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Normal Distribution

Master it step by step with milestone learning
probability
normal distribution
continuous distribution
gaussian distribution
68-95-99.7 rule
Try this concept yourself
한국어
What you will be able to do
Foundations
  • • Understand the basic properties of the normal distribution and the 68-95-99.7 rule
  • • Recognize real-life situations where the normal distribution applies
  • • Explain the symmetry and bell shape of the normal distribution
Deeper study
  • • Understand the normal probability density function and standardize a variable
  • • Compute probabilities using the standard normal table
  • • Apply the normal distribution in fields such as quality control and exam scoring
Real-world practice
  • • Understand the central limit theorem and the multivariate normal distribution
  • • Use variants and extensions of the normal distribution appropriately
  • • Carry out normal-distribution-based quality control and statistical inference in practice
At a glance

The normal distribution is a symmetric, bell-shaped continuous probability distribution centered on its mean μ, with probability density function f(x) = (1/√(2πσ²))e^(−(x−μ)²/2σ²); it is completely determined by two parameters, the mean μ and the standard deviation σ.

You can verify this hands-on in the NumLM interactive lab (Korean).

한국어

Learning roadmap

Click a step to view its content

Foundations
Definitions and core principles
About 30 min
Open
Deeper study
Derivations and worked examples
About 40 min
Open
Real-world practice
Real cases and advanced applications
About 50 min

Progress (completed levels)

0%

Basic: What Is the Normal Distribution?

Difficulty 2/5
About 30 min

The normal distribution is a symmetric, bell-shaped probability distribution and the one you will meet most often in nature. It is symmetric about its mean and has a special property known as the 68-95-99.7 rule.

Key points
  • A symmetric, bell-shaped distribution
  • Mean, median, and mode are all equal
  • Symmetric about the mean
  • The 68-95-99.7 rule applies
A simple example

If adult male height follows a normal distribution with mean 175 cm and standard deviation 7 cm, then 68% of men are between 168 and 182 cm, 95% between 161 and 189 cm, and 99.7% between 154 and 196 cm.

Check your understanding

Answer:

About 68% of the data lies within the mean ± 1 standard deviation. This is the first part of the 68-95-99.7 rule.

Answer:

Because many natural phenomena follow it, and because the central limit theorem says that sample means converge to a normal distribution regardless of the population's shape.

The probability density function of the normal distribution N(μ,σ²) is f(x) = (1/√(2πσ²))e^(-(x-μ)²/2σ²). Any normal distribution can be converted to the standard normal distribution N(0,1) by standardizing, which is what lets you compute probabilities from a single table.

Mathematical formula

Here μ is the mean, σ is the standard deviation, π is pi, and e is Euler's number. Z is the standardized random variable, which follows the standard normal distribution.

Learn through examples
Example 1: Analyzing Exam Scores

Math scores on a national college-entrance exam (the Korean CSAT) follow a normal distribution with mean 65 and standard deviation 15

N(65, 15²): probability of scoring 80 or higher
Z = (80-65)/15 = 1
P(X≥80) = P(Z≥1) = 1 - 0.8413 = 0.1587 (15.87%)

Answer:

The probability of scoring 80 or higher is about 15.9%.

A score of 80 sits exactly 1σ above the mean, so by the 68% rule, 32% of scores fall outside ±1σ and the upper half of that, about 16%, is the answer. When the Z-value lands on a round number you can do this in your head without a table. Bear in mind, though, that real exam scores are rarely perfectly normal because of the maximum-score ceiling and skewness, so the approximation error grows the further into the tail you go.

Example 2: Quality Control

Product weight follows a normal distribution with mean 500 g and standard deviation 10 g

N(500, 10²): probability of falling between 480 g and 520 g
Z₁ = (480-500)/10 = -2, Z₂ = (520-500)/10 = 2
P(480≤X≤520) = P(-2≤Z≤2) = 0.9545 (95.45%)

Answer:

About 95.5% of products fall within ±2σ.

Do not confuse the 95.45% at ±2σ with the 95% used for confidence intervals (±1.96σ); they are close but not the same. Flip it around and about 4.5% of products, roughly 1 in 22, fall outside 480–520 g. Setting specification limits at ±2σ means accepting a 4.5% defect rate, so to lower defects you have to shrink the process standard deviation itself, not adjust the inspection criteria.

Example 3: Interpreting Medical Data

Blood pressure follows a normal distribution with mean 120 mmHg and standard deviation 20 mmHg

N(120, 20²): probability of 140 mmHg or higher (the hypertension threshold)
Z = (140-120)/20 = 1
P(X≥140) = P(Z≥1) = 0.1587 (15.87%)

Answer:

About 15.9% of people meet the hypertension threshold.

In this distribution, 140 mmHg is exactly mean + 1σ, so 16% is the natural share above that point, and exceeding the threshold does not by itself mean the number of cases is "abnormally high." Lower the cutoff to 130 mmHg and P(Z ≥ 0.5) ≈ 30.9%, nearly doubling the affected group. Prevalence statistics depend on where you draw the line as much as on the distribution itself.

Case study: A Manufacturer's Quality-Control System
Background

An automotive parts manufacturer builds a quality-control system based on the normal distribution to monitor the diameter of engine cylinders and minimize the defect rate.

The problem

The plant must meet tight tolerance requirements while maintaining consistent quality in high-volume production.

The data

Target diameter: 85.0 mm, tolerance: ±0.3 mm, with actual measurements following N(85.0, 0.1²). Daily output of 10,000 units, with 500 units sampled for inspection

Method

Statistical quality control built on normal-distribution theory: Six Sigma methodology, real-time process monitoring with X̄-R control charts, and process capability index (Cp, Cpk) analysis

Solution and results

1) Quality specification: 84.7 mm ≤ diameter ≤ 85.3 mm
2) Defect rate: P(X < 84.7) + P(X > 85.3)
Z₁ = (84.7-85.0)/0.1 = -3, Z₂ = (85.3-85.0)/0.1 = 3
Defect rate = 2 × P(Z < -3) = 2 × 0.0013 = 0.27%
3) 6σ quality: 99.73% yield achieved
4) Process capability index: Cp = 0.3/(3×0.1) = 1.0 (adequate)
5) Real-time monitoring: X̄-R control charts track process stability

A defect rate of 0.27%, saving roughly 200 million KRW per year. Customer complaints fell 90%, rework fell 95%, and the process stability index improved from 1.33 to 1.67

Conclusion:

Quality control grounded in the normal distribution achieved 6σ quality, directly improving customer satisfaction and cutting costs. Statistical process control made it possible to prevent problems before they occurred.

References
  • Casella, G. & Berger, R. L. (2002), "Statistical Inference", 2nd ed., Duxbury Press, Pacific Grove — standard mathematical statistics text covering the normal density, standardization Z=(X−μ)/σ, and mean/variance properties
  • NIST/SEMATECH (2012), "e-Handbook of Statistical Methods" (NIST Handbook 151), https://www.itl.nist.gov/div898/handbook/ (DOI 10.18434/M32189) — authoritative open reference on normal-distribution properties and statistical quality-control practice, including process capability indices (Cp, Cpk) and control charts
  • Montgomery, D. C. (2019), "Introduction to Statistical Quality Control", 8th ed., Wiley — methodological basis for the quality-control case study: X̄-R control charts, Six Sigma, and process capability analysis
  • Stigler, S. M. (1986), "The History of Statistics: The Measurement of Uncertainty before 1900", Belknap Press of Harvard University Press, Cambridge, MA — authoritative history of how Gauss and Laplace established the normal distribution as the law of errors
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