Probability, Mean, Median & Range

By the end of this chapter you will:

1. Compute mean, median, mode, range from a list of numbers
2. Run mean problems in both directions (forward and inverse, including missing-term)
3. Compute probabilities for equally-likely AND weighted outcomes
4. Use tree diagrams for multi-stage probability
5. Optimize under median + range constraints ("smallest possible …")

9.0 The BC gap: data summary statistics

⚠️ BC G5 and G6 contain NO instruction on mean, median, mode, or range. Probability is touched lightly. Yet Gauss 7 puts at least one mean/median/range question on every paper.

This chapter is therefore mostly new material for a BC student.

9.1 The four summary statistics

Given a list of numbers $a_1, a_2, \ldots, a_n$:

Even-count median

For $\{3, 7, 9, 12\}$ (even count), median = average of the two middle: $(7 + 9) / 2 = 8$.

🔑 Always sort first. Median and range are about the sorted order, not the order given.

9.2 Mean — forward direction

$$\text{mean} = \frac{\text{sum of all items}}{\text{number of items}}$$

Example

Five students scored $78, 82, 91, 88, 76$. Find the mean.

Sum $= 415$. Mean $= 415 / 5 = \mathbf{83}$.

9.3 Mean — inverse directions

The mean formula has three quantities (sum, count, mean). Give any two; find the third.

2024 Q13 — inverse mean (find count)

"Eloise purchased a number of water hand pumps at mean price $85, total spent $765. How many pumps?"

$$\text{count} = \frac{765}{85} = \mathbf{9}$$

Answer (C). ✅

Example — missing term

Five numbers have mean $20$. Four of them are $15, 18, 22, 25$. Find the fifth.

Sum needed = $5 \cdot 20 = 100$. Known sum $= 80$. Fifth $= 20$.

Example — adding one changes the mean

Six numbers have mean $50$. A seventh is added and the mean becomes $52$. What is the seventh number?

Old sum = $6 \cdot 50 = 300$. New sum = $7 \cdot 52 = 364$. Seventh = $364 - 300 = \mathbf{64}$.

9.4 Median + range optimization — 2024 Q19

"Five different integers in a list have a median of $10$ and a range of $7$. What is the smallest possible integer?"

Sort $a_1 < a_2 < a_3 < a_4 < a_5$ (strictly distinct).

• Median = $a_3 = 10$
• Range = $a_5 - a_1 = 7$, so $a_5 = a_1 + 7$
• Since $a_3 = 10 \le a_5$, we need $a_1 + 7 \ge 10$, so $a_1 \ge 3$

Try $a_1 = 3$: $a_5 = 10$, but $a_5 > a_4 > a_3 = 10$ is impossible. Fail.
Try $a_1 = 4$: $a_5 = 11$, $a_4$ strictly between $10$ and $11$ — no integer. Fail.
Try $a_1 = 5$: $a_5 = 12$, $a_4 \in \{11\}$, $a_2 \in \{6, 7, 8, 9\}$ — OK.

Smallest is 5. Answer (B). ✅

🔑 "Smallest possible" optimization recipe:

1. Sort and assign variables.
2. Push the target variable to its extreme bound from the constraints.
3. Check feasibility (especially distinctness + integer-ness).
4. Increment if infeasible.

9.5 Probability — equally likely outcomes

$$P(\text{event}) = \frac{\text{favorable}}{\text{total}} \quad \text{(equally likely)}$$

2024 Q10

"From $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, $P(\text{divisible by 2 or 3 or both})$ = ?"

Favorable: $\{2, 3, 4, 6, 8, 9\}$ — 6 values. Total: 9.

$$P = \mathbf{\tfrac{6}{9}}$$

Answer (C). ✅

9.6 Probability — weighted outcomes (2024 Q17)

When outcomes are NOT equally likely (spinners with unequal slices, biased coins, regions of different sizes), use weights (areas, sizes) instead of counts.

"A spinner has $12$ unshaded sections and $3$ shaded sections, each unshaded being $3$ times the size of each shaded. P(shaded)?"

Let one shaded slice be $1$ unit. Each unshaded is $3$ units.

• Shaded total weight: $3 \cdot 1 = 3$
• Unshaded total weight: $12 \cdot 3 = 36$
• Total: $39$

$$P(\text{shaded}) = \frac{3}{39} = \mathbf{\frac{1}{13}}$$

Answer (D). ✅

⚠️ Trap: do NOT compute $P = \dfrac{3 \text{ shaded slices}}{15 \text{ total slices}} = \tfrac{1}{5}$. The slices are different sizes.

