Mathematics 8 · Handsworth 2026–27 · Unit 6

Data & Probability

Ms. Ras Mulinta
Handsworth Secondary
Guided notes

Data is how we make sense of the world — test scores, rainfall, how long the bus takes. Probability is how we reason about what might happen next — a coin, a dice game, the weather. In this unit you learn to describe a set of numbers with a single fair value (mean, median, mode), choose which one tells the truest story, and then work out the chance of two things happening together. Take your time. Each idea has a worked example, then a few problems to try. Answers are at the end of each section so you can check yourself right away.

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The toolkit words we will use

A measure of central tendency is one number that stands in for a whole set of data — the "typical" value. There are three. The mean is the average: add every value, then divide by how many there are. The median is the middle value once the data is in order. The mode is the value that appears most often. The range (how spread out the data is) is the largest value minus the smallest — useful, but not a centre.
In probability, an outcome is one thing that can happen; an event is the outcome (or outcomes) you care about. Theoretical probability is P = favourable outcomes / total outcomes. It is always between 0 (impossible) and 1 (certain), and you can write it as a fraction, a decimal, or a percent.

1 · Mean the average

To find the mean, add all the values, then divide by the number of values. The mean uses every number, so it gives a balanced picture — but a single very large or very small value (an outlier) can pull it off centre.
Worked example
Find the mean of 4, 7, 9, 6, 9.
Add: 4 + 7 + 9 + 6 + 9 = 35.
There are 5 values, so divide: 35 ÷ 5 = 7.
Mean = 7
Worked example — an outlier pulls the mean
Find the mean of 8, 9, 10, 9, 64.
Add: 8 + 9 + 10 + 9 + 64 = 100.   Divide by 5: 100 ÷ 5 = 20.
The mean is 20, yet four of the five values are near 9. The single big value (64) dragged the mean up, so the mean is not a fair "typical" here.
Mean = 20 (but watch the outlier)

Check yourself

  1. Find the mean of 5, 8, 11.
  2. Find the mean of 12, 15, 18, 15.
  3. Find the mean of 3, 3, 4, 10.
Answers
1. (5+8+11) ÷ 3 = 24 ÷ 3 = 8    2. (12+15+18+15) ÷ 4 = 60 ÷ 4 = 15    3. (3+3+4+10) ÷ 4 = 20 ÷ 4 = 5

2 · Median the middle value

To find the median, put the values in order first. If there is an odd number of values, the median is the one in the middle. If there is an even number, the median is the mean of the two middle values. The median ignores how extreme the outer values are, so it stays fair even when there is an outlier.
Worked example — odd number of values
Find the median of 12, 5, 9, 3, 8.
In order: 3, 5, 8, 9, 12. There are 5 values, so the middle (3rd) value is the median.
Median = 8
Worked example — even number of values
Find the median of 7, 2, 10, 4.
In order: 2, 4, 7, 10. There are 4 values, so the two middle ones are 4 and 7.
Median = mean of 4 and 7 = (4 + 7) ÷ 2 = 11 ÷ 2 = 5.5.
Median = 5.5

Check yourself

  1. Find the median of 6, 1, 9, 4, 7.
  2. Find the median of 20, 14, 11, 17.
  3. Find the median of 8, 8, 100 (notice the outlier).
Answers
4. In order 1, 4, 6, 7, 9 → middle is 6    5. In order 11, 14, 17, 20 → (14+17) ÷ 2 = 15.5    6. In order 8, 8, 100 → middle is 8 (the outlier barely matters)

3 · Mode the most frequent

The mode is the value that appears most often. A set can have one mode, more than one mode (if two values tie for most frequent), or no mode at all (if every value appears the same number of times). The mode is the only measure that works for data that is not numbers — like favourite colour or shoe size that comes in whole sizes.
Worked example
Find the mode of 3, 7, 7, 2, 7, 5.
Count how often each appears: 7 appears three times, more than any other value.
Mode = 7
Worked example — two modes
Find the mode of 4, 4, 6, 9, 9, 1.
Both 4 and 9 appear twice; nothing appears more often.
Modes = 4 and 9

Check yourself

  1. Find the mode of 2, 5, 5, 5, 8, 8.
  2. Find the mode of 10, 11, 12, 13.
  3. Shoe sizes sold: 7, 8, 8, 9, 8, 7. Which size is the mode?
Answers
7. 5 (appears three times)    8. No mode — every value appears once    9. Size 8 (appears three times)

4 · Choosing the right measure which one tells the truth?

All three are "averages," but they tell different stories. Pick the one that is fair for the data:
Mean — best when the data is fairly even with no extreme values. It uses every number. Median — best when there is an outlier or the data is skewed (like incomes or house prices), because it is not pulled by extremes. Mode — best for non-number data, or when you want the most common choice (most popular size, most frequent score).
Worked example — reasoning about the choice
Five part-time pay cheques: $120, $130, $125, $135, $900.
Mean = (120+130+125+135+900) ÷ 5 = 1410 ÷ 5 = $282.
Median: in order 120, 125, 130, 135, 900 → middle = $130.
The mean ($282) is higher than four of the five cheques because the $900 outlier pulls it up. The median ($130) better describes a typical cheque.
Median is the fair measure here.

