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Normal Distributions

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Textbook: Moore, McCabe, Craig, Introduction to the Practice of Statistics (IPS), 9th edition  •  Chapter: 1  •  Section: 4

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Field Value
Textbook Moore, McCabe, Craig, Introduction to the Practice of Statistics (IPS), 9th edition
Location Chapter 1, Looking at Data: Distributions; Section 1.4, Density Curves and Normal Distributions
Subsection Section 1.4, the Normal distributions and the 68-95-99.7 rule
Pages Page numbers pending faculty verification.
Course MATH380
Role The first named family of density curves in the course. In Section 1.4, the idealized density curve is the smooth model for a distribution; the Normal curves are the specific bell-shaped family that the rest of the course returns to constantly. The 68-95-99.7 rule taught here is the bridge to standardizing and z-scores in the next node, and the foundation for sampling distributions later in the course.

Try This First

A standardized reading test is built so that the scores follow a bell-shaped distribution with a center at 500 and a typical spread of 100 points. The picture below shows the bell, with tick marks placed every 100 points away from the center.

                       ____
                     /      \
                   /          \
                 /              \
               /                  \
            __/                      \__
        ___/                            \___
   ____/                                    \____
  |------|------|------|------|------|------|------|
  200    300    400    500    600    700    800
                       (center)

Before reading any definition, work these on paper.

  1. The bell is symmetric. What single score sits exactly at the center, with half of all test-takers scoring below it and half scoring above it?
  2. The tick marks are spaced one typical spread (100 points) apart. A rough rule says that almost everyone (about 95 out of 100 people) scores within two of those spacings of the center. Between which two scores does that put almost everyone?
  3. Predict, before checking: a score of 700 is two spacings above the center. Do you expect that score to be common or rare? Then keep reading.
Check your three answers
  1. The center score is 500. Because the curve is symmetric, the center is both the balance point and the middle value: half of all test-takers score below 500 and half score above it.
  2. Two spacings of 100 points on each side of 500 runs from $500 - 200 = 300$ to $500 + 200 = 700$. About 95 out of 100 test-takers score between 300 and 700.
  3. A score of 700 is rare. It sits at the upper edge of the range that holds about 95 percent of people, so only a small fraction of test-takers (about 2 to 3 out of 100) score that high or higher.

You have just read the two numbers that define a Normal distribution off a picture: its center (the mean) and its typical spread (the standard deviation). You also used a rough version of the 68-95-99.7 rule.


An Intuitive Introduction

A density curve is a smooth idealized picture of a distribution, drawn so that the total area beneath it equals 1 and so that the area above any interval equals the proportion of observations in that interval. The Normal curves are one particular family of density curves, and they are the most important family in the course.

A Normal curve has a single recognizable shape: a symmetric bell, highest at the center and tapering off smoothly toward both ends. What makes the Normal curves a family rather than a single curve is that two numbers are free to change. The first number, the mean, says where the center of the bell sits along the horizontal axis. Slide the mean up and the whole bell shifts to the right without changing shape. The second number, the standard deviation, says how wide the bell is. A small standard deviation gives a tall narrow bell with the data packed near the center; a large standard deviation gives a short wide bell with the data spread out.

Think of the curve as a process that takes a value on the horizontal axis and gives back a height. Two settings, the mean and the standard deviation, fully determine that process. Once those two numbers are fixed, the entire curve is fixed: its peak, its width, and the proportion of area in any region. No other information is needed. This is why a Normal distribution is often written in shorthand as $N(\mu, \sigma)$, naming only those two numbers.

The standard deviation has a feature worth noticing on the picture. Starting from the peak, the curve falls away steeply at first; then, as it nears the bottom, it changes from curving downward to curving outward. The point on each side where that change happens, called an inflection point, sits exactly one standard deviation away from the mean. So the standard deviation is not an abstract number for a Normal curve; it is a distance you can mark on the graph.


