Density Curves

16 min read

Quick Reference

Field	Value
Textbook	Moore, McCabe, Craig, Introduction to the Practice of Statistics (IPS), 9th edition
Chapter	1, Looking at Data: Distributions
Section	1.4, Density Curves and Normal Distributions
Subsection	Density curves, and the mean and median of a density curve
Pages	Page numbers pending faculty verification.

Try This First

Sketch a rough histogram of the exam scores below, then answer the two questions before reading on. No formula is needed yet.

Forty students scored, grouped into bins of width 10:

Score bin	Count
40 to 50	2
50 to 60	6
60 to 70	12
70 to 80	12
80 to 90	6
90 to 100	2

About what fraction of students scored 70 or higher? Estimate it by adding the counts in those bins and dividing by 40.
Imagine smoothing the tops of the bars into a single rounded curve. Where would that curve be tallest, and would the left half look about the same as the right half?

Check your estimate

Scores of 70 or higher cover the bins 70 to 80, 80 to 90, and 90 to 100: that is $12 + 6 + 2 = 20$ students, so $\frac{20}{40} = 0.5$, about half.

The smoothed curve is tallest over the 60 to 80 region (the bins with count 12) and is roughly symmetric: the left side from 40 mirrors the right side toward 100.

The smoothed curve carries the same information as the histogram. The fraction above is an area, and the curve has a tallest region, a center, and a shape, exactly as the bars do.

From Bars to a Smooth Curve

A histogram answers proportion questions with area. When the bars use a density scale (so that total area equals 1), the area of the bars over a range of scores is the fraction of students in that range. The estimate above, one half above 70, was an area read off the bars.

As the data set grows and the bins shrink, the jagged tops of the bars settle toward a single smooth curve. That limiting smooth curve is a density curve. It is an idealized model of the distribution: instead of describing the 40 students you measured, it describes the overall pattern that those students approximate.

The reading rule does not change. With a histogram you read a proportion as the area of the bars over a range. With a density curve you read the same proportion as the area under the curve over that same range. The bars become a curve; area stays the meaning.

Two pictures of the same idea, side by side:

  Histogram (density scale)              Density curve (smooth model)

  |        ___ ___                       |          .-""-.
  |     __|   |   |__                    |        .'      '.
  |   _|  |   |   |  |_                   |      .'          '.
  |  |    |   |   |    |                  |    .'              '.
  |__|____|___|___|____|__               |___.'________________'.___
     40  60  70  80  100                    40   60  70  80   100

  Area of bars over [70,100]             Area under curve over [70,100]
  = fraction scoring 70+                 = proportion scoring 70+

Translate between them in one sentence: a bar of width 10 and density-height $h$ contributes area $10h$, the proportion in that bin, and the smooth curve passes through the same region with the same total area of 1 underneath.

Prerequisite Hub

Builds on:

Skill	Why it is needed
Histograms	A density curve is the smooth limit of a histogram, so reading area and shape off bars comes first
Describing a distribution: shape, center, spread	Shape (symmetric or skewed) and center transfer directly to the curve
Measures of center: mean and median	The balance-point mean and the equal-areas median come straight from the data versions
Reading a graph and its axes	Area over a range is read off the horizontal axis, so axis literacy is assumed

Unlocks:

Skill	What this node gives it
Normal distributions	The Normal curve is one specific density curve, so its area-as-proportion meaning is set up here
Standardizing and z-scores	Standardizing rescales a density curve, which only makes sense once the curve and its area are understood

The Official Definition

Density curve.

A density curve is a curve that is always on or above the horizontal axis and has area exactly 1 underneath it. A density curve describes the overall pattern of a distribution. The area under the curve and above any range of values is the proportion of all observations that fall in that range.

Source: Moore, McCabe, Craig, IPS 9th edition, Section 1.4 (Density Curves and Normal Distributions); definition consistent across IPS editions.

Three parts of that definition, each with a reason it is there:

Part of the definition	What it means	Why it is required
Always on or above the horizontal axis	The curve never dips below 0	A negative proportion is impossible, so the curve cannot go below the axis
Area exactly 1 underneath	The total area equals 1	Every observation falls somewhere, so the whole proportion is 1, or 100 percent
Area over a range = proportion in that range	Pick any interval on the horizontal axis; the area above it is the fraction of observations there	This is how a density curve answers “what fraction is between $a$ and $b$”

A density curve is a model. The actual data have a histogram with real, slightly bumpy bars; the density curve is the smooth idealization. Areas under the curve are approximate descriptions of the real data, not exact counts.

