What is the difference between the empirical rule and the z-table?

The empirical rule gives quick approximate percentages (68, 95, 99.7) only for values exactly 1, 2, or 3 standard deviations from the mean. The z-table gives precise areas for ANY z-score, including decimals like 1.63. Use the empirical rule for fast estimates and the table for exact probabilities.

Does the standard normal table give area to the left or right?

The standard AP z-table gives the area to the LEFT of the z-score, which equals the percentile. To find area to the right, subtract the table value from 1. Always sketch and shade the region first so you combine areas correctly.

How do I know if I should use a normal model?

A normal model is reasonable when the distribution is roughly symmetric, unimodal, and bell-shaped with no strong skew or outliers. If a problem states the distribution is approximately normal, use it directly. Otherwise, check the shape from a graph or a stated description before applying z-scores.

Why do we convert to z-scores instead of using the original values?

Standardizing converts any normal distribution into the single standard normal distribution N(0,1), so one table works for every problem. z-scores also let you compare values measured on different scales or units on a common footing.

AP-STATS-1.10

U1.10 The Normal Distribution

Master the normal distribution for AP Statistics: apply the 68-95-99.7 rule, calculate z-scores, and find probabilities and percentiles with the standard normal table.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U1.10 The Normal Distribution, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U1.10 The Normal Distribution free

What this lesson covers

Many measurements in the real world — heights, test scores, measurement errors — pile up in a symmetric, bell-shaped pattern called the normal distribution. Because this shape is so common and so predictable, it becomes a powerful tool: once you know a distribution is roughly normal, you can answer questions like "What percent of values fall below 80?" or "What score marks the top 10%?"

In this lesson you will learn to recognize a normal model, use the quick 68-95-99.7 rule for benchmark values, standardize any raw value into a z-score, and then convert z-scores into exact probabilities and percentiles using the standard normal table. These skills recur throughout AP Statistics, especially in later units on sampling distributions and inference.

Recognizing the Normal Distribution

A normal distribution is a continuous, symmetric, bell-shaped model completely described by two numbers: its mean

\mu

and its standard deviation

\sigma

. We write

N(\mu, \sigma)

. The mean sits at the center, and because the curve is symmetric, the mean equals the median. The standard deviation controls the spread — a larger

\sigma

makes a wider, flatter bell.

The curve never touches the horizontal axis; it extends infinitely in both directions but gets very close to zero in the tails. The total area under the curve equals 1, which lets us treat areas as probabilities or proportions.

Not every unimodal, roughly symmetric distribution is normal. On the exam you may be asked whether a normal model is reasonable. Clues that a distribution is NOT normal include strong skew, obvious outliers, or multiple peaks. A common misconception is assuming any bell-ish histogram is normal — always check symmetry and whether the shape tapers smoothly in both tails.

Feature	Normal distribution
Shape	Symmetric, single peak, bell
Center	Mean = median
Parameters	$\mu$ (center), $\sigma$ (spread)
Total area	Equals 1

When a variable follows

N(\mu,\sigma)

, its inflection points (where the curve changes concavity) occur exactly one standard deviation from the mean, at

\mu - \sigma

and

\mu + \sigma

The 68-95-99.7 (Empirical) Rule

For any normal distribution, fixed percentages of the data fall within a set number of standard deviations from the mean. Approximately

68\%

of values lie within

1\sigma

of the mean, about

95\%

within

2\sigma

, and about

99.7\%

within

3\sigma

. This is called the empirical rule.

Interval	Approx. percent inside
$\mu \pm 1\sigma$	$68\%$
$\mu \pm 2\sigma$	$95\%$
$\mu \pm 3\sigma$	$99.7\%$

Because the curve is symmetric, you can split these regions. For example, since

68\%

lies within

1\sigma

, the remaining

32\%

is split between the two tails, so

16\%

lies above

\mu + 1\sigma

and

16\%

below

\mu - 1\sigma

.

