Probability DistributionsTopic #15 of 33

Normal Distribution

The bell curve: properties, standard normal distribution, z-scores, and the 68-95-99.7 rule.

Overview

The normal distribution (Gaussian distribution) is the most important continuous probability distribution. It describes many natural phenomena and is fundamental to statistical inference.

Probability Density Function

f(x)=1σ2πe(xμ)22σ2f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}

Parameters

  • μ\mu (mu): Mean (center of the distribution)
  • σ\sigma (sigma): Standard deviation (spread)

Notation

XN(μ,σ2)X \sim N(\mu, \sigma^2)

or

XNormal(μ,σ)X \sim \text{Normal}(\mu, \sigma)

Properties

PropertyValue
Meanμ\mu
Medianμ\mu
Modeμ\mu
Varianceσ2\sigma^2
Standard Deviationσ\sigma
Skewness0
Kurtosis (excess)0

The Bell Curve

        ╭───────╮
       ╱         ╲
      ╱           ╲
     ╱             ╲
────╱───────────────╲────
         μ
    ←  σ  →  ←  σ  →
  • Symmetric about the mean
  • Asymptotic to x-axis (never touches)
  • Total area under curve = 1

The Empirical Rule (68-95-99.7)

For normally distributed data:

P(μσ<X<μ+σ)68%P(\mu - \sigma < X < \mu + \sigma) \approx 68\% P(μ2σ<X<μ+2σ)95%P(\mu - 2\sigma < X < \mu + 2\sigma) \approx 95\% P(μ3σ<X<μ+3σ)99.7%P(\mu - 3\sigma < X < \mu + 3\sigma) \approx 99.7\%
RangeApproximate %
μ±1σ\mu \pm 1\sigma68.27%
μ±2σ\mu \pm 2\sigma95.45%
μ±3σ\mu \pm 3\sigma99.73%

Standard Normal Distribution

The standard normal has μ=0\mu = 0 and σ=1\sigma = 1:

ZN(0,1)Z \sim N(0, 1)

Standardization

Convert any normal to standard normal:

Z=XμσZ = \frac{X - \mu}{\sigma}

Convert back:

X=μ+ZσX = \mu + Z \cdot \sigma

Working with Normal Distributions

Finding Probabilities

  1. Standardize: Z=(Xμ)/σZ = (X - \mu) / \sigma
  2. Use Z-table or calculator
  3. Adjust for direction (left, right, between)

Finding Values from Probabilities

  1. Find ZZ from the probability
  2. Convert: X=μ+ZσX = \mu + Z \cdot \sigma

Z-Table Reference

ZZP(Z<z)P(Z < z)ZZP(Z<z)P(Z < z)
-3.00.00130.00.5000
-2.00.02281.00.8413
-1.00.15872.00.9772
-0.50.30853.00.9987

Examples

Example 1: Finding Probability

Test scores are N(70,10)N(70, 10). Find P(X<85)P(X < 85).

Z=857010=1.5Z = \frac{85 - 70}{10} = 1.5 P(Z<1.5)=0.9332P(Z < 1.5) = 0.9332

Example 2: Greater Than

Same distribution. Find P(X>60)P(X > 60).

Z=607010=1.0Z = \frac{60 - 70}{10} = -1.0 P(Z>1.0)=1P(Z<1.0)=10.1587=0.8413P(Z > -1.0) = 1 - P(Z < -1.0) = 1 - 0.1587 = 0.8413

Example 3: Between Two Values

Find P(65<X<80)P(65 < X < 80).

Z1=657010=0.5Z_1 = \frac{65 - 70}{10} = -0.5 Z2=807010=1.0Z_2 = \frac{80 - 70}{10} = 1.0 P(0.5<Z<1.0)=P(Z<1.0)P(Z<0.5)=0.84130.3085=0.5328P(-0.5 < Z < 1.0) = P(Z < 1.0) - P(Z < -0.5) = 0.8413 - 0.3085 = 0.5328

Example 4: Finding a Percentile

Find the score at the 90th percentile.

P(Z<z)=0.90z1.28P(Z < z) = 0.90 \quad \Rightarrow \quad z \approx 1.28 X=μ+Zσ=70+1.28×10=82.8X = \mu + Z \cdot \sigma = 70 + 1.28 \times 10 = 82.8

Linear Combinations

If XN(μX,σX2)X \sim N(\mu_X, \sigma_X^2) and YN(μY,σY2)Y \sim N(\mu_Y, \sigma_Y^2) are independent:

aX+bYN(aμX+bμY,a2σX2+b2σY2)aX + bY \sim N(a\mu_X + b\mu_Y, a^2\sigma_X^2 + b^2\sigma_Y^2)

Why So Important?

  1. Many natural phenomena are approximately normal
  2. Central Limit Theorem: sample means are normally distributed
  3. Basis for many statistical tests
  4. Convenient mathematical properties