Grade 12 Data Management (MDM4U)
MDM4U develops the mathematical tools needed to collect, analyze, and draw conclusions from data. The course sits at the intersection of mathematics, science, social science, and business. Its two major strands are counting and probability — which underpin everything from genetics to financial risk modelling — and statistical analysis, which provides the framework for scientific inference. The course culminates in a significant independent data management project that puts all of these skills to use on a question of your choice.
1. Counting: Permutations and Combinations
The Fundamental Counting Principle
Multiplication Principle
If event $A$ can occur in $m$ ways and, for each of these, event $B$ can occur in $n$ ways, then the total number of ways $A$ followed by $B$ can occur is $m \times n$.
This extends to any number of events: multiply the number of choices at each stage.
Factorial Notation
$n!$ (read "$n$ factorial") is the product of all positive integers from 1 to $n$: $n! = n \times (n-1) \times \cdots \times 2 \times 1$. By convention, $0! = 1$.
Permutations
Permutations
A permutation is an ordered arrangement of objects. The number of ways to arrange $r$ objects chosen from $n$ distinct objects is: $$_nP_r = P(n, r) = \frac{n!}{(n-r)!}$$
The number of arrangements of all $n$ objects is $n!$.
Permutations with repetition: if a set of $n$ objects has $a$ identical objects of one type, $b$ of another, etc., the number of distinct arrangements is $\dfrac{n!}{a!\, b!\, \cdots}$
Worked Example 1.1 — Permutations
In how many ways can a president, vice-president, and secretary be chosen from a club of 10 members?
Order matters (different positions). $_{{10}}P_3 = \dfrac{10!}{7!} = 10 \times 9 \times 8 = 720$ ways.
Worked Example 1.2 — Permutations with Repetition
How many distinct arrangements are there of the letters in the word MISSISSIPPI?
Total 11 letters: M appears once, I appears 4 times, S appears 4 times, P appears 2 times.
$$\frac{11!}{1!\, 4!\, 4!\, 2!} = \frac{39916800}{1 \times 24 \times 24 \times 2} = \frac{39916800}{1152} = 34650$$
Combinations
Combinations
A combination is a selection of objects where order does not matter. The number of ways to choose $r$ objects from $n$ distinct objects is: $$\binom{n}{r} = _nC_r = C(n,r) = \frac{n!}{r!(n-r)!}$$
Key identity: $\binom{n}{r} = \binom{n}{n-r}$ (choosing $r$ to include is the same as choosing $n-r$ to exclude).
Worked Example 1.3 — Combinations
A committee of 4 is to be selected from 6 women and 5 men, with exactly 2 women and 2 men. How many committees are possible?
Choose 2 from 6 women: $\binom{6}{2} = 15$. Choose 2 from 5 men: $\binom{5}{2} = 10$.
By the multiplication principle: $15 \times 10 = 150$ committees.
Pascal's Triangle and the Binomial Theorem
The entries of Pascal's triangle are the binomial coefficients $\binom{n}{r}$. The Binomial Theorem states: $$(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$$
2. Probability
Probability Rules
- $P(A) = \dfrac{n(A)}{n(S)}$ where $n(S)$ is the total number of equally likely outcomes
- $P(A') = 1 - P(A)$ (complement rule)
- $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ (addition rule)
- $P(A \cap B) = P(A) \times P(B|A)$ (multiplication rule)
- $P(B|A) = \dfrac{P(A \cap B)}{P(A)}$ (conditional probability)
- Events $A$ and $B$ are independent if $P(A \cap B) = P(A) \times P(B)$.
- Events $A$ and $B$ are mutually exclusive if $P(A \cap B) = 0$, so $P(A \cup B) = P(A) + P(B)$.
Worked Example 2.1 — Conditional Probability
In a survey of 200 students: 120 play sports, 80 play an instrument, and 40 do both. A student is chosen at random. Given that the student plays an instrument, what is the probability they also play sports?
$$P(\text{sports} | \text{instrument}) = \frac{P(\text{sports and instrument})}{P(\text{instrument})} = \frac{40/200}{80/200} = \frac{40}{80} = 0.5 = 50\%$$
Worked Example 2.2 — Using Combinations in Probability
A hand of 5 cards is dealt from a standard 52-card deck. Find the probability of getting exactly 3 aces.
Favourable: $\binom{4}{3} \times \binom{48}{2} = 4 \times 1128 = 4512$.
Total 5-card hands: $\binom{52}{5} = 2598960$.
$$P(\text{exactly 3 aces}) = \frac{4512}{2598960} \approx 0.001736 \approx 0.174\%$$
3. Probability Distributions
Discrete Probability Distributions
Discrete Random Variable
A discrete random variable $X$ takes on a countable number of distinct values. Its probability distribution assigns a probability $P(X = x)$ to each possible value, where $\displaystyle\sum_{\text{all }x} P(X = x) = 1$.
