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Description

Editorial

Gaussian Probability Density Evaluator

MEDIUM20 pts

The Gaussian distribution, commonly known as the bell curve or normal distribution, is arguably the most important probability distribution in all of statistics and machine learning. It describes the behavior of countless natural phenomena—from human heights and measurement errors to the distribution of intelligent behaviors in neural networks.

At the heart of the Gaussian distribution lies the Probability Density Function (PDF), which quantifies the relative likelihood of a continuous random variable taking on a specific value. For a Gaussian distribution with mean μ (mu) and standard deviation σ (sigma), the PDF at a point x is given by:

$$f(x; \mu, \sigma) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x - \mu}{\sigma}\right)^2}$$

Breaking Down the Formula:

1 / (σ√(2π)): This is the normalization constant that ensures the total area under the curve equals 1 (a fundamental requirement for any probability distribution)
(x - μ) / σ: This is the z-score, measuring how many standard deviations the point x is from the mean
e^(-½z²): The Gaussian kernel, which produces the characteristic bell shape—values closer to the mean have higher density, with symmetric exponential decay as you move away

Key Properties:

The distribution is symmetric around the mean (μ)
Approximately 68% of values fall within one standard deviation of the mean
Approximately 95% of values fall within two standard deviations
Approximately 99.7% of values fall within three standard deviations (the empirical rule)
The PDF value represents density, not probability—you must integrate over an interval to get actual probabilities

Your Task: Implement a function that computes the PDF value of a Gaussian distribution at a given point x, given the distribution's mean and standard deviation. Round your result to 5 decimal places for consistent floating-point comparison.

Important Note: The standard deviation (σ) will always be positive. A standard deviation of exactly 0 is mathematically undefined for the Gaussian distribution.

Example

Input

x = 16.0, mean = 15.0, std_dev = 2.04

Output

0.17342

Explanation

For this Gaussian distribution centered at μ = 15.0 with σ = 2.04:

Calculate the z-score: z = (16.0 - 15.0) / 2.04 = 0.4902
Compute the exponent: -0.5 × z² = -0.5 × 0.2403 = -0.1201
Evaluate the Gaussian kernel: e^(-0.1201) = 0.8868
Apply the normalization constant: 1 / (2.04 × √(2π)) = 0.1955
Multiply: 0.1955 × 0.8868 ≈ 0.17342

The point x = 16.0 is less than half a standard deviation from the mean, so the density is relatively high.

Example

Input

x = 0.0, mean = 0.0, std_dev = 1.0

Output

0.39894

Explanation

This is the standard normal distribution (mean = 0, standard deviation = 1). When x equals the mean:

The z-score is 0: (0.0 - 0.0) / 1.0 = 0
The exponent becomes 0: -0.5 × 0² = 0
The Gaussian kernel equals 1: e^0 = 1
The result is purely the normalization constant: 1 / √(2π) ≈ 0.39894

This is the maximum PDF value for the standard normal distribution, occurring exactly at the mean. The value 1/√(2π) ≈ 0.39894 is a fundamental constant in probability theory.

Example

Input

x = 10.0, mean = 10.0, std_dev = 2.0

Output

0.19947

Explanation

Evaluating the PDF at the mean of a Gaussian with σ = 2.0:

z-score: (10.0 - 10.0) / 2.0 = 0
Since x equals μ, this is the peak of the distribution
The result simplifies to: 1 / (2.0 × √(2π)) = 0.19947

Note that this maximum PDF value (0.19947) is exactly half of the standard normal's peak (0.39894), because the standard deviation is 2 instead of 1. Larger standard deviations create wider, flatter distributions with lower peak densities.

Accepted0/0·0% Acceptance

Constraints

-10⁶ ≤ x ≤ 10⁶
-10⁶ ≤ mean ≤ 10⁶
0 < std_dev ≤ 10⁶ (standard deviation is always positive)
Results should be rounded to exactly 5 decimal places
The function should handle very small PDF values that approach zero for extreme z-scores

Code

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x =

mean =

std_dev =

2.04

Gaussian Probability Density Evaluator

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