Median and mean of a density curve.

The median of a density curve is the equal-areas point, the point that divides the area under the curve in half. The mean of a density curve is the balance point, at which the curve would balance if made of solid material. The median and mean are the same for a symmetric density curve; the mean of a skewed curve is pulled away from the median toward the long tail.

Source: Moore, McCabe, Craig, IPS 9th edition, Section 1.4 (Density Curves and Normal Distributions).

Two faces of “center”, read two ways off the same curve:

Median, the equal-areas point. A vertical line drawn at the median splits the area under the curve into two pieces of size 0.5 each. Half the observations fall on each side.
Mean, the balance point. Imagine the region under the curve cut from a stiff, uniform sheet. The mean is the horizontal location where that shape would balance on a knife edge.

For a symmetric curve, the equal-areas point and the balance point land at the same place. For a skewed curve, a long tail carries area far out to one side, which tips the balance toward that tail, so the mean is pulled toward the long tail while the median stays nearer the bulk of the area.

To keep the two distinct, a mean and a median computed from a curve (a model) are written with their own symbols, $\mu$ for the mean and $M$ for the median, separate from the data versions $\bar{x}$ and the sample median, which are computed from actual numbers.

Worked Examples

Example 1: Area as a proportion (a uniform density curve)

Consider a density curve that is a flat horizontal line of height $0.5$ over the interval from $0$ to $2$, and height $0$ everywhere else.

Predict first. Before computing, predict the total area under this flat curve. It must equal 1 for the curve to be valid. Does a height of $0.5$ over a width of $2$ look like it will give area 1?

Check. The region under the curve is a rectangle of width $2$ and height $0.5$, so its area is

\[ \text{area} = \text{width} \times \text{height} = 2 \times 0.5 = 1. \]

The total area is 1, so this is a valid density curve.

Now find the proportion of observations between $0.5$ and $1.5$. That region is a rectangle of width $1.5 - 0.5 = 1$ and height $0.5$:

\[ \text{proportion} = 1 \times 0.5 = 0.5. \]

Half of all observations fall between $0.5$ and $1.5$.

By symmetry, both the median and the mean sit at the center of the interval, $x = 1$. The equal-areas point splits the width-2 rectangle into two width-1 rectangles, each of area $0.5$, and the rectangle balances at its midpoint.

Example 2: Mean versus median on a skewed curve

A density curve over the positive numbers rises quickly from 0, peaks near $x = 1$, then trails off with a long thin tail stretching out toward larger values (a right-skewed shape, like household incomes or repair costs).

Predict first. The bulk of the area sits on the left, near the peak, and a thin tail of area reaches far to the right. Predict which is larger, the mean or the median.

Check. The median is the equal-areas point. Because the area is packed near the peak, the equal-areas line lands close to the peak. The mean is the balance point, and the thin tail of area, though small, sits far out to the right, so its distance from the center tips the balance to the right. The mean is therefore pulled toward the long right tail.

Result: for this right-skewed curve, $\text{mean} > \text{median}$. The mean is pulled toward the long tail; the median is not.

A useful one-line rule, consistent with the definition: in a right-skewed curve the mean exceeds the median; in a left-skewed curve the mean is below the median; in a symmetric curve they coincide.

Example 3: Reading a proportion across two regions

Use the uniform curve from Example 1 (height $0.5$ on $[0, 2]$). Find the proportion of observations that are either below $0.4$ or above $1.6$.

Predict first. These are two thin strips at the two ends. Each has width $0.4$. Predict whether the combined proportion is closer to $0.1$, $0.4$, or $0.8$.

