Chapter 3: The Normal Distribution

AP Statistics · MathHub · 2026

Learning Objectives

3.1 Normal Curves

The Normal distribution is one of the most important probability models in statistics. It describes many real-world phenomena: heights, test scores, measurement errors, and more.

Properties of Normal Curves

Interactive: Adjust $\mu$ and $\sigma$ to see how the Normal curve changes shape and location

Figure 3.1 — Normal Distribution $N(\mu, \sigma)$ — Shape and Location

3.2 The 68-95-99.7 Rule

For any Normal distribution $N(\mu, \sigma)$, the Empirical Rule gives the percentage of data within 1, 2, and 3 standard deviations of the mean. (Reviewed from Chapter 2, now applied to Normal models.)

68-95-99.7 Rule (Empirical Rule)

68% of values lie within $\mu \pm \sigma$
95% of values lie within $\mu \pm 2\sigma$
99.7% of values lie within $\mu \pm 3\sigma$

Equivalently: 16% are below $\mu - \sigma$ (and 16% above $\mu + \sigma$); 2.5% are below $\mu - 2\sigma$; 0.15% are below $\mu - 3\sigma$.

Example 3.1 — Applying the Empirical Rule

Women's heights are approximately $N(64.5, 2.5)$ inches (mean 64.5 in, SD 2.5 in).

(a) What percent of women are between 59.5 and 69.5 inches?

59.5 = 64.5 − 2(2.5) = $\mu - 2\sigma$ and 69.5 = $\mu + 2\sigma$ → 95%

(b) What percent are shorter than 57 inches?

57 = 64.5 − 3(2.5) = $\mu - 3\sigma$. Below $\mu - 3\sigma$: 0.3%/2 = 0.15%

(c) Between what heights are the middle 68% of women?

$\mu \pm \sigma = 64.5 \pm 2.5$ → between 62 and 67 inches

TRY IT

SAT Math scores are $N(530, 115)$. (a) What percent score between 185 and 875? (b) What percent score above 760? (c) Between what scores are the middle 95%?

Show Answer
(a) 185=530−3(115)=μ−3σ; 875=μ+3σ → 99.7%
(b) 760=530+2(115)=μ+2σ. Above μ+2σ: 5%/2=2.5%
(c) μ±2σ = 530±230 → between 300 and 760

3.3 The Standard Normal Distribution

The standard Normal distribution is the special case $N(0,1)$ — mean 0, standard deviation 1. We convert any Normal value to a z-score to use the standard Normal table.

Standardizing: The Z-Score Formula

$$z = \frac{x - \mu}{\sigma}$$

The z-score tells us how many standard deviations $x$ is from $\mu$. After standardizing, all Normal distributions become $N(0,1)$, so we can use one table.

Reading the Standard Normal Table (z-table)

The standard Normal table gives the proportion of values at or below a given z-score (the area to the left of $z$ under the $N(0,1)$ curve). This proportion equals the percentile of the value.

A partial z-table (positive z-scores):

z.00.01.02.03.04.05.06.07.08.09
0.0.5000.5040.5080.5120.5160.5199.5239.5279.5319.5359
0.5.6915.6950.6985.7019.7054.7088.7123.7157.7190.7224
1.0.8413.8438.8461.8485.8508.8531.8554.8577.8599.8621
1.5.9332.9345.9357.9370.9382.9394.9406.9418.9429.9441
2.0.9772.9778.9783.9788.9793.9798.9803.9808.9812.9817
2.5.9938.9940.9941.9943.9945.9946.9948.9949.9951.9952
3.0.9987.9987.9987.9988.9988.9989.9989.9989.9990.9990

AP Exam Tip: The table gives area to the left. To find area to the right, compute $1 - \text{table value}$. To find area between two z-scores, compute (larger area) − (smaller area). Always draw a sketch first.

3.4 Normal Probability Calculations

Finding a Proportion (Area) — Forward Problem

  1. State the problem: what value(s) of $x$ and what distribution $N(\mu, \sigma)$?
  2. Draw a Normal curve and shade the region of interest.
  3. Standardize: compute $z = (x - \mu)/\sigma$.
  4. Use the z-table (or calculator) to find the area.
  5. State the answer in context.

