AP · The Normal Distribution · 14 min read · Updated 2026-05-10

The Normal Distribution — AP Statistics Study Guide

For: AP Statistics candidates sitting AP Statistics.

Covers: Definition of normal distributions, the empirical (68-95-99.7) rule, z-scores and the standard normal distribution, normal probability calculation, inverse normal calculations, and methods for assessing normality of one-variable data.

You should already know: How to calculate percentiles for one-variable data, how to calculate mean and standard deviation for quantitative data, properties of symmetric unimodal distributions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Statistics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is The Normal Distribution?

The normal distribution (also called the Gaussian distribution or bell curve) is the most widely used continuous probability distribution for quantitative one-variable data in statistics. It is defined entirely by two parameters: its mean $μ$ (center) and standard deviation $σ$ (spread), with standard notation $N (μ, σ)$ . By definition, normal distributions are symmetric, unimodal, bell-shaped, and never touch the horizontal axis, with a total area of 1 under the entire curve. Probability for a normal distribution corresponds to the area under the curve over an interval of values, since it is continuous, so the probability of any single exact value is 0.

In the AP Statistics CED, this topic falls within Exploring One-Variable Data (Unit 1), which accounts for 15-20% of the total AP exam score, with the normal distribution making up 4-6% of total exam points. It appears on both multiple choice (MCQ) as standalone questions and in free response (FRQ) as part of multi-step problems, and it is a foundational concept for nearly all units later in the course. Many naturally occurring quantitative variables (heights, test scores, measurement error) follow an approximately normal distribution, making it extremely useful for real-world data analysis.

2. The Empirical (68-95-99.7) Rule

The empirical rule is a quick approximation for the proportion of observations falling within 1, 2, or 3 standard deviations of the mean for any normal distribution. It only applies to approximately normal distributions, and it is the fastest way to solve problems where the bounds of the interval are exactly 1, 2, or 3 standard deviations from the mean. The rule states:

Approximately 68% of all observations fall within $μ \pm 1 σ$ (1 standard deviation of the mean)
Approximately 95% of all observations fall within $μ \pm 2 σ$ (2 standard deviations of the mean)
Approximately 99.7% of all observations fall within $μ \pm 3 σ$ (3 standard deviations of the mean)

Because the normal distribution is symmetric, any proportion of observations outside an interval centered at the mean is split evenly between the lower tail and upper tail of the distribution. This symmetry lets you quickly calculate proportions for one-tailed problems (e.g., the percentage of observations above $μ + 1 σ$ ) without needing a z-table or calculator.

Worked Example

The distribution of masses of adult male wild turkeys is approximately normal with mean $μ = 7.5$ kg and standard deviation $σ = 1.2$ kg. What percent of adult male wild turkeys weigh less than 5.1 kg?

Calculate how many standard deviations 5.1 kg is from the mean: $5.1 = 7.5 - (2 \times 1.2) = μ - 2 σ$ . So 5.1 kg is 2 standard deviations below the mean.
By the empirical rule, 95% of observations fall between $μ - 2 σ$ and $μ + 2 σ$ , so $100% - 95% = 5%$ of observations fall outside this interval.
By symmetry, half of the 5% is in the lower tail (below $μ - 2 σ$ ) and half is in the upper tail (above $μ + 2 σ$ ).
The percent of turkeys weighing less than 5.1 kg is $\frac{5%}{2} = 2.5%$ .

Exam tip: On AP MCQ, if the interval bounds are exactly 1, 2, or 3 standard deviations from the mean, the empirical rule is always faster than pulling out a z-table or running a calculator function, and it will give you the exact answer the question expects.

3. Z-Scores and the Standard Normal Distribution

The standard normal distribution is a specific normal distribution with mean $μ = 0$ and standard deviation $σ = 1$ , written as $N (0, 1)$ . To use a single set of probability values (or calculator function) for any normal distribution, we standardize any observation $x$ from $N (μ, σ)$ to a z-score, which measures how many standard deviations $x$ is from the mean, and in which direction.

The formula for a z-score for an individual observation is: $z = \frac{x - μ}{σ}$

A positive z-score means $x$ is above the mean, a negative z-score means $x$ is below the mean, and a z-score of 0 means $x$ equals the mean. Standardization lets you compare observations from different normal distributions with different units and scales (e.g., comparing an SAT score to an ACT score). On the AP exam, you will almost always be required to both calculate and interpret z-scores on FRQ, so context for your answer is critical.

