Normal Distribution Probability Density Function Equation, Derivation & Rearrangements

Core idea

Overview

The Normal Distribution Probability Density Function (PDF) describes a continuous random variable with a symmetric bell-shaped curve. Its height at a particular x-value depends on the mean and standard deviation, and the total area under the curve is 1.

When to use: Use this formula when modeling continuous data that clusters around a mean and forms a bell-shaped distribution.

Why it matters: It is one of the foundational distributions in statistics and appears in confidence intervals, hypothesis tests, and many real-world measurement processes.

Symbols

Variables

f(x | $μ$ , $σ^{2}$ ) = Probability density, x = Value of x, $μ$ = Mean, $σ^{2}$ = Variance, e = Euler's number

f (x ∣ μ, σ^{2})

Probability density

Variable

x

Value of x

Variable

μ

Mean

Variable

σ^{2}

Variance

Variable

e

Euler's number

Variable

π

Pi

Variable

Walkthrough

Derivation

Derivation of Normal Distribution Probability Density Function (PDF)

This derivation focuses on determining the normalization constant for the Normal Distribution's probability density function (PDF), assuming its characteristic bell-shaped exponential form. The process ensures that the total probability over the entire domain integrates to one.

The general functional form of the Normal Distribution's PDF is proportional to an exponential of a negative quadratic, specifically $e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}}$ . This form arises from various theoretical considerations, such as the Central Limit Theorem or the principle of maximum entropy.
The function must be a valid probability density function, meaning its integral over all real numbers must equal 1.

1

Start with the unnormalized form

We begin by assuming the core exponential shape of the Normal Distribution, where $μ$ is the mean, $σ^{2}$ is the variance, and $C$ is an unknown normalization constant. This constant is crucial to ensure that the total probability sums to one.

f (x) = C \cdot e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}}

Note: The exponential term ensures the bell shape and symmetry around the mean $μ$ . The term $2 σ^{2}$ in the denominator scales the spread of the distribution.

2

Apply the normalization condition

For $f (x)$ to be a valid probability density function, the area under its curve over the entire real line must be equal to 1. We set up the integral to reflect this condition.

\int_{- \infty}^{\infty} f (x) d x = 1 \int_{- \infty}^{\infty} C \cdot e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}} d x = 1

3

Simplify the integral using substitution

To evaluate the integral, we perform a substitution. By letting $u = \frac{x - μ}{2 σ}$ , the exponent simplifies to $- u^{2}$ . We also need to find the differential $d x$ in terms of $d u$ . The limits of integration remain from $- \infty$ to $\infty$ because as $x \to \pm \infty$ , $u \to \pm \infty$ .

Let u = \frac{x - μ}{2 σ} Then d u = \frac{1}{2 σ} d x ⟹ d x = 2 σ d u The integral becomes: C \int_{- \infty}^{\infty} e^{- u^{2}} \cdot 2 σ d u = 1

4

Utilize the Gaussian integral

The integral $\int_{- \infty}^{\infty} e^{- u^{2}} d u$ is a well-known result in calculus, often called the Gaussian integral or Euler-Poisson integral. Its value is $π$ .

\int_{- \infty}^{\infty} e^{- u^{2}} d u = π

Note: This integral is fundamental in many areas of mathematics and physics and is typically proven using techniques like polar coordinates for a double integral.

5

Solve for the normalization constant C

Substituting the value of the Gaussian integral into our equation from Step 3, we can now algebraically solve for the normalization constant $C$ . This constant ensures that the total probability is 1.

C \cdot 2 σ \cdot π = 1 C = \frac{1}{2 σ π} C = \frac{1}{2 π σ ^{2}}

6

Substitute C back into the PDF

Finally, by substituting the derived value of $C$ back into our initial unnormalized form, we obtain the complete and normalized probability density function for the Normal Distribution. This formula describes the probability distribution of a continuous random variable that is symmetrically distributed around its mean, forming a bell-shaped curve.

f (x ∣ μ, σ^{2}) = \frac{1}{2 π σ ^{2}} e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}}

Result

f (x ∣ μ, σ^{2}) = \frac{1}{2 π σ ^{2}} e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}}

Source: Any standard textbook on probability and statistics or advanced calculus.

Free formulas

Rearrangements

Solve for $x$

Make x the subject

x = μ \pm - 2 σ^{2} ln (f 2 π σ^{2})

Solve for x by isolating the exponential term using natural logarithms and rearranging the quadratic expression.

Difficulty: 4/5

Solve for $μ$

Make $μ$ the subject

μ = x \mp - 2 σ^{2} ln (f 2 π σ^{2})

Isolate the exponential component, apply logarithms, and rearrange to solve for the mean.

Difficulty: 3/5

Solve for $σ^{2}$

Make $σ^{2}$ the subject

σ^{2} = \frac{( x - μ ) ^{2}}{- W ( - 2 π f ^{2} ( x - μ ) ^{2} )}

Solve for variance by rearranging the transcendental equation using the Lambert W function or iterative methods.

