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Normal Distribution Probability Density Function (PDF)

The probability density function of a normal distribution describes the likelihood of a continuous random variable taking on a specific value based on its mean and variance.

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

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Core idea

Overview

This formula represents the classic bell-shaped Gaussian curve, where the peak is defined by the mean (μ) and the spread or width is controlled by the variance (σ²). It is the cornerstone of inferential statistics, as the Central Limit Theorem dictates that the sums of many independent random variables tend toward this distribution. The integral of this function over any interval represents the probability that the random variable falls within that range.

When to use: Use this to model physical, biological, or social phenomena where data points cluster around a central average with symmetric deviations.

Why it matters: It allows for the calculation of probabilities, hypothesis testing, and the estimation of parameters in nearly all scientific and engineering fields.

Symbols

Variables

x = Random Variable, $μ$ = Mean, $σ^{2}$ = Variance

x

Random Variable

Variable

μ

Mean

Variable

σ^{2}

Variance

Variable

Walkthrough

Derivation

Derivation of Normal Distribution Probability Density Function (PDF)

The normal distribution is derived from the requirement that the maximum likelihood estimator for a mean of independent observations is the arithmetic mean, leading to Gauss's functional equation.

The probability density function f(x) is dependent only on the distance from the mean.
The joint probability of independent observations is the product of their individual probabilities.
The function must be normalized such that the total area under the curve equals 1.

Formulating the Functional Equation

Assuming the most probable value for the mean is the arithmetic mean, the product of the densities must be a function of the sum of squares of the observations.

f (x_{1}) f (x_{2}) ... f (x_{n}) = ϕ (\sum x_{i}^{2})

Note: This is often referred to as Gauss's derivation based on the postulate of the arithmetic mean.

Solving via Logarithmic Differentiation

By taking the natural logarithm of both sides, the product transforms into a sum, which implies that the derivative must be linear, leading to the form f(x) = $C e^{a x^{2}}$ .

ln (f (x_{1})) + ln (f (x_{2})) + ... + ln (f (x_{n})) = ln (ϕ (\sum x_{i}^{2}))

Note: We identify 'a' as negative to ensure the function decays as |x| increases.

Determining Constants

We use the Gaussian integral identity to find the normalization constant C, ensuring the total probability integrates to 1.

\int_{- \infty}^{\infty} C e^{- k (x - μ)^{2}} d x = 1

Note: Recall that the integral of $e^{- x^{2}}$ is the square root of pi.

Final Normalization

Substituting the variance sigma-squared for the spread parameter yields the standard form of the normal PDF.

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

Note: This final form satisfies the property that the distribution is centered at mu with variance sigma-squared.

Result

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

Source: Gauss, C. F. (1809). Theoria motus corporum coelestium.

Free formulas

Rearrangements

Solve for $x$

Make x the subject

x = μ \pm - 2 σ^{2} ln (f (x ∣ μ, σ^{2}) 2 π σ^{2})

Isolate the variable x by taking the natural logarithm and performing algebraic operations.

Difficulty: 3/5

Solve for $μ$

Make $μ$ the subject

μ = x \mp - 2 σ^{2} ln (f 2 π σ^{2})

Solve for the mean by isolating the squared term inside the exponent.

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 the variance by using the Lambert W function or iterative methods as $σ^{2}$ appears in both base and exponent.

Difficulty: 4/5

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

Why it behaves this way

Intuition

Imagine a physical mountain range created by dropping sand on a flat surface. The peak (the mean) is where most of the sand gathers, and the height drops off exponentially as you move away from the center. The curve is a 'gravity-weighted' shape where the steepness of the slopes is controlled by the spread of the sand; a wide pile (large variance) is gentle, while a tall, skinny spike (small variance) is steep.

f (x ∣ μ, σ^{2})

Probability Density Function

The 'height' of the curve at any point x, representing the relative likelihood of encountering a specific value given the average and spread.

μ

Mean (Expected Value)

The central anchor point or the horizontal location of the peak of the bell curve.

σ^{2}

Variance

The 'width' factor; it dictates how broadly the data is scattered away from the center.

e

Euler's Number

Acts as the base for the decay, ensuring that the probability diminishes smoothly and predictably as we move away from the mean.

Signs and relationships

-(x - μ)²: The negative sign ensures the exponent is always negative or zero, creating a peak at the mean (where x=μ) and causing the function to decay toward zero as x moves away from the mean.
1 / sqrt(2πσ²): This is the 'normalization constant.' It ensures the total area under the entire curve is exactly 1, representing 100% total probability.

One free problem

Practice Problem

For a normal distribution with a mean (μ) of 0 and a variance (σ²) of 1, calculate the density f(x) at x = 0.

Random Variable0

Mean0

sigmaSq1

Solve for: $f (x ∣ μ, σ^{2})$

Hint: Recall that $e^{0}$ = 1 and the expression simplifies to 1/sqrt(2π).

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

The heights of adult men in a specific population, which cluster around an average height with a predictable standard deviation.

Study smarter

Tips

Remember that the total area under the curve is always exactly 1.
Use the standard normal distribution (Z-score) by setting μ=0 and σ=1 to simplify complex calculations.
Note that approximately 68%, 95%, and 99.7% of data fall within 1, 2, and 3 standard deviations of the mean, respectively.

Avoid these traps

Common Mistakes

Confusing standard deviation (σ) with variance (σ²).
Assuming the PDF value is a probability itself, rather than a density (the probability of an exact point is 0).

Common questions

Frequently Asked Questions

The normal distribution is derived from the requirement that the maximum likelihood estimator for a mean of independent observations is the arithmetic mean, leading to Gauss's functional equation.

Use this to model physical, biological, or social phenomena where data points cluster around a central average with symmetric deviations.

It allows for the calculation of probabilities, hypothesis testing, and the estimation of parameters in nearly all scientific and engineering fields.

Confusing standard deviation (σ) with variance (σ²). Assuming the PDF value is a probability itself, rather than a density (the probability of an exact point is 0).

The heights of adult men in a specific population, which cluster around an average height with a predictable standard deviation.

Remember that the total area under the curve is always exactly 1. Use the standard normal distribution (Z-score) by setting μ=0 and σ=1 to simplify complex calculations. Note that approximately 68%, 95%, and 99.7% of data fall within 1, 2, and 3 standard deviations of the mean, respectively.

References

Sources

Feller, W. (1968). An Introduction to Probability Theory and Its Applications.
Ross, S. M. (2014). A First Course in Probability.
Gauss, C. F. (1809). Theoria motus corporum coelestium.

Normal Distribution Probability Density Function (PDF)

Overview

Variables

Derivation

Formulating the Functional Equation

Solving via Logarithmic Differentiation

Determining Constants

Final Normalization

Rearrangements

Intuition

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Standard Normal Distribution

Frequently Asked Questions

Sources