Worked Example

The distribution of ACT composite scores is approximately normal with mean 21 and standard deviation 5. The distribution of SAT total scores is approximately normal with mean 1050 and standard deviation 150. A student scores 28 on the ACT and 1320 on the SAT. Which score is higher relative to their respective distributions? Interpret the z-score for the higher score.

Calculate the z-score for the ACT: $z_{A C T} = \frac{28 - 21}{5} = 1.4$ .
Calculate the z-score for the SAT: $z_{S A T} = \frac{1320 - 1050}{150} = \frac{270}{150} = 1.8$ .
Compare z-scores: 1.8 > 1.4, so the SAT score is higher relative to its distribution.
Interpretation: This student's SAT total score is 1.8 standard deviations above the mean SAT score for all test-takers.

Exam tip: Always interpret z-scores with three key components: the observation, the number of standard deviations from the mean, and the direction (above/below) to earn full credit on FRQ; omitting any of these will cost you a point.

4. Normal Probabilities and Inverse Normal Calculations

For any normal distribution, we can calculate the proportion of observations falling in any interval, or find the observation corresponding to a given percentile (called an inverse normal calculation). For probability, we find the cumulative area under the normal curve: the area to the left of a given $x$ is the proportion of observations less than $x$ . For an interval between $x_{1}$ and $x_{2}$ (where $x_{1} < x_{2}$ ), the proportion is the area to the left of $x_{2}$ minus the area to the left of $x_{1}$ . For the proportion greater than $x$ , subtract the area to the left of $x$ from 1.

On the AP exam, you are allowed to use your calculator's built-in normalcdf (for probability) and invNorm (for inverse normal) functions directly with $μ$ and $σ$ , so you do not need to standardize by hand if you do not want to. However, you must clearly label the distribution and the probability you are calculating to earn full credit on FRQ. For inverse normal, you start with a given proportion (percentile) and solve for the corresponding $x$ value, which is commonly used to set cutoffs for exams or quality control.

Worked Example

The distribution of wait times for customers at a coffee shop is approximately normal with mean 3.2 minutes and standard deviation 0.8 minutes. (a) What is the probability a randomly selected customer waits between 2 and 4 minutes? (b) What wait time is the 90th percentile, meaning 90% of customers wait less than this time?

Part (a): Define the distribution: $X \sim N (μ = 3.2, σ = 0.8)$ . We want $P (2 < X < 4)$ .
Calculate z-scores: $z_{1} = \frac{2 - 3.2}{0.8} = - 1.5$ , $z_{2} = \frac{4 - 3.2}{0.8} = 1.0$ .
Find cumulative areas: $P (Z < 1.0) = 0.8413$ , $P (Z < - 1.5) = 0.0668$ . Subtract to get the interval probability: $0.8413 - 0.0668 = 0.7745$ , so the probability is approximately 0.77.
Part (b): We need the x value for the 90th percentile. Use invNorm: $x = in v N or m (0.9, 3.2, 0.8) \approx 4.22$ minutes. So the 90th percentile wait time is approximately 4.2 minutes.

Exam tip: When asked for probability on FRQ, always write down the distribution (e.g., $N (3.2, 0.8)$ ) and what you are calculating before giving your final answer; this earns you the method point even if you enter the wrong numbers into your calculator.

5. Assessing Normality

A key AP Statistics skill is determining whether a given data set is approximately normal, which is required for most inference procedures later in the course. There are two common methods for assessing normality tested on the AP exam:

Empirical Rule Check: Calculate the proportion of observations in the data set that fall within 1, 2, and 3 standard deviations of the mean. If the proportions are close to 68%, 95%, and 99.7% respectively, the data is approximately normal. If the proportions are very different, it is not.
Normal Probability Plot (Q-Q Plot): A plot that compares the observed data values to the values we would expect if the data were exactly normal. If the points on the plot lie approximately along a straight line, the data is approximately normal. Curvature away from the straight line indicates non-normality: upward curvature (points bend above the line on the right end) indicates right skew, and downward curvature indicates left skew.

