such as a curve in the data.
The context should support the model. A quadratic for the height of a ball predicts negative heights after it lands, so the domain must be restricted to the time of flight. A linear model for the population of a city predicts negative values far in the future. State the domain over which the model is valid and the reason. Interpret parameters in context with units: in a model for taxi cost, the gradient is the cost per kilometre, and the intercept is the fixed fee.
A piecewise model uses different rules on different intervals, written with a brace and conditions such as x less than or equal to 10. Tax rates, parking fees and mobile plans are common examples. To evaluate it, choose the rule that matches the value of x. To graph it, plot each piece on its own interval, with open and closed endpoints to show which rule applies at the boundary. For continuity, check that the pieces meet at the boundary. When comparing models, prefer the simplest one that fits well and makes sense in context.
Worked example
A taxi charges 3.50 dollars plus 1.20 dollars per km. Write the model and find the cost of 10 km.
Linear model: C = 1.2d + 3.5.
Substitute d = 10.
C = 12 + 3.5.
C = 15.50 dollars.
Practice problem and solution
A model is C = 2x + 5 for 0 to 10 items and C = 3x - 5 for more than 10 items. Find C for x = 12.
x = 12 is more than 10, so C = 3 x 12 - 5 = 31.
Mental model: Pick the simplest model that fits and makes sense.
Common trap: Using the wrong piece at the boundary.
13. Differentiation in context: rates and optimisation
Learning goal: Differentiate polynomials and interpret the gradient as a rate of change.
The derivative of a function is its rate of change. For f(x) = ax^n, the derivative is f'(x) = a n x^(n-1). So the derivative of 3x^4 is 12x^3, and the derivative of a constant is zero. Differentiate each term separately. Before differentiating, rewrite expressions with roots and reciprocals as powers. The derivative at a point gives the gradient of the tangent there, which is the instantaneous rate of change. In context, the units of the derivative are the units of y divided by the units of x, such as metres per second.
To interpret, state what is changing and how fast. If C(x) is the cost in dollars of producing x items, then C'(x) is the rate of change of cost in dollars per item, and C'(100) = 4 means that at 100 items the cost is increasing at 4 dollars per extra item. A negative derivative shows a decrease. When a question gives a velocity function, the derivative is acceleration. Write sentences that include the numbers, units and the point at which the rate is measured.
The tangent to y = f(x) at x = a has gradient f'(a) and passes through (a, f(a)). Its equation is y - f(a) = f'(a)(x - a). The normal is perpendicular to the tangent, with gradient -1 / f'(a). With technology, you can also find the derivative at a point from the graph function, but write the derivative algebraically when the course asks you to show the working. Use the calculator to check your answer by graphing the tangent against the curve.
Stationary points occur where f'(x) = 0. To classify them, use the sign of f'(x) on either side, or the sign of f''(x). A maximum has f'(x) changing from positive to negative; a minimum has it changing from negative to positive. In an optimisation problem, such as maximising the area of a garden with fixed fencing, form a function of one variable, differentiate, set the derivative to zero, solve and interpret. Check the result by graphing the function. State the answer in the context with its units, and check that the solution lies within the allowed domain.
Worked example
Find the gradient of y = x^3 - 4x at x = 2.
dy/dx = 3x^2 - 4.
At x = 2: 3 x 4 - 4.
= 12 - 4.
Gradient = 8.
Practice problem and solution
Find the derivative of f(x) = 5x^3 at x = 2. Enter the value.
f'(x) = 15x^2. At x = 2: 15 x 4 = 60.
Mental model: Interpret the derivative with units.
Common trap: Dropping the units from a rate of change.
14. Integration and the trapezoidal rule
Learning goal: Find areas by integration and estimate areas from data with the trapezoidal rule.
Integration is the reverse of differentiation. For a power, the integral of x^n is x^(n+1) / (n + 1) + C, for n not equal to -1. The constant C appears in an indefinite integral. For a definite integral from a to b, evaluate the antiderivative at b and at a and subtract. The constant cancels, so it is not needed. A definite integral gives the signed area between the curve and the x-axis, so regions below the axis contribute negative values.
To find the total area when the curve crosses the axis, split the integral at the crossing points and take the positive value of each part. With a graphing calculator, you can find the definite integral of a function numerically, and use it to check hand work. To find the area between two curves, integrate the upper curve minus the lower curve between the intersections. A sketch of the region helps you decide the order and the limits, and prevents sign errors.
When a function is known only through a table of values, estimate the area with the trapezoidal rule. For n strips of equal width h, the area is approximately (h / 2) x [first y + last y + 2 x (sum of the other y values)]. Each strip is a trapezoid of area h x (the average of its two heights). The rule overestimates the area when the curve is concave up, and underestimates when it is concave down. Increasing the number of strips improves the estimate.
