Lesson 5: The Normal Distribution

(a) Warm-Up

A college admissions officer is staring at two students' files. One scored 1280 on the SAT. The other scored 28 on the ACT. The SAT and ACT are completely different tests on completely different scales — the SAT runs to 1600, the ACT only to 36. So here's the question that decides who looks stronger: which score is more impressive?

You can't compare 1280 to 28 directly; that's comparing apples to elevators. But you can ask: how far above average is each student, measured in standard deviations? That single idea — counting how many standard deviations a value sits from the mean — is the engine behind nearly everything in this lesson.

Most of the variables you'll meet in AP Stats (heights, test scores, battery lifetimes, reaction times) pile up in a familiar bell shape: lots of values near the middle, fewer out at the extremes. When a variable follows that bell shape, one model unlocks every "what proportion…?" question you can throw at it. Let's build it.

(b) Core Concept

Density curves: area is everything

Before the bell curve, meet its parent idea. A density curve is a smooth curve drawn over a histogram that describes the overall pattern of a distribution. It has two non-negotiable properties:

1. The curve is always on or above the horizontal axis (it never dips negative).
2. The total area under the curve equals exactly 1.

That second rule is the whole point. Because the total area is 1, the area under the curve over any interval equals the proportion of values that fall in that interval. Area = proportion. If 23% of the area sits to the left of some value, then 23% of the data is below that value. A density curve is an idealized description of a distribution — it smooths over the bumps of a real histogram so we can do clean math.

The normal model: N(μ, σ)

The most important density curve in statistics is the normal curve — the symmetric, bell-shaped one. A normal distribution is completely described by just two numbers, so we write it as N(μ, σ):

• μ (mu) is the mean, which sits at the center and at the peak (and equals the median, by symmetry).
• σ (sigma) is the standard deviation, which controls the spread — the distance from the center to the inflection points where the curve switches from bending down to bending out.

[GRAPH: A symmetric bell-shaped normal curve over a horizontal axis. Center peak labeled μ. Vertical dashed lines at μ−σ, μ−2σ, μ−3σ, μ+σ, μ+2σ, μ+3σ. The region between μ−σ and μ+σ is shaded, labeled "68% of the data." Axis ticks read μ−3σ, μ−2σ, μ−σ, μ, μ+σ, μ+2σ, μ+3σ.]

A larger σ makes the curve shorter and wider; a smaller σ makes it taller and narrower. But every normal curve has the same shape, which is why one rule works for all of them.

The 68–95–99.7 (Empirical) Rule

For any normal distribution, the proportion of data within a set number of standard deviations of the mean is fixed:

• ≈ 68% of values fall within 1 standard deviation of the mean (between μ−σ and μ+σ)
• ≈ 95% of values fall within 2 standard deviations (between μ−2σ and μ+2σ)
• ≈ 99.7% of values fall within 3 standard deviations (between μ−3σ and μ+3σ)

Because the curve is symmetric, you can slice these further. Outside ±2σ lies 100% − 95% = 5%, split evenly into 2.5% in each tail.

Worked empirical-rule example. Resting heart rates for healthy adults are approximately N(72, 8) beats per minute. What proportion of adults have a resting heart rate between 56 and 88 bpm?

Notice 56 = 72 − 2(8) and 88 = 72 + 2(8). So 56 and 88 are exactly 2 standard deviations below and above the mean. By the 95 part of the rule, about 95% of healthy adults have a resting heart rate between 56 and 88 bpm.

Push it further: what proportion is above 88? Above +2σ is half of the leftover 5%, so about 2.5%.

The empirical rule is your fast, no-calculator sanity check. But it only works for "nice" values landing on whole standard-deviation marks. For everything else, we standardize.

Standardizing: the z-score

The z-score answers the Warm-Up question directly: how many standard deviations is a value from the mean?

z = (x − μ) / σ

A positive z-score means the value is above the mean; negative means below; z = 0 means exactly average. A z-score has no units — it's pure "standard deviations from the mean" — which is precisely what lets us compare an SAT score to an ACT score.