AP almost always asks for interpretation of a given normal probability plot, rather than asking you to construct one from scratch.

Worked Example

A researcher collects data on the distance 100 golfers hit a driver, and creates a normal probability plot of the distances. The points on the plot follow a straight line very closely, except for three very large distances that bend far above the straight line on the right end of the plot. Is the distribution of driving distances approximately normal? What shape does it have?

Recall the interpretation rule: A straight line indicates approximate normality, and curvature indicates non-normality with a specific skew.
The large values bend above the straight line, which means the large values are larger than we would expect for a normal distribution — this is upward curvature, which indicates right skew.
Conclusion: The distribution is not approximately normal; it is right-skewed due to the few very long driving distances.

Exam tip: If a problem says "is it reasonable to assume this distribution is approximately normal?" you must reference either the empirical rule check or the normal probability plot, not just that it is symmetric and unimodal — shape alone is not sufficient.

6. Common Pitfalls (and how to avoid them)

Wrong move: Using the empirical rule for a distribution that is not stated to be normal. Why: Students memorize the empirical rule and automatically use it any time they see a mean and standard deviation, even for explicitly skewed distributions. Correct move: Only use the empirical rule if the problem explicitly states the distribution is approximately normal; for non-normal distributions, state that the empirical rule cannot be used.
Wrong move: Interpreting a z-score without mentioning it is measured in standard deviations, or omitting direction from the mean. Why: Students often just state the numerical z-score and skip the context required for FRQ points. Correct move: Use the template: "[Observation value] is [z] standard deviations [above/below] the mean of [variable name]" for every interpretation.
Wrong move: Subtracting the higher cumulative probability from the lower, resulting in a negative probability for an interval. Why: Students mix up which z-score corresponds to which bound of the interval. Correct move: Always label z-scores with their corresponding x-values, and remember that higher x = higher z = higher cumulative probability, so always subtract left bound cumulative from right bound cumulative.
Wrong move: Confusing the standard deviation of an individual observation with the standard error of a sample mean, adding an unnecessary $n$ denominator to the z-score formula. Why: Students confuse this topic with later sampling distribution content and incorrectly apply the standard error formula early. Correct move: For an individual observation from a normal distribution, use $z = \frac{x - μ}{σ}$ , no $n$ in the denominator.
Wrong move: Interpreting upward curvature on a normal probability plot as left skew. Why: Students mix up the direction of curvature and skew. Correct move: Memorize: Right skew = upward curve (large values are larger than expected for normal), left skew = downward curve (large values are smaller than expected for normal).
Wrong move: Claiming a distribution is normal just because it is symmetric and unimodal. Why: Many symmetric unimodal distributions are not normal. Correct move: Always use either the empirical rule check or a normal probability plot to confirm normality; shape alone is not sufficient evidence.

7. Practice Questions (AP Statistics Style)

Question 1 (Multiple Choice)

The distribution of the amount of coffee brewed by a commercial coffee machine into 12-ounce pots is approximately normal with mean 11.7 ounces and standard deviation 0.25 ounces. Approximately what percent of pots have less than 11.2 ounces of coffee? A) 2.5% B) 5% C) 16% D) 32%

Worked Solution: First, calculate how many standard deviations 11.2 ounces is from the mean: $z = \frac{11.2 - 11.7}{0.25} = - 2$ , so 11.2 ounces is 2 standard deviations below the mean. By the empirical rule, 95% of observations are within 2 standard deviations of the mean, so 5% are outside. Half of that 5% is in the lower tail, so 2.5% of pots have less than 11.2 ounces. The correct answer is A.

Question 2 (Free Response)

The distribution of birth weights of full-term babies in the US is approximately normal with mean 3500 grams and standard deviation 500 grams. (a) A birth weight below 2500 grams is considered low birth weight. What proportion of full-term babies are born with low birth weight? (b) Doctors want to identify the heaviest 5% of full-term births for additional monitoring. What birth weight marks the cutoff for this group, rounded to the nearest 10 grams? (c) A student claims that because the standard deviation is 500 grams, half of all full-term babies weigh between 3000 grams and 4000 grams. Is the student correct? Explain.

