Lesson 15: Probability and Statistics: Distributions and the Normal Curve

Data becomes information when it is summarised honestly and modelled well. This lesson covers three tools: conditional probability from two-way tables (what fraction of a subgroup has a property), the normal distribution with its empirical rule, and z-scores, which convert any measurement into "how many standard deviations from the mean" — the universal currency of statistics. Probability here builds on counting; Lesson 16 picks up the counting itself.

Two-way tables and conditional probability

A two-way table cross-tabulates two categorical variables. The key skill is reading conditional probabilities: P(A | B) = (count in both A and B)/(count in B) — the denominator shrinks to the given subgroup.

Worked Example 1 — language enrolments. A school surveys 200 students about language study:

Read the cells as counts: 30 students take both languages, 90 take only Spanish, 50 take only French, and 30 (bottom right of the inner block) take neither — 30 + 50 + 90 + 30 = 200 ✓. The margins give the single-language totals: Spanish 120, French 80, and 170 students take at least one language.

• P(takes French) = 80/200 = 0.40.
• P(takes French | takes a language) = 80/170 ≈ 0.47 — conditioning on the 170 language-takers raises the probability.
• P(takes both | takes French) = 30/80 = 0.375.
• Are "takes Spanish" and "takes French" independent? Independence would require P(both) = P(Spanish)·P(French) = 0.60 · 0.40 = 0.24, but the actual P(both) = 30/200 = 0.15. Not equal — not independent (in fact students here tend to pick one language or the other). Also note the contrast with disjoint: disjoint events have P(both) = 0; independent events have P(both) = product. Events cannot be both (unless one has probability 0).

The normal distribution and the empirical rule

A normal distribution is the symmetric bell curve fixed by its mean μ and standard deviation σ. For normally distributed data, the empirical rule says approximately:

• 68% of values lie within 1σ of the mean (μ − σ to μ + σ)
• 95% lie within 2σ
• 99.7% lie within 3σ

Halving the tails turns these into one-sided facts: 2.5% lie above μ + 2σ, 16% lie above μ + 1σ, and so on. The rule is a model: apply it only when the distribution is plausibly bell-shaped — incomes and reaction times are skewed, and forcing the rule onto them fabricates precision.

z-scores and percentiles

The z-score z = (x − μ)/σ standardises any value. Positive z means above the mean; z = 2 means two standard deviations up.

Worked Example 2 — test scores. Scores are normal with μ = 72, σ = 8. Amara scores 84: z = (84 − 72)/8 = 1.5. A z of 1.5 corresponds to about the 93rd percentile — she outperformed roughly 93% of students. (Percentile is not "percent correct": 93rd percentile is a rank, not a mark.) Her classmate scores 60: z = (60 − 72)/8 = −1.5, about the 7th percentile — the symmetry of the curve makes the two one lookup. What percentage score between 64 and 88? Those are μ ± 2σ... precisely, 64 = μ − σ and 88 = μ + 2σ, so the range spans z from −1 to 2: the empirical rule gives 68%/2 + 95%/2 = 34% + 47.5% = 81.5%.

Worked Example 3 — expected value. A raffle sells 1,000 tickets at $1; the single prize is $400. The expected value of a ticket is (400 · 1/1000) + (0 · 999/1000) − 1 = 0.40 − 1 = −$0.60: on average each ticket loses 60 cents. Expected value is the long-run average, not a prediction for one ticket — you will win $399 net or lose $1, never lose exactly 60 cents. It is nonetheless the right tool for comparing games, insurance and lotteries.

Modelling: quality control on the bottling line

A machine fills bottles with a mean of 500 mL and standard deviation 4 mL, normally distributed. Regulations flag bottles under 492 mL. z = (492 − 500)/4 = −2, and the empirical rule puts about 2.5% of bottles below μ − 2σ — roughly 25 bottles in every 1,000 flagged. The plant manager wants under 0.15% flagged (about 3σ territory): the fill mean would need to rise so that 492 = μ − 3σ, i.e. μ = 492 + 12 = 504 mL, costing 4 extra mL per bottle. Statistics becomes a business trade-off: the price of fewer rejects is giving away product. Every quality standard in manufacturing is negotiated exactly here, between the cost of the tail and the cost of the mean.