Prerequisite Hub

Builds On

Skill Why it is needed
Density Curves A Normal curve is a particular density curve. The reader must already know that the total area under a density curve is 1 and that area over an interval represents the proportion of observations there.
Measures of Center: Mean and Median The mean $\mu$ is located at the center of a Normal curve and equals the median. The reader must understand mean and median as measures of center to place the center of the bell.
Measures of Spread: Variance and Standard Deviation The standard deviation $\sigma$ controls the spread of a Normal curve and is read off the graph as the distance to the inflection points. The reader must know what a standard deviation measures.

Unlocks

Skill What it adds
Standardizing and z-Scores Standardizing converts any Normal distribution into the single standard Normal distribution $N(0, 1)$, so that one table or one calculation handles every Normal curve. The 68-95-99.7 rule learned here is the first special case.

Cross-Course Prerequisites

None. The density-curve, center, and spread ideas are all in the earlier MATH380 nodes above.


The Official Definition

Normal distribution. The Normal distributions are described by a special family of bell-shaped, symmetric density curves called Normal curves. A particular Normal distribution is completely specified by two numbers, its mean $\mu$ and its standard deviation $\sigma$. The mean is located at the center of the symmetric curve and is the same as the median; the standard deviation $\sigma$ controls the spread of the curve, and is the distance from the center to the change-of-curvature (inflection) points on either side.

(Moore, McCabe, Craig, IPS 9th edition, Section 1.4, Density Curves and Normal Distributions; corroborated by OpenStax Introductory Statistics 2e, 6.1 The Standard Normal Distribution, https://openstax.org/books/introductory-statistics-2e/pages/6-1-the-standard-normal-distribution.)

Three features inside this definition are easy to skim past and worth holding onto.

Reading the Two Numbers Off a Curve

What you want to know What it is Where to find it on the picture
Where the distribution is centered The mean $\mu$ The horizontal position of the peak, which is also the line of symmetry.
How spread out the distribution is The standard deviation $\sigma$ The horizontal distance from the peak out to either inflection point.

Changing $\mu$ slides the whole bell left or right without reshaping it. Changing $\sigma$ stretches or squeezes the bell without moving its center. Translating between the two numbers and the picture of the curve is the core skill for reading a Normal distribution.


The 68-95-99.7 Rule

The single most useful fact about every Normal distribution is that the proportion of area in a band around the mean depends only on how many standard deviations wide that band is, never on the particular values of the mean or the standard deviation.

The 68-95-99.7 rule (empirical rule for Normal distributions). In any Normal distribution with mean $\mu$ and standard deviation $\sigma$:

  • approximately 68 percent of the observations fall within $\sigma$ of the mean $\mu$ (that is, between $\mu - \sigma$ and $\mu + \sigma$);
  • approximately 95 percent of the observations fall within $2\sigma$ of $\mu$ (between $\mu - 2\sigma$ and $\mu + 2\sigma$);
  • approximately 99.7 percent of the observations fall within $3\sigma$ of $\mu$ (between $\mu - 3\sigma$ and $\mu + 3\sigma$).

(Moore, McCabe, Craig, IPS 9th edition, Section 1.4, Density Curves and Normal Distributions: the 68-95-99.7 rule.)

The rule is the same for every Normal distribution because the only thing that changes from one Normal curve to another is the scale on the horizontal axis. Measuring distance in units of $\sigma$ removes that scale, so the proportions are identical.

A picture keeps the three bands straight.

                    ___
                  /     \
                /         \
              /             \
            /                 \
          /                     \
       __/                        \__
   ___/   |       |       |        \___
  |-------|-------|-------|-------|-------|
 mu-3s  mu-2s   mu-s     mu     mu+s   mu+2s  mu+3s

  |<-------------- 68% -------------->|   within 1 sigma
  |<------------------- 95% ------------------->|  within 2 sigma
  |<------------------------ 99.7% ----------------------->| within 3 sigma

Two consequences follow directly from the symmetry of the curve and are used constantly.

These splits turn the three headline percentages into the proportion above or below any point that lands on a whole number of standard deviations from the mean.