Check. The left strip runs from $0$ to $0.4$, width $0.4$, height $0.5$, so its area is $0.4 \times 0.5 = 0.2$. The right strip runs from $1.6$ to $2$, also width $0.4$, so its area is $0.2$ as well. The two strips do not overlap, so add them:

\[ 0.2 + 0.2 = 0.4. \]

The proportion below $0.4$ or above $1.6$ is $0.4$. The estimate closest to this was $0.4$. The complementary proportion, between $0.4$ and $1.6$, is $1 - 0.4 = 0.6$, which matches a width-$1.2$ rectangle of height $0.5$ directly: $1.2 \times 0.5 = 0.6$.

Common Misconceptions

Common misconception

the height of a density curve is the proportion at that point. The vertical reading off a density curve is a density, not a proportion. A proportion is always an area over a range, never a single height. On the uniform curve of Example 1 the height is $0.5$ everywhere on $[0,2]$, yet the proportion of observations at any single exact value is effectively zero, because a single point has no width and so encloses no area. Predict then check: the height at $x = 1$ is $0.5$, but the proportion in the zero-width slice at $x = 1$ is $0$, since area equals height times width and the width is $0$. Read a proportion by finding the area over an interval, and remember a density height can even exceed 1 (a tall narrow curve still has total area 1).

Common misconception

a density curve is a literal picture of the data, so each bump is a specific student. A density curve is a smooth model of the overall pattern, not a plot of individual observations. The actual data form a histogram with bumpy bars; the curve smooths those bumps away. Reading the curve as a literal picture leads to claims like “the curve touches the axis at 100, so the highest score is exactly 100”, which the model does not promise. Predict then check: a Normal density curve never actually reaches the horizontal axis, yet no observation is infinitely large, because the curve is an idealization whose tails approximate, rather than record, the rare extreme cases.

Practice Problems

Level 1 Total Area Check

A proposed density curve is a flat horizontal line of height $0.25$ over the interval from $0$ to $4$, and $0$ elsewhere. Is the total area under it equal to 1, so that it is a valid density curve?

Thought Process

The region under a flat line is a rectangle. Its area is width times height. A valid density curve has total area exactly 1.

Show Answer

The rectangle has width $4 - 0 = 4$ and height $0.25$, so its area is

\[ 4 \times 0.25 = 1. \]

The total area is 1, so this is a valid density curve.

Level 2 Proportion Over a Range

Using the valid density curve from Level 1 (height $0.25$ on $[0, 4]$), find the proportion of observations between $1$ and $3$.

Thought Process

The proportion over a range is the area under the curve over that range. Find the width of the interval, then multiply by the height.

Show Answer

The interval from $1$ to $3$ has width $3 - 1 = 2$, and the height is $0.25$, so

\[ \text{proportion} = 2 \times 0.25 = 0.5. \]

Half of the observations fall between $1$ and $3$.

Level 2 Median by Equal Areas

For the height $0.25$ on $[0, 4]$ curve, where is the median (the equal-areas point)?

Thought Process

The median splits the area into two halves of $0.5$ each. The curve is flat and symmetric on $[0, 4]$, so look for the point with equal area on each side.

Show Answer

A vertical line at $x = 2$ splits the rectangle into two pieces, each of width $2$ and height $0.25$, so each has area $2 \times 0.25 = 0.5$.

The median is $x = 2$. Because the curve is symmetric, the mean (balance point) is also at $x = 2$.

Level 3 Two Separate Regions

For the height $0.25$ on $[0, 4]$ curve, find the proportion of observations that are below $1$ or above $3.5$.

Thought Process

These are two non-overlapping strips, one at each end. Find each area, then add them, because they do not overlap.

Show Answer

Left strip: width $1 - 0 = 1$, height $0.25$, area $1 \times 0.25 = 0.25$.

Right strip: width $4 - 3.5 = 0.5$, height $0.25$, area $0.5 \times 0.25 = 0.125$.

The strips do not overlap, so add:

\[ 0.25 + 0.125 = 0.375. \]

The proportion below $1$ or above $3.5$ is $0.375$.

Level 4 Mean Versus Median on a Skewed Curve

A density curve for repair costs rises sharply from 0, peaks at a small dollar value, then has a long thin tail stretching toward large costs (right-skewed). Which statement is correct?

(A) The mean equals the median, because both describe center.

(B) The mean is less than the median, because most costs are small.

(C) The mean is greater than the median, because the long right tail pulls the balance point right.

(D) The median is greater than the mean, because the median always sits at the peak.