9.7 Multi-stage probability — tree diagrams

For sequential events: multiply along a branch, add across branches.

Example — two draws without replacement

A bag has 3 red, 2 blue balls. Draw 2 without replacement. P(both red)?

$$P = \underbrace{\tfrac{3}{5}}_{\text{1st red}} \cdot \underbrace{\tfrac{2}{4}}_{\text{2nd red, given 1st red}} = \tfrac{3}{10}$$

Example — "at least one" via complement

Same bag. P(at least one red in two draws)?

Complement = "no red" = both blue.

$$P(\text{no red}) = \tfrac{2}{5} \cdot \tfrac{1}{4} = \tfrac{1}{10}$$

$$P(\text{at least one red}) = 1 - \tfrac{1}{10} = \tfrac{9}{10}$$

🔑 "At least one" is the canonical signal to use the complement.

9.8 Mode — when does it matter?

Mode shows up rarely on Gauss (BC ignores it too), but be ready:

• Unique mode: one value appears more often than any other
• Bimodal: two values tied for most frequent
• No mode: every value appears the same number of times

Gauss flavor: "$\{a, b, 5, 7, 5\}$ has a unique mode of $5$. What are the possible values of $a + b$?" — forces $a, b$ to each appear $\le 1$ time and not equal $5$.

9.9 Trap Alerts ⚠️

1. "Different integers" = strictly distinct. No ties allowed.
2. Median needs sorting first. Don't read the median off an unsorted list.
3. Even-count median is the AVERAGE of the two middles, not the lower or upper one.
4. Weighted ≠ equal slices. When sizes differ, weights replace counts.
5. Without vs with replacement changes the second draw's denominator.
6. The mode might not exist or might tie — don't assume uniqueness.

9.10 Mnemonic

"Sum-over-count for mean, sort-and-pick for median, frequency for mode, max-minus-min for range — and complement for 'at least'."

Practice Set (10 problems, mixed difficulty)

1. (Part A) Find the mean of $\{4, 7, 8, 10, 11\}$.
2. (Part A) Find the median of $\{12, 5, 9, 3, 7\}$.
3. (Part A) Find the range of $\{15, 22, 8, 30, 12, 18\}$.
4. (Part A) A coin is flipped twice. P(at least one heads)?
5. (Part B) Six numbers have mean $25$. After removing one, the mean of the remaining 5 is $24$. What was the removed number?
6. (Part B) A spinner has 4 equal red sections and 1 blue section. P(red on 2 spins)?
7. (Part B) Find the median of $\{2, 4, 4, 6, 8, 9\}$.
8. (Part C) Five different positive integers have mean $10$ and median $9$. What is the largest possible value?
9. (Part C) Three numbers have mode $12$ and mean $14$. The smallest is at least $10$. What are the three numbers if uniquely determined?
10. (Part C) A bag has 5 red and 3 blue balls. Two are drawn without replacement. P(same color)?

Answers: 1) 8; 2) sort → $\{3,5,7,9,12\}$, median 7; 3) $30 - 8 = 22$; 4) $1 - 1/4 = 3/4$; 5) sum was 150, after removal is 120 → removed = 30; 6) $(4/5)^2 = 16/25$; 7) (4+6)/2 = 5; 8) Sum $= 50$. $a_1 < a_2 < a_3 = 9 < a_4 < a_5$. Minimize $a_1 + a_2 + a_4$ to maximize $a_5$. Min: $a_1 = 1, a_2 = 2$ (distinct, < 9), $a_4 = 10$ (min > 9). Sum so far $= 1+2+9+10 = 22$, so $a_5 = 28$; 9) Mode $12$ unique → at least two $12$s. Three numbers $\{a, 12, 12\}$ with $a \ne 12$. Mean $14$ → sum $42$ → $a = 18$. But $a$ smallest $\ge 10$, and we'd need $a$ smallest so $a \le 12$ → $a \in [10, 12)$. But $a = 18 > 12$ contradicts smallest. So $\{12, 12, b\}$ with $b > 12$, smallest is $12 \ge 10$ ✓. $b = 18$. Numbers: $\{12, 12, 18\}$; 10) $P(\text{RR}) + P(\text{BB}) = \frac{5 \cdot 4}{8 \cdot 7} + \frac{3 \cdot 2}{8 \cdot 7} = \frac{20 + 6}{56} = \frac{26}{56} = \frac{13}{28}$.

End of chapter. Next: Test Strategy — 60-Minute Playbook.