Worked Solution: (a) Let $X$ = birth weight of a randomly selected full-term baby, so $X \sim N (3500, 500)$ . We want $P (X < 2500)$ . The z-score is $z = \frac{2500 - 3500}{500} = - 2$ . The cumulative probability for $z = - 2$ is approximately 0.0228, so the proportion of low birth weight babies is approximately 0.023. (b) We need the 95th percentile of the distribution, since the heaviest 5% are above the 95th percentile. Using inverse normal: $in v N or m (0.95, 3500, 500) \approx 4322$ grams, which rounds to 4320 grams. The cutoff weight is 4320 grams. (c) The student is incorrect. 3000 grams is 1 standard deviation below the mean, and 4000 grams is 1 standard deviation above the mean. By the empirical rule, 68% of observations fall within 1 standard deviation of the mean, so 68% of full-term babies (not 50%) weigh between 3000 and 4000 grams.

Question 3 (Application / Real-World Style)

A bottling company produces 2-liter bottles of soda, and the actual volume of soda in the bottles is approximately normally distributed with mean 2.01 liters and standard deviation 0.02 liters. The company labels any bottle with volume less than 2.00 liters as underfilled and reprocesses it. What percentage of bottles will need to be reprocessed? Interpret your answer in context.

Worked Solution: Let $X$ = volume of a randomly selected bottle, so $X \sim N (μ = 2.01, σ = 0.02)$ . We want $P (X < 2.00)$ . Calculate the z-score: $z = \frac{2.00 - 2.01}{0.02} = - 0.5$ . The cumulative probability for $z = - 0.5$ is approximately 0.3085. This means approximately 31% of all bottles produced by the company are underfilled, so the company can expect to reprocess about 3 out of every 10 bottles it produces.

8. Quick Reference Cheatsheet

Category	Formula	Notes
General Normal Distribution	$N (μ, σ)$	Continuous, symmetric, unimodal; total area under curve = 1; defined by mean $μ$ and standard deviation $σ$
Standard Normal Distribution	$N (0, 1)$	Mean 0, standard deviation 1; used for standardized z-scores
Z-score (individual observation)	$z = \frac{x - μ}{σ}$	Positive $z$ = $x$ above mean; negative $z$ = $x$ below mean
Empirical Rule	68% within $μ \pm 1 σ$ 95% within $μ \pm 2 σ$ 99.7% within $μ \pm 3 σ$	Only applies to approximately normal distributions
Probability between $x_{1}$ and $x_{2}$	$P (x_{1} < X < x_{2}) = Φ (z_{2}) - Φ (z_{1})$	$Φ (z)$ = cumulative area to the left of $z$ ; $x_{1} < x_{2}$ so always subtract lower from higher
Probability greater than $x$	$P (X > x) = 1 - Φ (z)$	Always subtract cumulative area from 1 for "greater than" problems
Inverse Normal (find x for p-th percentile)	$x = μ + z_{p} σ$	$z_{p}$ = z-score for p-th percentile of the standard normal
Normal Probability Plot Interpretation	N/A	Straight line = approximately normal; upward curve = right skew; downward curve = left skew

9. What's Next

Mastery of the normal distribution is a foundational requirement for almost every unit that follows in AP Statistics. Immediately after this topic in Unit 1, you will use z-scores from the normal distribution to compare observations from different distributions with different units and scales. Later in the course, the normal distribution is the backbone of sampling distributions, confidence intervals, and hypothesis testing for means and proportions. Without a solid understanding of how to calculate normal probabilities, standardize observations, and assess normality, you will not be able to correctly carry out inference later in the course. The normal distribution also appears in many advanced contexts, from describing residuals in linear regression to checking inference conditions.

Follow-on topics to review next:

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The Normal Distribution — AP Statistics Study Guide

1. What Is The Normal Distribution?

2. The Empirical (68-95-99.7) Rule

Worked Example

3. Z-Scores and the Standard Normal Distribution

Worked Example

4. Normal Probabilities and Inverse Normal Calculations

Worked Example

5. Assessing Normality

Worked Example

6. Common Pitfalls (and how to avoid them)

7. Practice Questions (AP Statistics Style)

Question 1 (Multiple Choice)

Question 2 (Free Response)

Question 3 (Application / Real-World Style)

8. Quick Reference Cheatsheet

9. What's Next

More study guides