In context, the definite integral of a rate of change gives the total change. If R(t) is the rate of water flowing into a tank in litres per minute, then the integral of R from 0 to 10 is the number of litres that flow in during those 10 minutes. Include the units: the units of the integral are the units of y multiplied by the units of x. When asked for the area under a graph given by data, name the method (trapezoidal rule), show the formula with the numbers and state that the answer is an estimate.
Worked example
Values of y at x = 0, 2, 4 are 3, 5, 9. Estimate the area using the trapezoidal rule.
h = 2.
Area = (h / 2)[first + last + 2 x middle].
= 1 x [3 + 9 + 2 x 5].
Area = 22.
Practice problem and solution
Find the definite integral of 2x from x = 0 to x = 3. Enter the value.
The integral of 2x is x^2. Evaluate: 9 - 0 = 9.
Mental model: State the formula and the numbers.
Common trap: Forgetting that areas below the axis are negative.
15. Solving with technology: show the setup
Learning goal: Present calculator work so that method marks are awarded and answers can be checked.
In the technology papers, a correct answer without working earns less than a correct answer with a clear setup. Marks are given for the method, so write the equation or command you used and the inputs that you entered, then the output. For a probability, write the distribution and the probability statement, such as X ~ N(50, 4^2) and P(X < 54). For an equation, write the equation before using the solver. For a regression, state the type of model that you chose and the values of its parameters.
Use correct notation. Do not write calculator syntax, such as 'normalcdf', as the only communication; write the mathematical statement first. Give the final answer to three significant figures unless another form is requested, but keep more digits in intermediate steps. Use the units of the question. Where a sketch is requested, label the axes and the key points, such as intercepts, turning points and asymptotes, with their coordinates. A sketch need not be to scale but must show the features.
Graphing tools can solve equations by finding intersections, and find maxima, minima and zeros. When you use the graph to solve, write what you graphed, the window if it affects the answer and the coordinates of the point read. Check the window shows all relevant features, since a poor window can hide a second solution. When the question asks for all solutions in an interval, count the intersections in the interval and confirm there are no more by changing the window.
Finally, check the answer for reasonableness: a probability must be between 0 and 1, a length must be positive and the units must make sense. Compare with an estimate. If a solution contradicts the context, such as negative time, discard it and say why. If the calculator gives an error or unexpected output, check the mode, the brackets and the inputs. Do not round intermediate values, and avoid using stored rounded values from previous parts. A short concluding sentence in context shows understanding and satisfies the communication criterion.
Worked example
X ~ N(50, 4^2). Using technology, P(X < 54) is 0.841. Write a complete answer.
State the distribution: X ~ N(50, 4^2).
Write the statement: P(X < 54).
Quote the result: 0.841 to three significant figures.
Conclude: the probability that X is less than 54 is 0.841.
Practice problem and solution
A probability computed with technology is 0.3412. Round it to three significant figures.
The fourth significant figure is 2, so round down: 0.341.
Mental model: Show the setup and the output.
Common trap: Rounding intermediate values.
16. The exploration as a modelling cycle
Learning goal: Plan a mathematical exploration around a question you can model and finish.
The exploration is an internal assessment in IB Mathematics: applications and interpretation. According to the IB subject brief it is compulsory, worth 20 percent of the final grade at both standard and higher level, with a recommended 15 hours of class time. Students choose their own topic, and the course supports modelling, so it is natural to use the cycle of choosing a real situation, making assumptions, building a model, testing it against data and reflecting on its limits. The report should be clear enough for a peer to follow.
A good starting point is a question you care about and that you can answer with mathematics at your level, such as how the length of a sports season affects results or how fast a hot drink cools. Check that you can obtain data from a source you can cite, or collect it yourself. A question that is too wide, like 'the mathematics of music', leads to a list of facts; a focused question like 'how well does a sinusoidal model fit daily temperatures in my city' leads to analysis.
Build the model in steps. State your assumptions and why they are reasonable. Choose a mathematical tool that fits the data, such as regression, a sinusoidal model or a statistical test, and justify the choice. Use technology and show the setup. Present results clearly, with tables and graphs that are labelled and referenced in the text. Use accurate notation, and define variables. Show some mathematics that goes beyond routine work, and explain it in your own words.
Evaluation turns a calculation into an exploration. Compare the model's predictions with data or with reality, and report the size of the errors. Discuss limitations, such as the range of data, the assumptions you made and the sample size. Suggest what you would do next. Reflect on what the results mean in the context of your question. Keep the report to a manageable length, with a clear structure and a bibliography. Avoid padding, and do not copy material without acknowledging the source.
Worked example
A student wants to explore how quickly a drink cools. Suggest a focused question.
Choose a measurable quantity: temperature over time.
Choose a model: exponential decay toward room temperature.
State a question: how well does an exponential model fit the cooling of a cup of tea?
Plan to test it with measured data and report errors.
Practice problem and solution
What percentage of the final grade is the exploration worth according to the IB subject brief? Enter the number.
The IB subject brief lists the exploration at 20 percent.
Mental model: Choose a question you can finish.
Common trap: Copying a source with no analysis of your own.