Density Values Versus Probabilities

Why Height and Probability Are Different

A normal curve is often drawn with a vertical axis labelled “density.” The curve’s height at a value $x$ tells us how concentrated the distribution is near $x$, but it is not the probability that a random variable equals $x$. For a continuous normal variable, probabilities are represented by areas over intervals. This distinction matters whenever a curve’s height is read or reported: a height and a probability have different meanings and, in general, different units.

Assume that $X$ follows a normal model $X\sim N(\mu,\sigma^2)$, where $\mu$ is the mean and $\sigma>0$ is the standard deviation. Its density function is

f(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right).

The curve described by $f(x)$ is nonnegative, and its total area is 1. That total area is what allows areas under the curve to represent probabilities. The height $f(x)$, by contrast, is the value of the density function at one location. It is not itself an area.

Definition: For a continuous random variable $X$ with density $f$, the probability that $X$ falls between $a$ and $b$, where $a<b$, is the area under the density curve from $a$ to $b$: $$ P(a\leq X\leq b)=\int_a^b f(x)\,dx. $$ The density height $f(x)$ is not $P(X=x)$. For a continuous normal variable, $P(X=x)=0$ for every single value $x$.

Including or excluding an interval endpoint makes no difference for a continuous normal variable. For example, $P(a<X<b)$ and $P(a\leq X\leq b)$ are equal: each endpoint has probability zero. This does not mean values near the endpoints are impossible; it means that a single exact point contributes no area.

A Density Height Is Probability per Unit

The units help make the distinction clear. If $X$ is measured in minutes, then $f(x)$ has units of “per minute.” Multiplying a density by a small interval width in minutes produces a dimensionless quantity, which can be a probability. An interval probability itself has no measurement units.

For a short interval near $x$, the density may be used to approximate the area if the curve changes little across that interval. In that case, area is approximately height multiplied by width. This approximation is not a rule that converts every height directly into a probability: the interval width is essential, and a wider interval generally covers more area.

Worked Example: A Density Height at a Commute Time

Suppose, hypothetically, that a commuter’s travel time $X$ is modelled as normal with mean 30 minutes and standard deviation 5 minutes. Find the density height at 35 minutes and explain what it means.

Check the model and inputs. We are assuming a continuous normal model for travel time, with $\mu=30$ minutes and $\sigma=5$ minutes. The standard deviation is positive, as required for the normal density.

Substitute $x=35$ into the density formula:

f(35)=\frac{1}{5\sqrt{2\pi}}\exp\left(-\frac{(35-30)^2}{2(5)^2}\right) =\frac{1}{5\sqrt{2\pi}}\exp(-0.5) \approx 0.04839\ \text{per minute}.

This is the height of the density curve at 35 minutes. It is not a 4.839% probability, and it is not the probability that the commute takes exactly 35 minutes. Under this continuous model, $P(X=35)=0$. The height indicates how much density there is locally, per minute, around 35 minutes; to obtain a probability, we need an interval and its area.

Point Probabilities and the Area Argument

The zero probability at a single value follows directly from the area definition. A single point has zero width, so integrating a finite density over that point gives zero. More formally, because a normal density is continuous, it remains bounded in a sufficiently small interval around any fixed value $x$. If the interval has width $h$, its area is at most $h$ times a finite upper bound for the density there. As $h$ approaches zero, that upper bound multiplied by $h$ approaches zero.

There is also a useful local interpretation of density. For a small positive width $h$, the average density across the interval from $x$ to $x+h$ is its area divided by its width:

\frac{P(x\leq X\leq x+h)}{h} =\frac{1}{h}\int_x^{x+h} f(t)\,dt.

As $h$ shrinks toward zero, continuity of $f$ means its values across that interval get arbitrarily close to $f(x)$. Therefore the average density approaches the density at $x$:

\lim_{h\to 0^+}\frac{P(x\leq X\leq x+h)}{h}=f(x).

This limit explains why density can be understood as probability per unit width locally. It does not turn density into the probability of a point: the ratio divides an interval probability by a width, and the point probability itself remains zero.

Finding an Interval Probability from Area

For a normal variable, an interval probability is the integral of the normal density over the specified interval. When an exact antiderivative is not available in elementary functions, numerical methods can approximate the area. The examples below use Simpson’s rule with one panel:

\int_a^b f(x)\,dx\ \approx\ \frac{b-a}{6}\left[f(a)+4f\left(\frac{a+b}{2}\right)+f(b)\right].

This formula uses the density at the two endpoints and the midpoint to approximate the area. The normal density is smooth, so for the short intervals below this gives a close approximation. The result is an interval probability, not a density height.