Check yourself

  1. A class's quiz scores are all close together with no outliers. Which measure best describes a typical score?
  2. A store wants to know the most-requested T-shirt size. Which measure should it use?
  3. One house on a street sold for far more than the others. Mean or median for a "typical" sale price?
Answers
10. Mean — even data, no outliers    11. Mode — it wants the most common size    12. Median — the one high sale would pull the mean up unfairly

5 · Theoretical probability of one event favourable over total

Theoretical probability is what should happen by reasoning, before you try it. P(event) = favourable outcomes / total outcomes. Every probability is between 0 and 1: 0 means it cannot happen, 1 means it is certain. Simplify the fraction like any other.
Worked example — a die
Roll one fair six-sided die. Find P(rolling a 5).
Favourable outcomes: just one (the 5). Total outcomes: 6.
P(5) = 1⁄6
Worked example — more than one favourable outcome
Roll one fair die. Find P(an even number).
Favourable: 2, 4, 6 → three outcomes. Total: 6.
P = 3⁄6 = 1⁄2.
P(even) = 1⁄2

Check yourself

  1. One die: P(rolling a 3).
  2. One die: P(rolling a number greater than 4).
  3. A bag has 3 red and 5 blue marbles. P(drawing red)?
Answers
13. 1⁄6    14. greater than 4 means 5 or 6 → 2⁄6 = 1⁄3    15. 3 red of 8 total → 3⁄8

6 · Two independent events multiply the probabilities

Two events are independent when the first does not change the second — like flipping a coin and rolling a die, or spinning a spinner twice. For independent events, the probability that both happen is the product: P(A and B) = P(A) × P(B). This multiplication rule is the heart of the unit.
Worked example — coin and die
Flip a fair coin and roll a fair die. Find P(heads and a 4).
P(heads) = 1⁄2.   P(rolling a 4) = 1⁄6.
Multiply: 1⁄2 × 1⁄6 = 1⁄12.
P(heads and 4) = 1⁄12
Worked example — two spins
A spinner has 4 equal sections: red, blue, green, yellow. Spin it twice. Find P(green then green).
P(green) = 1⁄4 each spin, and the spins are independent.
Multiply: 1⁄4 × 1⁄4 = 1⁄16.
P(green, green) = 1⁄16

Check yourself

  1. Flip two coins. P(heads and heads)?
  2. Roll two dice. P(a 6 on the first and a 6 on the second)?
  3. Spinner with 4 equal sections. P(red then blue)?
Answers
16. 1⁄2 × 1⁄2 = 1⁄4    17. 1⁄6 × 1⁄6 = 1⁄36    18. 1⁄4 × 1⁄4 = 1⁄16

7 · Showing all the outcomes tables & tree diagrams

Multiplying gives the answer fast, but a table or a tree diagram shows why — it lists every possible combined outcome. A table works well for two events; a tree diagram works for two or more steps. Count the favourable rows/branches over the total to confirm the multiplication.
Outcome table for flipping a coin (H, T) and rolling a die (1–6). Each cell is one equally likely combined outcome — there are 2 × 6 = 12 in all.
 123456
HH1H2H3H4H5H6
TT1T2T3T4T5T6
Reading the table
P(heads and 4): the cell H4 is one outcome out of 12.
P(H4) = 1⁄12 — matches 1⁄2 × 1⁄6.
Tree diagram for flipping two coins. Each flip branches into H or T; follow a path to read one outcome.
Flip 1 Flip 2 Outcome ┌─ H ───── HH ┌─ H ───┤ │ └─ T ───── HT Start │ ┌─ H ───── TH └─ T ───┤ └─ T ───── TT
Reading the tree
There are 4 equally likely paths: HH, HT, TH, TT.
P(both heads) = the one HH path out of 4 = 1⁄4.
P(HH) = 1⁄4 — matches 1⁄2 × 1⁄2.

Check yourself

  1. Using the coin-and-die table, P(tails and an even number)?
  2. Using the two-coin tree, P(exactly one head)?
Answers
19. Tails with 2, 4, or 6 → T2, T4, T6 = 3 cells of 12 = 3⁄12 = 1⁄4    20. HT and TH = 2 paths of 4 = 2⁄4 = 1⁄2

If you want to go further

The multiplication rule and the table are really the same idea seen two ways. A table for two events has (rows) × (columns) cells — that product is the total number of outcomes, which is why dividing one favourable cell by the total gives the same fraction as multiplying the two separate probabilities. Try building a table for a coin and a four-section spinner (2 × 4 = 8 cells) and confirm one outcome is 1⁄8.

Mathematics 8 · Unit 6 — Data & Probability · Ms. Ras Mulinta · Handsworth Secondary 2026–27 · BC Math 8 content: measures of central tendency (mean, median, mode) and when each is appropriate; theoretical probability with two independent events (tables, tree diagrams, multiplication). Math 8 is assessed on the BC proficiency scale. Pairs with the Unit 6 practice set and answer key.