Example 3.2 — Finding a Proportion

IQ scores are $N(100, 15)$. What proportion of people have IQ above 118?

Step 1: $\mu = 100$, $\sigma = 15$, want $P(X > 118)$.

Step 2: (Sketch — shade right tail above 118.)

Step 3 — Standardize: $z = (118 - 100)/15 = 18/15 = 1.20$

Step 4 — Table: $P(Z \le 1.20) = 0.8849$

Step 5: $P(X > 118) = 1 - 0.8849 = \mathbf{0.1151}$

About 11.5% of people have IQ above 118.

Example 3.3 — Area Between Two Values

Adult male blood pressure is $N(125, 14)$ mmHg. What percent have pressure between 111 and 139?

Standardize both:
$z_1 = (111 - 125)/14 = -14/14 = -1.00$
$z_2 = (139 - 125)/14 = 14/14 = 1.00$

Area between: $P(-1 \le Z \le 1) = P(Z \le 1) - P(Z \le -1)$
$= 0.8413 - 0.1587 = \mathbf{0.6826}$

About 68.3% — confirming the Empirical Rule ($\mu \pm \sigma$).

Interactive: Find the shaded area — drag the left and right z-score bounds

Figure 3.2 — Normal Probability: Shaded Area Between Two Z-Scores

Finding a Value Given a Percentile — Inverse Problem

  1. State the percentile (area to the left).
  2. Look up the z-score in the body of the z-table (reverse lookup).
  3. Unstandardize: $x = \mu + z\sigma$.
  4. State the answer in context.

Example 3.4 — Inverse Normal (Percentile → Value)

AP Calculus scores are approximately $N(2.9, 1.1)$ (on a 1–5 scale). What score is at the 90th percentile?

Step 1: Need $x$ such that $P(X \le x) = 0.90$.

Step 2: Look up 0.90 in z-table → $z \approx 1.28$.

Step 3 — Unstandardize: $x = 2.9 + (1.28)(1.1) = 2.9 + 1.408 = \mathbf{4.31}$

A score of about 4.3 is at the 90th percentile for AP Calculus.

TRY IT

Heights of adult males are $N(70, 3)$ inches. (a) What percent are between 64 and 73 inches? (b) A man is at the 25th percentile — what is his height?

Show Answer
(a) z₁=(64−70)/3=−2.00; z₂=(73−70)/3=1.00
P(−2 ≤ Z ≤ 1) = 0.8413 − 0.0228 = 0.8185 (about 81.9%)

(b) 25th percentile: z ≈ −0.67 (from table: 0.2514)
x = 70 + (−0.67)(3) = 70 − 2.01 ≈ 68.0 inches

Interactive: Given a percentile (shaded area), find the corresponding z-score — adjust the area slider

Figure 3.3 — Inverse Normal: Finding a Value from a Percentile

3.5 Assessing Normality

We can assess whether a dataset is approximately Normal using:

AP Exam Tip: On the AP exam, you often need to check the Large Counts or Normality condition before performing inference. For sample means, the Central Limit Theorem (Ch 9) handles this. For now, state "approximately Normal" only when the data graph looks symmetric and bell-shaped, or when the problem states Normality.

Practice Problems

1

Scores on a standardized test are $N(500, 100)$. Using the Empirical Rule, find the percent of scores (a) between 300 and 700, (b) above 600, (c) below 400.

Show Solution
(a) 300=500−2(100)=μ−2σ; 700=μ+2σ → 95%
(b) 600=μ+σ. Above μ+σ: (100%−68%)/2 = 16%
(c) 400=μ−σ. Below μ−σ: 16%
2

Find the z-score for each value, given $N(85, 12)$: (a) $x = 97$ (b) $x = 73$ (c) $x = 85$

Show Solution
(a) z=(97−85)/12=1.00
(b) z=(73−85)/12=−12/12=−1.00
(c) z=(85−85)/12=0 (at the mean)
3

Newborn weights are $N(7.5, 1.1)$ lb. What percent of newborns weigh less than 6 pounds?