Worked Examples

Example 1: Reading the Two Numbers and Marking the Bands

A Normal distribution models the heights of adult women in a population, with mean $\mu = 64$ inches and standard deviation $\sigma = 2.5$ inches, written $N(64, 2.5)$.

Predict first. The center of the bell is 64 inches. One standard deviation is 2.5 inches, so the inflection points sit at $64 - 2.5 = 61.5$ and $64 + 2.5 = 66.5$ inches. Most heights should cluster within a few inches of 64.

Step 1: Mark one, two, and three standard deviations on each side.

Distance from mean Low side High side
$1\sigma$ $64 - 2.5 = 61.5$ $64 + 2.5 = 66.5$
$2\sigma$ $64 - 5 = 59$ $64 + 5 = 69$
$3\sigma$ $64 - 7.5 = 56.5$ $64 + 7.5 = 71.5$

Step 2: Attach the percentages. By the 68-95-99.7 rule, about 68 percent of women have heights between 61.5 and 66.5 inches, about 95 percent between 59 and 69 inches, and about 99.7 percent between 56.5 and 71.5 inches.

Read it back. Almost every adult woman in this model (99.7 percent) falls between 56.5 and 71.5 inches. That matches the prediction that heights cluster tightly around 64. The numbers 64 and 2.5 alone produced this entire description.

Example 2: Proportion Below a Point Using Symmetry

Scores on a placement exam follow $N(500, 100)$: mean 500, standard deviation 100. What proportion of test-takers score below 600?

Predict first. The point 600 is one standard deviation above the mean. More than half the area lies below 600, because 600 is above the center, so the answer should be larger than 50 percent but not close to 100 percent.

Step 1: Locate 600 in standard-deviation units. Since $600 = 500 + 100 = \mu + \sigma$, the point 600 is exactly one standard deviation above the mean.

Step 2: Use the 68 percent band and symmetry. The 68-95-99.7 rule says 68 percent of scores fall between $\mu - \sigma = 400$ and $\mu + \sigma = 600$. By symmetry, half of that 68 percent, which is 34 percent, lies between the mean 500 and the point 600.

Step 3: Add the lower half. Half of all area, 50 percent, lies below the mean 500. Add the 34 percent between 500 and 600:

\[ 50\% + 34\% = 84\%. \]

Read it back. About 84 percent of test-takers score below 600. This is larger than 50 percent and well short of 100 percent, exactly as predicted, because 600 sits one standard deviation above the center.

Example 3: Proportion in a Tail

A machine fills bottles so that the fill volume follows $N(500, 5)$: mean 500 milliliters, standard deviation 5 milliliters. What proportion of bottles contain more than 510 milliliters?

Predict first. The point 510 is two standard deviations above the mean. Two standard deviations out is far, so the proportion beyond it should be small, a couple of percent at most.

Step 1: Locate 510 in standard-deviation units. Since $510 = 500 + 10 = 500 + 2(5) = \mu + 2\sigma$, the point 510 is two standard deviations above the mean.

Step 2: Use the 95 percent band. The rule says 95 percent of bottles fall within two standard deviations of the mean, between 490 and 510 milliliters. So 5 percent fall outside that band.

Step 3: Split the outside area between the two tails. By symmetry the 5 percent outside the band splits evenly: 2.5 percent below 490 and 2.5 percent above 510.

\[ \frac{5\%}{2} = 2.5\%. \]

Read it back. About 2.5 percent of bottles contain more than 510 milliliters. The answer is small, as predicted, because 510 is two standard deviations above the center, near the upper edge of where almost all bottles fall.

Example 4: Same Rule, Different Curves

Two Normal distributions describe two manufacturing lines. Line A has $N(200, 4)$ and Line B has $N(50, 4)$, both measured in grams. For each line, between which two values does the middle 95 percent of output fall?

Predict first. The two lines have the same standard deviation but different means. The 95 percent bands should have the same width (they share $\sigma = 4$) but be centered at different places.