The rule is fast but only works for values at whole-number multiples of

\sigma

. If a question asks about a value that is

1.4

standard deviations above the mean, you must use z-scores and the table instead. On the exam, the empirical rule is ideal for mental-math or no-calculator reasoning, such as "What percent of observations exceed a value two standard deviations below the mean?" Answer:

2.5\% + 95\% = 97.5\%

.

A frequent error is forgetting that the leftover percentage is split evenly between two tails. Always draw the bell, mark

\mu

, then step out by

\sigma

increments and label the regions before answering.

Standardizing: Computing z-Scores

A z-score measures how many standard deviations a value lies above or below the mean. The formula is

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

A positive z-score means the value is above the mean; a negative z-score means below. A z-score of

0

is exactly average. Standardizing removes the units, so a z-score of

1.5

means "1.5 standard deviations above average" whether the data are heights in cm or scores in points.

z-scores let you compare values from different distributions. If Maria scored

z = 1.2

on a biology test and

z = 0.8

on a chemistry test, she performed relatively better in biology, even if her raw chemistry score was numerically higher.

Standardizing also transforms any normal distribution

N(\mu, \sigma)

into the standard normal distribution

N(0, 1)

, which has mean

0

and standard deviation

1

. This single reference distribution is what the standard normal table describes, so every normal probability problem reduces to computing a z-score first.

To reverse the process — going from a z-score back to a raw value — solve the formula for

x

x = \mu + z\sigma

This is essential for percentile problems where you know the desired area and want the cutoff value. Keep track of signs carefully: a value below the mean has a negative z, so

x = \mu + z\sigma

correctly subtracts when

z

is negative.

Finding Probabilities and Percentiles with the Table

The standard normal table (the z-table) gives the area to the LEFT of a given z-score — that is, the proportion of values below it. This left area is a percentile. For example, a z-score of

1.00

corresponds to a table value of about

0.8413

, meaning roughly

84.13\%

of values fall below that point.

To find probabilities, follow a consistent routine. First, sketch the curve and shade the region you want. Second, standardize the boundary value(s) into z-scores. Third, look up the left-area for each z. Fourth, combine the areas to match your shaded region.

Question type	How to combine table areas
$P(X < x)$	Use table value directly
$P(X > x)$	$1 - (\text{table value})$
$P(a < X < b)$	(table for $b$ ) $-$ (table for $a$ )

For percentile (inverse) problems, work backward: find the area in the table body closest to your target percentile, read off the corresponding z-score, then convert with

x = \mu + z\sigma

. For instance, the 90th percentile corresponds to about

z = 1.28

.

The biggest exam pitfall is confusing "area to the left" with "area to the right." Always sketch first. On free-response questions, you must show the z-score calculation and clearly state the probability — a bare number with no work loses credit even if correct.

Key terms

Normal distribution.: A symmetric, bell-shaped continuous distribution defined by its mean $\mu$ and standard deviation $\sigma$ , written $N(\mu, \sigma)$ , with total area 1 under the curve.
Empirical rule (68-95-99.7).: For a normal distribution, about 68%, 95%, and 99.7% of values fall within 1, 2, and 3 standard deviations of the mean, respectively.
z-score.: A standardized value $z = \frac{x-\mu}{\sigma}$ giving the number of standard deviations a data value lies above (positive) or below (negative) the mean.
Standard normal distribution.: The normal distribution with mean $0$ and standard deviation $1$ , denoted $N(0,1)$ , obtained by standardizing any normal variable.
Standard normal table.: A reference table giving the area (proportion) to the left of a given z-score under the standard normal curve.
Percentile.: The percentage of observations that fall at or below a given value; equal to the area to the left under a normal curve.
Inverse normal calculation.: Working backward from a known area or percentile to find the corresponding z-score and then the raw value using $x = \mu + z\sigma$ .