Expected value: $E(X) = \mu = \displaystyle\sum x \cdot P(X = x)$
Variance: $\text{Var}(X) = \sigma^2 = \displaystyle\sum (x - \mu)^2 P(X = x)$; Standard deviation: $\sigma = \sqrt{\text{Var}(X)}$
Binomial Distribution
Binomial Distribution
If an experiment consists of $n$ independent trials each with probability $p$ of success, the probability of exactly $x$ successes is: $$P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}, \quad x = 0, 1, 2, \ldots, n$$
Mean: $\mu = np$ Variance: $\sigma^2 = np(1-p)$ Std dev: $\sigma = \sqrt{np(1-p)}$
Worked Example 3.1 — Binomial Distribution
A multiple choice test has 10 questions, each with 4 options. If a student guesses randomly on all questions, find: (a) the probability of getting exactly 4 correct, and (b) the mean and standard deviation of the number correct.
Here $n = 10$, $p = 0.25$, $q = 0.75$.
(a) $P(X = 4) = \binom{10}{4}(0.25)^4(0.75)^6 = 210 \times 0.003906 \times 0.1780 \approx 0.1460$ or about 14.6%.
(b) $\mu = np = 10 \times 0.25 = 2.5$. $\sigma = \sqrt{10 \times 0.25 \times 0.75} = \sqrt{1.875} \approx 1.37$.
Hypergeometric Distribution
Hypergeometric Distribution
When sampling without replacement from a population of $N$ items containing $K$ successes, the probability of exactly $x$ successes in a sample of size $n$ is: $$P(X = x) = \frac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}}$$
Mean: $\mu = \dfrac{nK}{N}$
4. Organisation of Data and Sampling
Types of Data and Variables
Data are either qualitative (categorical) or quantitative (numerical). Quantitative data can be discrete (countable, e.g., number of siblings) or continuous (measurable, e.g., height).
Sampling Methods
| Method | Description | Strength |
|---|---|---|
| Simple random sampling | Every sample of size $n$ has equal probability of being chosen | No bias in selection |
| Stratified random | Population divided into strata; random sample taken from each | Ensures representation of subgroups |
| Systematic sampling | Every $k$th member selected after a random start | Easy to implement |
| Cluster sampling | Population divided into clusters; some clusters selected in full | Cost-efficient for large populations |
| Convenience sampling | Easily accessible individuals selected | Cheap and fast |
Types of Bias
Valid statistical inference requires identifying and minimizing bias:
- Sampling bias: sample does not represent the population (e.g., voluntary response samples, undercoverage)
- Response bias: systematic tendency to give incorrect responses (e.g., social desirability bias)
- Non-response bias: those who do not respond differ from those who do
- Leading questions: question wording that suggests a preferred answer
5. Statistical Analysis: Normal Distribution
The Normal Distribution
A continuous random variable $X$ follows a normal distribution with mean $\mu$ and standard deviation $\sigma$, written $X \sim N(\mu, \sigma^2)$, if its probability density function is: $$f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$
The curve is bell-shaped and symmetric about $\mu$. All probabilities correspond to areas under this curve.
Standardization and the $z$-Score
$z$-Score
To find probabilities, convert to the standard normal distribution $Z \sim N(0, 1)$: $$z = \frac{x - \mu}{\sigma}$$
The $z$-score tells you how many standard deviations $x$ is above (positive) or below (negative) the mean.
The 68-95-99.7 Rule (Empirical Rule)
- Approximately 68% of data falls within $1\sigma$ of the mean: $P(\mu - \sigma < X < \mu + \sigma) \approx 0.6827$
- Approximately 95% within $2\sigma$: $P(\mu - 2\sigma < X < \mu + 2\sigma) \approx 0.9545$
- Approximately 99.7% within $3\sigma$: $P(\mu - 3\sigma < X < \mu + 3\sigma) \approx 0.9973$
Worked Example 5.1 — Normal Distribution Probability
The heights of adult males in a city are approximately normally distributed with $\mu = 175$ cm and $\sigma = 8$ cm. Find the probability that a randomly selected male is between 163 cm and 191 cm tall.
Step 1 Convert to $z$-scores: $$z_1 = \frac{163 - 175}{8} = \frac{-12}{8} = -1.5 \qquad z_2 = \frac{191 - 175}{8} = \frac{16}{8} = 2.0$$
Step 2 Using standard normal tables (or technology): $P(Z < 2.0) = 0.9772$ and $P(Z < -1.5) = 0.0668$.
Step 3 $P(-1.5 < Z < 2.0) = 0.9772 - 0.0668 = 0.9104$.
About 91% of adult males in this city are between 163 cm and 191 cm tall.