Common mistakes

• Using the whole table as the denominator for a conditional probability. P(French | takes a language) divides by 170, not 200. The phrase after "given" (or the subgroup described) owns the denominator.
• Confusing disjoint with independent. Disjoint events (no overlap) are strongly *dependent* — knowing one occurred tells you the other did not. Independent events overlap exactly as the product rule dictates.
• z = (μ − x)/σ. The order matters for the sign: above the mean must give positive z. A quick plausibility check ("this score is above average, so z > 0") catches the inversion.
• Applying the empirical rule to skewed data. Heights within one sex: fine. Household incomes: the rule would promise negative incomes within 2σ — the model's failure, announced by arithmetic.
• Reading a percentile as a percentage score. The 93rd percentile means 93% of the reference group scored at or below — it says nothing about how many questions were answered correctly.

Exam Experience

1. A score of 90 on a test with μ = 75 and σ = 5 has z-score

• (A) 15
• (B) 3
• (C) 0.3
• (D) −3

Answer: (B). z = (90 − 75)/5 = 15/5 = 3. (A) is the raw difference — standardising means dividing by σ; (D) reverses the subtraction.

2. For a normal distribution, the percentage of data within one standard deviation of the mean is approximately

• (A) 50%
• (B) 95%
• (C) 68%
• (D) 99.7%

Answer: (C). The empirical rule's first band: 68% within 1σ, 95% within 2σ (B), 99.7% within 3σ (D).

3. In a two-way table of 300 voters, 120 support a proposal; of the 180 voters under 40, 90 support it. P(support | under 40) is

• (A) 90/300 = 0.30
• (B) 90/180 = 0.50
• (C) 120/180 ≈ 0.67
• (D) 90/120 = 0.75

Answer: (B). Condition on the under-40 row: the denominator is 180, and the count who both support and are under 40 is 90. (A) is the unconditional joint probability's cousin — it ignores the condition; (D) conditions on the wrong variable.

Constructed response. Adult men's heights are approximately normal with μ = 68 in and σ = 3 in. (a) What percentage of men are between 65 in and 71 in tall? (b) What percentage are taller than 74 in? (c) Find the height marking the 97.5th percentile, explaining your use of the empirical rule. (d) Two men are chosen independently at random. Estimate the probability that both are taller than 74 in, and state the assumption you used.

Scoring guide. (a) 1 point: 65 = μ − σ and 71 = μ + σ — within 1σ, so ≈ 68%. (b) 1 point: 74 = μ + 2σ; the upper tail beyond 2σ is (100 − 95)/2 = 2.5%. (c) 2 points: 1 for recognising the 97.5th percentile leaves 2.5% above it, matching the μ + 2σ boundary; 1 for the height 68 + 2·3 = 74 in. (d) 2 points: 1 for independence: 0.025 · 0.025 = 0.000625 ≈ 0.06%; 1 for stating the assumption — the two choices are independent (random sampling, one man's height does not affect the other's), which is what licences multiplying.

Going deeper: computing σ, and a bridge to binomial probability

Standard deviation from raw data. The scores 4, 8, 6, 10, 7 have mean 7. Deviations: −3, 1, −1, 3, 0. Squared: 9, 1, 1, 9, 0 — sum 20. The (population) variance is 20/5 = 4, so σ = 2. Interpretation: a typical score sits about 2 points from the mean — and note the squaring makes σ punish outliers: replace the 10 by 16 and the mean becomes 8.2 while σ jumps to about 4.1; the mean moved 1.2, the spread more than doubled. (Sample standard deviation divides by n − 1 instead of n; know which your calculator's σ and s buttons report.)

From normal to binomial. Flip a fair coin 10 times: the number of heads is a binomial distribution — discrete, not normal, but bell-shaped already. P(exactly 6 heads) = C(10, 6)/2^10 = 210/1,024 ≈ 20.5% (combinations from Lesson 16 doing the counting). With 100 flips the binomial histogram hugs a normal curve with μ = 50 and σ = 5 so closely that z-scores estimate its probabilities well — the deep reason the normal distribution rules statistics: it is the limit shape of repeated random trials, not merely a pretty curve. The empirical rule you used in this lesson is that limit shape, put to work.

Association, read from the same table

Return to Lesson 15's language table. The unconditional probability of taking French is P(French) = 80/200 = 0.40. Condition on taking Spanish: P(French | Spanish) = 30/120 = 0.25. The conditional probability is lower than the unconditional one — taking Spanish makes French *less* likely — a negative association, which is exactly the dependence detected earlier by the product test (0.15 ≠ 0.24). The two diagnostics always agree: events are independent precisely when conditioning changes nothing, P(A | B) = P(A). In survey analysis this comparison is the first thing computed: if the conditional percentages differ across groups, there is a story in the data; if they match (within sampling noise), there is independence, and the story is that there is no story.