Back to our two students. SAT scores are about N(1050, 200); ACT scores are about N(21, 5.7).

SAT 1280:  z = (1280 − 1050) / 200 = 230 / 200  = 1.15
ACT 28:    z = (28 − 21) / 5.7     = 7 / 5.7    ≈ 1.23

The ACT student sits 1.23 standard deviations above average versus 1.15 for the SAT student. The ACT score is (slightly) more impressive. Same scale, fair fight.

From z-score to proportion: normalcdf

Once you can standardize, you can find the exact proportion below, above, or between any values. On the TI-84, the workhorse is:

normalcdf(lower, upper, μ, σ) — returns the proportion of a normal distribution between lower and upper.

The catch is the bounds for open-ended tails. "Below" has no real lower limit, and "above" has no real upper limit. We feed the calculator a number so extreme it acts like infinity:

• For a left tail ("less than x"): normalcdf(−1E99, x, μ, σ)
• For a right tail ("greater than x"): normalcdf(x, 1E99, μ, σ)

(On the TI-84, type 1E99 using the 2nd → , (EE) key, not the caret. −1E99 is negative one followed by EE 99.)

Worked z-score-to-proportion example. A brand of AA battery has lifetimes that are approximately N(500, 50) minutes. What proportion of batteries last longer than 540 minutes?

By hand (the z-score story):

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z = (540 − 500) / 50 = 40 / 50 = 0.80

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We want the area to the right of z = 0.80.

On the calculator:

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TI-84: normalcdf(540, 1E99, 500, 50)

Output: 0.2119

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About 21.2% of batteries last longer than 540 minutes. (You can also feed the standardized version, normalcdf(0.80, 1E99, 0, 1), and get the same 0.2119 — using μ and σ directly just skips a step.)

That's the full toolkit for proportions. Now we run it backward.

From percentile to value: invNorm

Sometimes you know the proportion and want the value. "What battery lifetime marks the top 5%?" "What SAT score is the 90th percentile?" These are inverse normal problems, and they use a different command:

invNorm(area, μ, σ) — returns the value with that much area (proportion) to its left.

The single most important rule: invNorm always works with the area to the LEFT of the value you want. If a problem gives you a right-tail proportion, subtract from 1 first.

Worked inverse example. For the SAT, N(1050, 200), what score is the 90th percentile?

The 90th percentile has 0.90 of the area to its left — already a left area, so no subtraction needed.

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TI-84: invNorm(0.90, 1050, 200)

Output: 1306.31

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A score of about 1306 marks the 90th percentile. A student at the 90th percentile scored higher than 90% of test-takers.

Is a normal model even reasonable?

A normal model is an assumption, not a law of nature — and using it on the wrong data gives confidently wrong answers. Before you trust normalcdf, ask: is the distribution roughly symmetric and bell-shaped, with no strong skew or major outliers? A histogram or dotplot that's heavily right-skewed (incomes, home prices) or clearly bimodal is not well described by a normal model, and the empirical rule won't hold for it. Many real variables — heights, standardized test scores, measurement errors — are close enough to normal that the model works beautifully. Just don't apply it blindly.

(c) Worked Examples

Example 1 (Empirical rule — easy)

Problem. The heights of adult women in the U.S. are approximately N(64, 2.7) inches. Between what two heights do the middle 95% of women fall?

Strategy. The middle 95% is captured by ±2σ (the 95 part of the empirical rule).

Solution.

μ − 2σ = 64 − 2(2.7) = 64 − 5.4 = 58.6
μ + 2σ = 64 + 2(2.7) = 64 + 5.4 = 69.4

Interpretation. About 95% of adult women are between 58.6 and 69.4 inches tall.

Example 2 (Proportion between two values — medium)

Problem. Using the same model, N(64, 2.7) for women's heights, what proportion of women are between 60 and 66 inches tall?

Strategy. These aren't whole-σ marks, so the empirical rule won't pin it down. Standardize both endpoints, then use normalcdf.