Worked Example 5.2 — Finding a Value from a Percentile
Using the same height distribution ($\mu = 175$, $\sigma = 8$), find the height that corresponds to the 90th percentile.
Step 1 Find the $z$-score such that $P(Z < z) = 0.90$. From tables: $z \approx 1.28$.
Step 2 Solve for $x$: $x = \mu + z\sigma = 175 + 1.28(8) = 175 + 10.24 = 185.24$ cm.
The 90th percentile is approximately 185.2 cm.
6. Confidence Intervals and Hypothesis Testing
Confidence Intervals for a Population Proportion
Confidence Interval for a Proportion
If $\hat{p}$ is the sample proportion from a large random sample of size $n$, then a $C$% confidence interval for the true population proportion $p$ is: $$\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$
where $z^*$ is the critical value from the standard normal distribution corresponding to confidence level $C$:
- 90% confidence: $z^* = 1.645$
- 95% confidence: $z^* = 1.960$
- 99% confidence: $z^* = 2.576$
The margin of error is $E = z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$.
Worked Example 6.1 — Confidence Interval
In a random survey of 400 Ontario students, 248 said they prefer streaming music over radio. Construct a 95% confidence interval for the true proportion of Ontario students who prefer streaming.
Step 1 $\hat{p} = \dfrac{248}{400} = 0.62$. $z^* = 1.96$.
Step 2 Margin of error: $E = 1.96\sqrt{\dfrac{0.62 \times 0.38}{400}} = 1.96\sqrt{\dfrac{0.2356}{400}} = 1.96 \times 0.02426 \approx 0.0475$.
Step 3 CI: $(0.62 - 0.0475,\; 0.62 + 0.0475) = (0.5725,\; 0.6675)$.
We are 95% confident the true proportion of Ontario students who prefer streaming is between 57.3% and 66.8%.
Introduction to Hypothesis Testing
Structure of a Hypothesis Test
- State hypotheses: the null hypothesis $H_0$ (status quo, no effect) and the alternative $H_a$ (what you're trying to show).
- Choose significance level $\alpha$ (commonly 0.05 or 0.01). This is the maximum acceptable probability of a Type I error (rejecting a true $H_0$).
- Compute test statistic. For a proportion: $z = \dfrac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}}$
- Find the $p$-value (probability of obtaining a result at least as extreme as observed, assuming $H_0$ is true).
- Decision: if $p\text{-value} \leq \alpha$, reject $H_0$. Otherwise, fail to reject $H_0$.
Worked Example 6.2 — Hypothesis Test for a Proportion
A coin manufacturer claims that a coin is fair ($p = 0.5$). You flip it 200 times and get 115 heads. At the $\alpha = 0.05$ level, is there evidence the coin is biased?
Hypotheses $H_0: p = 0.5$ vs. $H_a: p \neq 0.5$ (two-tailed test).
Test statistic $\hat{p} = \dfrac{115}{200} = 0.575$. $$z = \frac{0.575 - 0.5}{\sqrt{0.5 \times 0.5 / 200}} = \frac{0.075}{0.03536} \approx 2.12$$
$p$-value For $z = 2.12$ in a two-tailed test: $p\text{-value} = 2 \times P(Z > 2.12) = 2 \times 0.0170 = 0.034$.
Decision Since $0.034 < 0.05$, we reject $H_0$. There is statistically significant evidence at the 5% level that the coin is biased.
7. Culminating Data Management Project Guidance
About the Culminating Project
The MDM4U culminating project (often worth 20–30% of the final mark) requires you to independently design and execute a complete statistical investigation. It typically spans several weeks and integrates every strand of the course. Here is a framework for success.
Project Stages
- Choose a research question. It must be measurable and interesting. Good questions are specific: not "What do students like?" but "Is there a statistically significant relationship between the number of hours of sleep and academic performance among Grade 12 students at my school?"
- Identify your population and sampling method. Justify your choice. If you use a survey, explain how you will minimize bias. Calculate the required sample size to achieve a desired margin of error: $n \geq \left(\dfrac{z^* \cdot 0.5}{E}\right)^2$ (using $\hat{p} = 0.5$ as a conservative estimate when $p$ is unknown).
- Collect your data. Use either primary data (your own survey, experiment, or observation) or a reputable secondary source. Document your collection method, date, and any limitations.
- Organise and display your data. Use appropriate representations: frequency tables, histograms, box plots, scatter plots, or stem-and-leaf plots. The choice depends on your data type and research question.
- Calculate descriptive statistics. Include measures of central tendency (mean, median, mode) and dispersion (range, standard deviation, interquartile range). Identify outliers using the $1.5 \times IQR$ rule.
- Perform inferential statistics. Depending on your question: construct a confidence interval, perform a hypothesis test, compute a correlation coefficient, or use the binomial distribution to model a count variable.