Step 1: Apply the two-standard-deviation band to each line. The middle 95 percent runs from $\mu - 2\sigma$ to $\mu + 2\sigma$.

\[ \text{Line A: } 200 - 2(4) = 192 \quad \text{to} \quad 200 + 2(4) = 208. \]

\[ \text{Line B: } 50 - 2(4) = 42 \quad \text{to} \quad 50 + 2(4) = 58. \]

Read it back. Line A holds its middle 95 percent between 192 and 208 grams; Line B between 42 and 58 grams. Both bands are 16 grams wide, because both lines share $\sigma = 4$. Only the center moved, which confirms that the standard deviation sets the width of the band and the mean sets its location. The 68-95-99.7 rule produced the same proportions for two different curves.


Common Misconceptions

Common misconception

every bell-shaped or roughly symmetric distribution is Normal. Symmetry and a single peak are necessary for a Normal curve, but they are not enough. Many distributions are symmetric and unimodal without being Normal; a Normal curve has one exact mathematical shape determined by its mean and standard deviation, including the specific 68-95-99.7 proportions. Predict-then-check: sketch a symmetric triangle (a tent shape) centered at 0. It is symmetric and single-peaked, yet the proportion within one standard deviation is not 68 percent, so it is not Normal. Reading the bell as a literal picture of “any hump” rather than the specific Normal shape leads to applying the 68-95-99.7 rule where it does not hold.

Common misconception

the standard deviation changes the center, or the mean changes the spread. The two numbers do separate jobs. The mean $\mu$ slides the bell left or right and never changes its width. The standard deviation $\sigma$ stretches or squeezes the bell and never moves its center. Predict-then-check: compare $N(0, 1)$ with $N(0, 3)$. They share the same center, 0, but the second is three times as wide. Now compare $N(0, 1)$ with $N(5, 1)$. Same width, but the second is shifted five units to the right. Changing one number does not do the other number’s job.

Common misconception

the 68-95-99.7 percentages are spread out so that 95 percent are added beyond the 68 percent. The three percentages describe nested bands measured from the same center, not separate slices laid end to end. The 95 percent band contains the 68 percent band inside it; it does not sit beside it. Predict-then-check: in $N(0, 1)$, the 68 percent runs from $-1$ to $1$, and the 95 percent runs from $-2$ to $2$. The first interval is entirely inside the second. The extra area, from $-2$ to $-1$ and from $1$ to $2$, is what raises 68 percent up to 95 percent; it is not stacked on top as a separate 95 percent. The bands grow outward from the mean.


Practice Problems

Level 1 Naming the Two Numbers

A Normal distribution is written $N(120, 15)$. State its mean and its standard deviation, and give the value of the inflection point on each side of the center.

Show Answer

The mean is $\mu = 120$ and the standard deviation is $\sigma = 15$. The inflection points sit one standard deviation from the center, at $120 - 15 = 105$ and $120 + 15 = 135$.

Level 2 Applying the 68 Percent Band

Resting heart rates for a group of adults follow $N(70, 8)$ beats per minute. Between which two heart rates does the middle 68 percent of this group fall?

Thought Process

The middle 68 percent of a Normal distribution falls within one standard deviation of the mean, from $\mu - \sigma$ to $\mu + \sigma$.

Show Answer

One standard deviation from the mean runs from $70 - 8 = 62$ to $70 + 8 = 78$. The middle 68 percent of the group has resting heart rates between 62 and 78 beats per minute.

Level 3 Proportion Below a Whole-Sigma Point

The lengths of a manufactured part follow $N(40, 2)$ millimeters. What proportion of parts are shorter than 38 millimeters?

Thought Process

First find how many standard deviations 38 is from the mean. Then use the 68 percent band split by symmetry, and subtract from the 50 percent that lies below the mean.

Show Answer

Since $38 = 40 - 2 = \mu - \sigma$, the point 38 is one standard deviation below the mean. The 68 percent band runs from 38 to 42, so 34 percent lies between 38 and the mean 40. The proportion below the mean is 50 percent. Subtract the 34 percent that lies between 38 and 40:

\[ 50\% - 34\% = 16\%. \]

About 16 percent of parts are shorter than 38 millimeters.