Show Solution
z=(6−7.5)/1.1=−1.5/1.1≈−1.36
Table: P(Z ≤ −1.36) ≈ 0.0869 (about 8.7%)
4

Lifetime of a light bulb is $N(1200, 80)$ hours. Find the probability a bulb lasts between 1100 and 1350 hours.

Show Solution
z₁=(1100−1200)/80=−1.25 → table: 0.1056
z₂=(1350−1200)/80=1.875 → table ≈ 0.9699
P(1100 ≤ X ≤ 1350) = 0.9699 − 0.1056 = 0.8643 (≈86.4%)
5

AP Biology scores are $N(2.8, 1.0)$. What score corresponds to the (a) 75th percentile? (b) 10th percentile?

Show Solution
(a) 75th pctile: z≈0.67
x=2.8+(0.67)(1.0)=3.47
(b) 10th pctile: z≈−1.28
x=2.8+(−1.28)(1.0)=1.52
6

Resting heart rate is $N(72, 10)$ bpm. (a) What percent have rate above 90? (b) Between what rates are the middle 90% of people?

Show Solution
(a) z=(90−72)/10=1.80 → P(Z > 1.80)=1−0.9641=0.0359 (3.6%)
(b) Middle 90% → 5% each tail → z=±1.645
72±(1.645)(10) → between 55.6 and 88.5 bpm
7

Two students compare scores. Student A got 82 on a test with $N(75, 8)$. Student B got 72 on a test with $N(65, 5)$. Who performed better relative to their class?

Show Solution
A: z=(82−75)/8=0.875
B: z=(72−65)/5=1.40
Student B's z-score is higher → Student B performed better relative to their class (at the 91.9th vs 80.9th percentile).
8

(AP Free Response Style) Annual rainfall in a city is approximately $N(32, 5)$ inches. (a) Find $P(25 \le X \le 40)$. (b) The city issues a drought warning when annual rainfall is below the 15th percentile. What is the drought warning threshold?

Show Solution
(a) z₁=(25−32)/5=−1.40 → 0.0808; z₂=(40−32)/5=1.60 → 0.9452
P = 0.9452−0.0808=0.8644 (≈86.4%)

(b) 15th pctile: z≈−1.04
x=32+(−1.04)(5)=32−5.2=26.8 inches

📋 Chapter Summary

Normal Distribution Properties

Normal Distribution

Bell-shaped, symmetric, described by mean $\mu$ and standard deviation $\sigma$. Notation: $N(\mu, \sigma)$. The curve extends infinitely in both directions.

68-95-99.7 Rule

In any Normal distribution: ~68% within $\pm 1\sigma$, ~95% within $\pm 2\sigma$, ~99.7% within $\pm 3\sigma$ of the mean.

z-score

$z = \dfrac{x - \mu}{\sigma}$ — number of standard deviations from the mean. Standardizes values to $N(0,1)$.

Standard Normal Table

Table of $P(Z \leq z)$ for $Z \sim N(0,1)$. Use to find areas (probabilities) by looking up the z-score.

Using the Normal Table

  1. Convert to z-score: $z = (x-\mu)/\sigma$
  2. Look up $P(Z \leq z)$ in the standard Normal table
  3. For $P(Z \geq z)$: compute $1 - P(Z \leq z)$
  4. For $P(a \leq Z \leq b)$: compute $P(Z \leq b) - P(Z \leq a)$
  5. Working backwards: given probability, find $z$, then $x = \mu + z\sigma$

📘 Key Terms

Normal DistributionA symmetric, bell-shaped distribution completely described by its mean $\mu$ and standard deviation $\sigma$.
z-score$(x-\mu)/\sigma$ — standardized score telling how many SDs a value is above or below the mean.
Standard NormalThe Normal distribution with $\mu = 0$ and $\sigma = 1$, written $N(0,1)$ or $Z$.
68-95-99.7 RuleEmpirical rule: 68%, 95%, 99.7% of Normal data fall within 1, 2, 3 standard deviations of the mean.
PercentileThe value below which a given percent of observations fall. The 90th percentile has 90% of data below it.
Normality AssessmentCheck using a histogram (bell-shaped?), a Normal probability plot (linear?), or the 68-95-99.7 rule.
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