The Normal Distribution and the Empirical Rule
Describe the properties of the normal distribution (symmetric, bell-shaped, parameterized by μ and σ). Apply the 68-95-99.7 rule. Determine whether a dataset is approximately normal using graphical and numerical checks.
A lesson of AP Statistics, in Exploring One-Variable Data, about 10 minutes.
The Normal Distribution and Z-Scores
By the end of this lesson, you will be able to calculate and interpret z-scores, apply the 68-95-99.7 rule, use the normal distribution to find probabilities and percentiles, and assess normality.
The Normal Distribution
Normal distribution: a continuous probability distribution that is:
- Bell-shaped and symmetric about the mean
- Described completely by two parameters: mean $\mu$ and standard deviation $\sigma$
- Notation: $N(\mu, \sigma)$
Why the normal distribution matters:
- Many real-world variables are approximately normally distributed
- The Central Limit Theorem makes it central to inference
- It's the foundation of many statistical tests
The 68-95-99.7 Rule (Empirical Rule)
For any Normal distribution $N(\mu, \sigma)$:
- Approximately 68% of observations fall within 1 standard deviation of the mean: $(\mu - \sigma, \mu + \sigma)$
- Approximately 95% of observations fall within 2 standard deviations: $(\mu - 2\sigma, \mu + 2\sigma)$
- Approximately 99.7% of observations fall within 3 standard deviations: $(\mu - 3\sigma, \mu + 3\sigma)$
Example: SAT Math scores are approximately $N(530, 100)$. What percent score between 430 and 630?
$430 = 530 - 100 = \mu - \sigma$; $630 = 530 + 100 = \mu + \sigma$
By the empirical rule: approximately 68%.
Z-Scores (Standardized Scores)
Z-score: measures how many standard deviations an observation is from the mean.
$$z = \frac{x - \mu}{\sigma}$$
- $z > 0$: observation is above the mean
- $z < 0$: observation is below the mean
- $z = 0$: observation equals the mean
Interpretation: "This observation is $z$ standard deviations above/below the mean."
Example: In a distribution with $\mu = 70$ and $\sigma = 15$:
$x = 85 \Rightarrow z = \dfrac{85-70}{15} = 1.0$ → 1 standard deviation above mean