Worked example

Adult resting heart rates are approximately normal with mean

\mu = 72

beats per minute and standard deviation

\sigma = 8

beats per minute. (a) What proportion of adults have a resting heart rate above 85 bpm? (b) What heart rate marks the 25th percentile?

Part (a): First sketch the normal curve centered at 72 and shade the region above 85. Standardize the boundary:

z = \frac{85 - 72}{8} = \frac{13}{8} = 1.625 \approx 1.63

Look up

z = 1.63

in the standard normal table: the area to the left is about

0.9484

. Since we want the area to the RIGHT (above 85), subtract from 1:

P(X > 85) = 1 - 0.9484 = 0.0516

So about

5.2\%

of adults have a resting heart rate above 85 bpm.

Part (b): The 25th percentile means 25% of the area lies to the left. Search the body of the table for an area near

0.2500

; the closest z-score is about

z = -0.67

. Now convert back to a raw value:

x = \mu + z\sigma = 72 + (-0.67)(8) = 72 - 5.36 = 66.64

So the 25th percentile is about

66.6

bpm. About one-quarter of adults have a resting heart rate below this value. Notice the negative z-score correctly places the answer below the mean of 72.

Practice questions

Scores on a placement exam are approximately normal with mean 500 and standard deviation 100. Using the empirical rule, approximately what percent of students score between 400 and 700?

68%
81.5%
95%
97.5%

Answer: 81.5%

The value 400 is one standard deviation below the mean (

z=-1

) and 700 is two standard deviations above (

z=+2

). By the empirical rule, from the mean to

-1\sigma

covers

34\%

, and from the mean to

+2\sigma

covers

47.5\%

(half of 95%). Adding:

34\% + 47.5\% = 81.5\%

. A common wrong answer is 68% or 95%, which would only apply to symmetric intervals.

The weights of a certain fruit are approximately normal with mean 150 grams and standard deviation 20 grams. A fruit weighs 180 grams. Calculate its z-score and interpret what the z-score means in context.

Answer: z = 1.5, meaning the fruit weighs 1.5 standard deviations above the mean weight.

Apply the z-score formula:

z = \frac{180 - 150}{20} = \frac{30}{20} = 1.5

. Interpretation must be in context: this fruit's weight is 1.5 standard deviations above the average weight of 150 grams. A full-credit AP response names the direction (above), the magnitude (1.5 standard deviations), and the variable in context (weight of the fruit).

Heights of a population are normal with mean 65 inches and standard deviation 3 inches. What height corresponds to the 90th percentile?

Answer: About 68.8 inches

The 90th percentile has area 0.9000 to its left. Searching the table, the closest z-score is about

z = 1.28

. Convert to a raw value:

x = 65 + (1.28)(3) = 65 + 3.84 = 68.84

inches. So roughly 90% of the population is shorter than about 68.8 inches. Students often stop at the z-score; you must convert back to the original units to answer the question.

FAQ

What is the difference between the empirical rule and the z-table?: The empirical rule gives quick approximate percentages (68, 95, 99.7) only for values exactly 1, 2, or 3 standard deviations from the mean. The z-table gives precise areas for ANY z-score, including decimals like 1.63. Use the empirical rule for fast estimates and the table for exact probabilities.
Does the standard normal table give area to the left or right?: The standard AP z-table gives the area to the LEFT of the z-score, which equals the percentile. To find area to the right, subtract the table value from 1. Always sketch and shade the region first so you combine areas correctly.
How do I know if I should use a normal model?: A normal model is reasonable when the distribution is roughly symmetric, unimodal, and bell-shaped with no strong skew or outliers. If a problem states the distribution is approximately normal, use it directly. Otherwise, check the shape from a graph or a stated description before applying z-scores.
Why do we convert to z-scores instead of using the original values?: Standardizing converts any normal distribution into the single standard normal distribution $N(0,1)$ , so one table works for every problem. z-scores also let you compare values measured on different scales or units on a common footing.

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