Difficulty: 5/5

Solve for $f$

Make f the subject

f = \frac{1}{2 π σ ^{2}} e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}}

The function is already expressed as f(x | Î¼, ÏƒÂ²), so this is the trivial case.

Difficulty: 1/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Why it behaves this way

Intuition

Imagine a perfectly symmetrical bell-shaped curve, like a pile of sand dropped from a single point. The highest point of the 'bell' is exactly at the center, representing the most common or average value. As you move away from the center in either direction, the height of the curve gradually decreases, indicating that values further from the average are less likely. The total area under this bell curve always equals 1, representing 100% of all possible outcomes. The width and height of the bell are determined by how spread out the data is.

f (x ∣ μ, ϒ^{2})

Probability Density Function (PDF) value at x

This is the 'height' of the bell curve at a specific point 'x'. It tells you how concentrated the probability is around that particular value. A higher value means 'x' is more likely to occur, but it's not a probability itself (probabilities are areas under the curve).

x

Random Variable

This is the specific value or outcome you are interested in. It's the point on the horizontal axis of the bell curve for which you want to know the probability density.

Î¼ (mu)

Mean (Expected Value)

This is the center of the distribution, the average value. It's where the bell curve peaks, indicating the most probable outcome. It determines the location of the curve on the x-axis.

ϒ² (sigma squared)

Variance

This measures how spread out the data is from the mean. A larger variance means the data points are more dispersed, resulting in a wider, flatter bell curve. It's the square of the standard deviation.

e

Euler's Number

A fundamental mathematical constant (approximately 2.718) that appears in many natural growth and decay processes. In this equation, it's crucial for shaping the exponential decay of probability density as you move away from the mean.

Ϭ (pi)

Pi

Another fundamental mathematical constant (approximately 3.14159). Its presence, often with 'e', is common in formulas for continuous distributions, particularly as part of the normalization factor.

Signs and relationships

-: The negative sign in the exponent ensures that as 'x' moves further away from the mean 'Î¼' (in either direction), the value of the exponent becomes more negative. This causes 'e' raised to that power to become smaller, correctly reflecting that values further from the mean have lower probability density.
(x-Î¼)Â²: Squaring the difference `(x-Î¼)` serves two key purposes: 1) It makes the deviation from the mean always positive, ensuring the curve is symmetric around the mean. 2) It penalizes larger deviations from the mean more heavily than smaller ones, contributing to the characteristic bell shape where values far from the mean drop off rapidly.
1 / âˆš(2Ï€ÏƒÂ²): This entire term is a normalization constant. Its purpose is to ensure that the total area under the probability density curve integrates to exactly 1. This is a fundamental requirement for any probability distribution, as the sum of all possible probabilities must be 1.

One free problem

Practice Problem

For a standard normal distribution with mean 0 and variance 1, find the probability density at x = 0.

Value of x0

Mean0

Variance1

Solve for: $f (x ∣ μ, σ^{2})$

Hint: At the mean, the exponential term becomes 1.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

The distribution of adult height in a population is often approximated by a normal distribution.

Study smarter

Tips

The curve is symmetric around the mean.
The total area under the curve is always 1.
A larger standard deviation makes the curve wider and flatter.
This formula gives density, not direct probability.

Avoid these traps

Common Mistakes

Treating the PDF height as a probability.
Forgetting the 1/(sigma * sqrt(2pi)) normalizing factor.
Confusing variance with standard deviation.
Using the formula for non-continuous outcomes.

Keep going

Related Formulas

Common questions

Frequently Asked Questions

This derivation focuses on determining the normalization constant for the Normal Distribution's probability density function (PDF), assuming its characteristic bell-shaped exponential form. The process ensures that the total probability over the entire domain integrates to one.

Use this formula when modeling continuous data that clusters around a mean and forms a bell-shaped distribution.

It is one of the foundational distributions in statistics and appears in confidence intervals, hypothesis tests, and many real-world measurement processes.

Treating the PDF height as a probability. Forgetting the 1/(sigma * sqrt(2pi)) normalizing factor. Confusing variance with standard deviation. Using the formula for non-continuous outcomes.

The distribution of adult height in a population is often approximated by a normal distribution.

The curve is symmetric around the mean. The total area under the curve is always 1. A larger standard deviation makes the curve wider and flatter. This formula gives density, not direct probability.

References

Sources

Britannica Editors (2026) 'Standard normal distribution' Encyclopaedia Britannica.
NIST/SEMATECH e-Handbook of Statistical Methods, normal distribution section.
NIST Special Publication 1017, standard normal distribution references.
Standard probability and statistics textbooks
Any standard textbook on probability and statistics or advanced calculus.

Normal Distribution Probability Density Function

Overview

Variables

Derivation

Start with the unnormalized form

Apply the normalization condition

Simplify the integral using substitution

Utilize the Gaussian integral

Solve for the normalization constant C

Substitute C back into the PDF

Rearrangements

Intuition

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Normal Distribution PDF

Frequently Asked Questions

Sources