Solution.

z₆₀ = (60 − 64) / 2.7 = −4 / 2.7  ≈ −1.48
z₆₆ = (66 − 64) / 2.7 =  2 / 2.7  ≈  0.74

TI-84: normalcdf(60, 66, 64, 2.7)
Output: 0.7013

Interpretation. About 70.1% of adult women are between 60 and 66 inches tall. (Sanity check: the interval spans roughly from −1.5σ to +0.7σ, so a value around 0.70 is exactly what we'd expect.)

Example 3 (Inverse normal — "find the cutoff")

Problem. A manufacturer wants to advertise a minimum battery life that 95% of its batteries beat. Lifetimes are N(500, 50) minutes. What lifetime should appear on the package?

Strategy. "95% beat it" means 95% lie above the cutoff, so only 5% lie below it. invNorm needs the LEFT area, which is 0.05.

Solution.

TI-84: invNorm(0.05, 500, 50)
Output: 417.76

Interpretation. About 418 minutes. If they advertise 418 minutes, roughly 95% of batteries will exceed that claim. (Trap avoided: feeding 0.95 would give 582 minutes — the top cutoff, the opposite of what's wanted.)

Example 4 (Comparing z-scores across different distributions — AP-style)

Problem. Maria scored 1280 on the SAT, N(1050, 200). Her friend Devon scored 28 on the ACT, N(21, 5.7). Colleges treat both as equivalent admissions data. Whose performance is stronger, and approximately what percentile did each reach?

Strategy. Standardize each score to compare on a common scale, then convert each z-score to a percentile with normalcdf (left tail).

Solution.

Maria (SAT):  z = (1280 − 1050) / 200 = 1.15
Devon (ACT):  z = (28 − 21) / 5.7    ≈ 1.23

Percentiles (proportion scoring at or below):
  normalcdf(−1E99, 1.15, 0, 1) = 0.8749   → ~87th percentile
  normalcdf(−1E99, 1.23, 0, 1) = 0.8907   → ~89th percentile

Interpretation. Devon's z-score (1.23) is higher than Maria's (1.15), so Devon's performance is slightly stronger, landing around the 89th percentile versus Maria's 87th. The raw scores (28 vs. 1280) are uncomparable, but z-scores put them on one honest scale.

(d) Common Mistakes

1. Forgetting to standardize (or comparing raw values across distributions). Students compare 1280 to 28 directly, or plug a raw value into a z-table without converting. Fix: whenever distributions differ, convert to z-scores first — z = (x − μ) / σ — so everything is in the same "standard deviations from the mean" currency.

2. Swapping invNorm and normalcdf. Use normalcdf when you have a value and want a proportion; use invNorm when you have a proportion and want a value. Fix: ask "what does the question give me, and what does it want back?" Given x → want area? normalcdf. Given area → want x? invNorm.

3. Using the wrong tail bound. Writing normalcdf(540, 500, 50) (only three arguments, no real upper bound) or using 0 instead of 1E99. Fix: a right tail needs a huge upper bound: normalcdf(540, 1E99, 500, 50). A left tail needs a huge negative lower bound: normalcdf(−1E99, x, μ, σ).

4. Feeding invNorm a right-tail area. For "the top 10%," students type invNorm(0.10, …) and get the bottom cutoff. Fix: invNorm always uses area to the LEFT. Top 10% means 0.90 to the left, so use invNorm(0.90, …).

5. Mixing up "% below" and "% above." Reporting the left-tail area when the question asked "what proportion exceed…". Fix: the two always sum to 1. If normalcdf gives you the area below and the question wants above, subtract from 1 (or reverse the bounds).

(e) Practice Problems

Assume each variable is approximately normal. MC items have exactly 4 choices.

(Interpretation) The 25th percentile of IQ scores, N(100, 15), is about 89.9. Write a sentence interpreting this value in context.

(Interpretation) A student computes normalcdf(72, 1E99, 69, 2.8) = 0.142 for adult male heights. Explain in one sentence what 0.142 means here.