Level 4 Proportion in a Middle Band

Battery lifetimes follow $N(500, 50)$ hours. A battery is graded “standard” if its lifetime is between 450 and 600 hours.

(a) Express 450 and 600 as distances from the mean in standard deviations. (b) Find the proportion of batteries graded “standard.”

Thought Process

Locate each endpoint in standard-deviation units. The two endpoints are different distances from the mean, so handle each side separately using the 68 and 95 percent bands and symmetry, then add the two pieces around the mean.

Show Answer

(a) Lower endpoint: $450 = 500 - 50 = \mu - \sigma$, so 450 is one standard deviation below the mean. Upper endpoint: $600 = 500 + 100 = \mu + 2\sigma$, so 600 is two standard deviations above the mean.

(b) Work from the mean outward on each side.

Below the mean: the 68 percent band gives 34 percent between $\mu - \sigma = 450$ and the mean 500.

Above the mean: the 95 percent band gives 47.5 percent (half of 95 percent) between the mean 500 and $\mu + 2\sigma = 600$.

Add the two pieces:

\[ 34\% + 47.5\% = 81.5\%. \]

About 81.5 percent of batteries are graded “standard.”

Level 5 Reason From a Known Proportion Back to the Numbers

A Normal distribution of exam scores is known to have about 2.5 percent of scores above 90 and about 2.5 percent of scores below 60.

(a) Use the 68-95-99.7 rule to explain why 60 and 90 must each be two standard deviations from the mean. (b) Find the mean $\mu$ and the standard deviation $\sigma$ of this distribution. (c) Construct a different question on this same distribution whose answer is 84 percent, and give that answer. Justify it.

Thought Process

A tail of 2.5 percent on each side means 95 percent lies in the middle band, so the two cutoffs are at $\mu - 2\sigma$ and $\mu + 2\sigma$. The mean is the midpoint of the two cutoffs, and the distance from the mean to either cutoff is $2\sigma$. For part (c), find a point one standard deviation above the mean, where the proportion below is 84 percent.

Show Answer

(a) A 2.5 percent tail on each side leaves 95 percent of the area in the middle. By the 68-95-99.7 rule the central 95 percent band runs from $\mu - 2\sigma$ to $\mu + 2\sigma$, so its two edges, 60 and 90, must each be two standard deviations from the mean.

(b) The mean is the midpoint of the two cutoffs:

\[ \mu = \frac{60 + 90}{2} = 75. \]

The distance from the mean to either cutoff is $2\sigma$. Using the upper cutoff, $90 - 75 = 15 = 2\sigma$, so:

\[ \sigma = \frac{15}{2} = 7.5. \]

The distribution is $N(75, 7.5)$.

(c) One valid question: “What proportion of scores are below $82.5$?” The point $82.5 = 75 + 7.5 = \mu + \sigma$ is one standard deviation above the mean. By symmetry, 34 percent lies between the mean and $\mu + \sigma$, and 50 percent lies below the mean, so the proportion below 82.5 is $50\% + 34\% = 84\%$. (Other questions work; the key is a cutoff one standard deviation above the mean.)


Mastery Checklist

Novice (Level 1-2):

Competent (Level 3-4):

Proficient (Level 5):


Connections

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Real-world connections:


Resources

Resource What it covers
Moore, McCabe, Craig, Introduction to the Practice of Statistics (IPS), 9th edition, Chapter 1, Section 1.4, Density Curves and Normal Distributions The primary text for Normal distributions. Normal curves, the mean and standard deviation that define them, and the 68-95-99.7 rule. Page numbers pending faculty verification.
OpenStax Introductory Statistics 2e, 6.1 The Standard Normal Distribution An open-source companion treatment of the Normal distribution and its standardized form, with worked examples and exercises.


Last updated: 2026-06-16