Thought Process

The median is the equal-areas point; with the area packed near the peak, it sits near the peak. The mean is the balance point; a thin tail far to the right tips the balance toward the tail. Compare the two.

Show Answer

Answer: (C) The mean is greater than the median.

Analysis:

(A) Incorrect: mean and median coincide only for a symmetric curve, and this curve is skewed.
(B) Incorrect: this reverses the direction. The small costs hold the median back, but the long right tail pulls the mean up, so the mean is the larger of the two.
(C) Correct: the equal-areas median stays near the bulk of the area at the peak, while the balance-point mean is pulled out toward the long right tail.
(D) Incorrect: the median is not pinned to the peak, and here the mean exceeds the median.

Key insight: for a right-skewed curve, mean $>$ median. The mean follows the tail; the median follows the area.

Level 5 Building a Valid Triangular Density Curve

A density curve is a triangle that starts at the point $(0, 0)$, rises in a straight line to a peak at $(2, h)$, then falls in a straight line back to $(4, 0)$, and is $0$ outside $[0, 4]$. First find the height $h$ that makes this a valid density curve. Then find the proportion of observations between $0$ and $2$, and state the median.

Thought Process

The region under the curve is a triangle with base on $[0, 4]$ and peak height $h$. A valid density curve has total area 1, so set the triangle area to 1 and solve for $h$. For the proportion on $[0, 2]$, use symmetry. For the median, use equal areas, again with symmetry.

Show Answer

Find $h$. The triangle has base $4 - 0 = 4$ and height $h$, so its area is

\[ \text{area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times h = 2h. \]

For a valid density curve this must equal 1, so $2h = 1$, which gives

\[ h = 0.5. \]

Proportion between 0 and 2. The triangle is symmetric about its peak at $x = 2$. The left half, over $[0, 2]$, is a triangle of base $2$ and height $0.5$, with area

\[ \frac{1}{2} \times 2 \times 0.5 = 0.5. \]

So the proportion of observations between $0$ and $2$ is $0.5$.

Median. The left half holds area $0.5$, exactly half the total area. The equal-areas point is therefore at $x = 2$, so the median is $2$. By symmetry the mean (balance point) is also at $x = 2$.

Mastery Checklist

Novice (Level 1-2):

I can state the two defining properties of a density curve: it stays on or above the axis, and the total area underneath is exactly 1
I can find a proportion as the area under a flat (uniform) density curve over a range
I can locate the median of a symmetric curve as the equal-areas point

Competent (Level 3-4):

I can add areas over two separate, non-overlapping regions to get a combined proportion
I can explain why the mean is the balance point and the median is the equal-areas point
I can decide, from a skewed curve, whether the mean is above or below the median, and why

Proficient (Level 5):

I can solve for the height that makes a triangular curve a valid density curve, then read proportions and the median off it
I can explain why a density height is not a proportion, and why a single point encloses no area
I can describe a density curve as a smooth model of a histogram rather than a literal plot of the data

Connections

Looking back:

Histograms: a density curve is the smooth limit of a density-scaled histogram, so area-as-proportion starts there
Describing a distribution: shape and center read off the curve the same way they read off the data
Mean and median: the balance point and the equal-areas point are the curve versions of these data ideas

Looking ahead:

Normal distributions: the Normal curve is one specific, symmetric density curve, and its area gives proportions through the 68-95-99.7 rule
Standardizing and z-scores: standardizing rescales a density curve so that areas can be looked up against a common Normal model

Real-world connections:

Repair costs, household incomes, and insurance claims are right-skewed, so their mean sits above their median
A density curve lets a proportion question (what fraction is above a cutoff) become an area question, which is the same move used for every probability later in the course

Resources

Textbook: Moore, McCabe, Craig, Introduction to the Practice of Statistics (IPS), 9th edition, Chapter 1, Section 1.4 (Density Curves and Normal Distributions). The density-curve subsection, with the mean and median of a density curve.
OpenStax Introductory Statistics 2e: Chapter 6, The Normal Distribution, introduction. Open-access motivation for moving from a histogram to a smooth continuous model.

Previous	Up	Next
Measures of Center: Mean and Median	Skills Index	Normal Distributions

Last updated: 2026-06-16