NORMAL DISTRIBUTION

Tags: #◇Computer Science #◇STT101A #◇Conceptual #◇Normal Distribution

---

◇Overview

The Normal (Gaussian) Distribution, denoted $X \sim \mathcal{N}(\mu, \sigma^2)$, is the most important continuous probability distribution in statistics. Its bell-shaped probability density function is symmetric around the mean $\mu$, and by the Central Limit Theorem (CLT), the sum or sample mean of many independent random variables converges toward a Normal distribution.

◇Key Definitions

• Probability Density Function (PDF):
  $f(x) = \frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{(x - \mu)^2}{2\sigma^2}\right), \quad -\infty < x < \infty$
• Standard Normal Distribution ($Z \sim \mathcal{N}(0, 1)$): Standardized via the Z-score transformation $Z = \frac{X - \mu}{\sigma}$, with CDF $\Phi(z) = P(Z \le z)$ satisfying symmetry $\Phi(-z) = 1 - \Phi(z)$.
• Empirical ($68\text{--}95\text{--}99.7$) Rule:
  • $P(\mu - 1\sigma \le X \le \mu + 1\sigma) \approx 0.6827$
  • $P(\mu - 2\sigma \le X \le \mu + 2\sigma) \approx 0.9545$
  • $P(\mu - 3\sigma \le X \le \mu + 3\sigma) \approx 0.9973$

◇Worked Example

Problem: Network round-trip latency follows $X \sim \mathcal{N}(\mu = 50\text{ ms}, \sigma^2 = 100\text{ ms}^2)$, so $\sigma = 10\text{ ms}$. Find $P(30 \le X \le 60)$ given $\Phi(1.0) = 0.8413$ and $\Phi(2.0) = 0.9772$.

Solution:

1. Convert bounds to $Z$-scores: $z_1 = \frac{30 - 50}{10} = -2.0$ and $z_2 = \frac{60 - 50}{10} = +1.0$.
2. $P(-2.0 \le Z \le 1.0) = \Phi(1.0) - \Phi(-2.0) = \Phi(1.0) - (1 - \Phi(2.0)) = 0.8413 - (1 - 0.9772) = 0.8413 - 0.0228 = 0.8185$.

◇References & Navigation

• Previous Topic: ◇Hyper-geometric Distribution
• Next Topic: ◇Hypothesis Testing
• Related Notes: ◇Probability Mass Function