- Interpret results and write conclusions. Connect your statistical findings back to your research question. Acknowledge sources of error and bias. Suggest how the study could be improved or extended.
Sample Research Question and Analysis Plan
Research question: Among Ontario Grade 12 students, is the proportion who plan to attend university significantly higher than 60%?
Hypotheses: $H_0: p = 0.60$ vs. $H_a: p > 0.60$ (one-tailed test).
Sample: Stratified random sample of $n = 300$ students across three schools (stratify by school to ensure representation). Minimum sample size for $E = 0.05$ at 95%: $n \geq \left(\dfrac{1.96 \times 0.5}{0.05}\right)^2 = (19.6)^2 = 384.16$, so 385 students is the minimum.
Data collection: anonymous online survey to reduce social desirability bias.
Analysis: compute $\hat{p}$, the test statistic $z$, the $p$-value, and a 95% confidence interval for $p$.
8. Practice Problems
Problem 1 — Permutations
In how many ways can 8 athletes finish in 1st, 2nd, and 3rd place in a race (assuming no ties)?
Show Solution
$_8P_3 = \dfrac{8!}{5!} = 8 \times 7 \times 6 = 336$ ways.
Problem 2 — Combinations
How many different 5-card hands can be dealt from a standard 52-card deck?
Show Solution
$\binom{52}{5} = \dfrac{52!}{5!\,47!} = \dfrac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} = 2{,}598{,}960$ hands.
Problem 3 — Probability
Two events $A$ and $B$ satisfy $P(A) = 0.4$, $P(B) = 0.5$, and $P(A \cup B) = 0.7$. Find $P(A | B)$.
Show Solution
$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.4 + 0.5 - 0.7 = 0.2$.
$P(A | B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.2}{0.5} = 0.4$.
Since $P(A | B) = P(A) = 0.4$, events $A$ and $B$ are independent.
Problem 4 — Binomial Distribution
A biased coin has $P(\text{heads}) = 0.6$. It is flipped 8 times. Find the probability of getting at most 2 heads.
Show Solution
$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.
$P(X=0) = (0.4)^8 \approx 0.000655$
$P(X=1) = 8(0.6)(0.4)^7 \approx 0.007864$
$P(X=2) = \binom{8}{2}(0.6)^2(0.4)^6 = 28 \times 0.36 \times 0.004096 \approx 0.04129$
$P(X \leq 2) \approx 0.000655 + 0.007864 + 0.04129 \approx 0.0498$ or about 5%.
Problem 5 — Normal Distribution
Scores on a standardized test follow a normal distribution with $\mu = 72$ and $\sigma = 9$. What score is needed to be in the top 5%?
Show Solution
Top 5% means $P(X > x) = 0.05$, so $P(X < x) = 0.95$. From tables, $z \approx 1.645$.
$x = 72 + 1.645 \times 9 = 72 + 14.8 = 86.8$. A score of at least 87 is needed.
Problem 6 — Confidence Interval
In a sample of 500 Canadians, 320 said they support a carbon tax. Construct a 99% confidence interval for the true proportion of Canadians who support a carbon tax.
Show Solution
$\hat{p} = \dfrac{320}{500} = 0.64$. $z^* = 2.576$.
$E = 2.576\sqrt{\dfrac{0.64 \times 0.36}{500}} = 2.576 \times 0.02147 \approx 0.0553$.
CI: $(0.64 - 0.0553,\; 0.64 + 0.0553) = (0.585,\; 0.695)$.
We are 99% confident the true proportion is between 58.5% and 69.5%.
Problem 7 — Hypothesis Test
A fast food chain claims that 70% of its customers are satisfied. A consumer group surveys 150 customers and finds that 96 are satisfied. At $\alpha = 0.05$, test whether the true proportion is less than 70%.
Show Solution
$H_0: p = 0.70$ vs. $H_a: p < 0.70$ (left-tailed).
$\hat{p} = \dfrac{96}{150} = 0.64$. $z = \dfrac{0.64 - 0.70}{\sqrt{0.70 \times 0.30 / 150}} = \dfrac{-0.06}{0.03742} \approx -1.604$.
$p\text{-value} = P(Z < -1.604) \approx 0.054 > 0.05$. Fail to reject $H_0$. There is insufficient evidence to conclude that fewer than 70% of customers are satisfied.
Problem 8 — Permutations with Restriction
In how many ways can the letters of the word EQUATION be arranged if the vowels must always be kept together?
Show Solution
EQUATION has 5 vowels (E, U, A, I, O) and 3 consonants (Q, T, N). Treating the vowel block as one unit: 4 units (vowel block + Q + T + N) can be arranged in $4! = 24$ ways. The 5 vowels within the block can be arranged in $5! = 120$ ways.
Total: $24 \times 120 = 2880$ arrangements.