Worked Example: Probability of a Short Interval

Suppose, hypothetically, that a measurement $X$ is modelled as normal with mean 70 units and standard deviation 4 units. Approximate the probability that $X$ lies between 69.9 and 70.1 units.

Check the model and inputs. We assume a continuous normal model, with $\mu=70$ units and $\sigma=4$ units. The interval has width $70.1-69.9=0.2$ units. We need the area across that interval, not the density at one point.

Using the density formula, the midpoint height is $f(70)\approx 0.0997356$ per unit. At either endpoint, the distance from the mean is 0.1 unit, so $f(69.9)=f(70.1)\approx 0.0997044$ per unit. Applying Simpson’s rule:

\begin{aligned} P(69.9\leq X\leq 70.1) &\approx \frac{0.2}{6}\left(0.0997044+4(0.0997356)+0.0997044\right)\\ &\approx 0.019945. \end{aligned}

The probability is approximately 0.01995, or about 1.995%, for a measurement between 69.9 and 70.1 units. The midpoint density, about 0.09974 per unit, is not this probability. Multiplying that height by the interval width gives a close quick approximation, $0.0997356\times0.2\approx0.019947$, because the density changes very little across this narrow interval.

Density Can Be Greater Than One

A probability must be between 0 and 1, but a density height does not have to be. A narrow distribution can have a high peak while still having total area 1. The curve’s width and height work together: a greater height can be balanced by a smaller width. This is another reason not to interpret a density as a probability.

Worked Example: A Density Height Above One

Suppose, hypothetically, that a small part’s thickness $X$, measured in millimetres, follows a normal model with mean 5 mm and standard deviation 0.1 mm. Find the density height at 5 mm, then estimate the probability that thickness is between 4.99 mm and 5.01 mm.

Check the model and inputs. We assume a continuous normal model for thickness, with $\mu=5$ mm and $\sigma=0.1$ mm. This positive standard deviation makes the density defined. The requested interval has width 0.02 mm.

At the mean, the exponential factor is 1, so the density height is

f(5)=\frac{1}{0.1\sqrt{2\pi}}\approx 3.989\ \text{per millimetre}.

This height is greater than 1, which is entirely valid because it is a density, not a probability. At either endpoint, the distance from the mean is 0.01 mm, giving $f(4.99)=f(5.01)\approx 3.970$ per millimetre. Simpson’s rule gives

\begin{aligned} P(4.99\leq X\leq 5.01) &\approx \frac{0.02}{6}\left(3.970+4(3.989)+3.970\right)\\ &\approx 0.07966. \end{aligned}

Thus the probability of a thickness in this 0.02 mm interval is about 0.07966, or 7.966%. The density height is about 3.989 per millimetre, not a probability of 3.989. The interval area is much smaller than the height because the interval is narrow.

Common Mistakes and Why the Distinction Matters

Confusing height with probability can lead to numerical answers that are impossible or have the wrong units. In particular, the claim “the probability of exactly 35 minutes is 0.04839” would confuse a density measured per minute with a probability. The correct point probability is zero; the density describes the local concentration of probability near 35 minutes.

Wrong reasoning: “The curve height at $x$ is the probability of $x$.” The correction is that $f(x)$ is a density height. A probability comes from integrating $f$ over an interval.
Wrong reasoning: “A density above 1 cannot be valid.” The correction is that density is not bounded like probability. Its units are inverse measurement units, and its total area—not each height—is 1.
Wrong reasoning: “Multiply any height by any interval width to get an exact probability.” The correction is that this is generally an approximation when the density is nearly constant across a short interval. The exact probability is the area integral.
Wrong reasoning: “A point has probability zero, so values near it cannot occur.” The correction is that zero probability applies to one exact value. An interval around that value can have positive area and positive probability.

These distinctions are useful whenever a normal curve is used to describe measurements. They keep the meaning, units, and scale of a reported result clear: a density tells how concentrated the curve is at a location, while an area tells how much probability lies across a range.

Key takeaway: For a continuous normal variable, $P(X=x)=0$ for each exact value, while $P(a\leq X\leq b)$ is the area under the density from $a$ to $b$. Density has inverse measurement units; probability is dimensionless.

Check Your Understanding

Answer each question by distinguishing a density height from an area or probability.

A normal density at 12 seconds is 0.08 per second. What does this number describe, and what does it not describe?
For a continuous normal variable, what is $P(X=4.5)$? Explain why this does not imply that values near 4.5 cannot occur.
A density height is 2.4 per centimetre. Is this possible? Explain what determines whether the full density is a valid model.
Why can multiplying a density height by an interval width give a useful approximation to an interval probability for a sufficiently short interval?
A student reports a probability of 0.15 for an interval of width 0.5 units after multiplying a constant-looking density height of 0.30 per unit by the width. What assumption about the density across the interval is implicit in this calculation?