The Normal Distribution | AP Stats Unit 1 Study Guide

Quick Summary

This guide covers the properties and applications of the Normal distribution, the most important distribution in statistics. You will learn to describe a Normal distribution using its mean and standard deviation, apply the Empirical Rule for quick approximations, and calculate z-scores to standardize values. By the end, you will be able to use your calculator to find probabilities and percentiles for any Normal distribution, a foundational skill for the rest of the course.

Key Concepts

The Normal distribution is a continuous probability distribution that is fundamental to statistics. It describes how data for many natural phenomena (like height, weight, or IQ scores) are distributed.

Properties of a Normal Distribution:
- It is symmetric, single-peaked, and bell-shaped.
- The mean (μ), median, and mode are all equal and located at the exact center of the distribution.
- The distribution is defined by two parameters: the mean μ (mu), which sets the center, and the standard deviation σ (sigma), which determines the spread or variability.
- The notation N(μ, σ) is used to describe a Normal distribution with mean μ and standard deviation σ.
- The total area under the curve is exactly 1 (or 100%).
[Image: A bell-shaped curve labeled "Normal Distribution." The peak is labeled with μ. The points μ+σ and μ-σ are marked at the inflection points of the curve.]
The Empirical Rule (The 68-95-99.7 Rule):
This rule is a useful approximation for any distribution that is roughly bell-shaped and symmetric.
- Approximately 68% of observations fall within 1 standard deviation of the mean (μ ± 1σ).
- Approximately 95% of observations fall within 2 standard deviations of the mean (μ ± 2σ).
- Approximately 99.7% of observations fall within 3 standard deviations of the mean (μ ± 3σ).
[Image: A Normal distribution curve with the percentages of the Empirical Rule clearly marked. The central section between μ-σ and μ+σ is labeled "68%". The section between μ-2σ and μ+2σ is labeled "95%". The section between μ-3σ and μ-3σ is labeled "99.7%".]
Z-Scores (Standardizing Values):
A z-score measures the number of standard deviations an observation (x) is from the mean (μ). It standardizes values, allowing us to compare observations from different Normal distributions.
- Formula: $z = (x - μ) / σ$
- A positive z-score means the observation is above the mean.
- A negative z-score means the observation is below the mean.
- A z-score of 0 means the observation is exactly the mean.
The Standard Normal Distribution:
This is a special Normal distribution with a mean of 0 and a standard deviation of 1, denoted as N(0, 1). When we calculate a z-score, we are converting our original distribution into the standard Normal scale.
Finding Probabilities and Proportions:
The area under a Normal curve corresponds to a proportion or probability. While the Empirical Rule works for values exactly 1, 2, or 3 standard deviations from the mean, we need technology for all other values.
- We use a calculator function ( $n or ma l c df$ ) to find the area (probability) between any two boundaries on a Normal distribution.
- For example, to find the proportion of observations less than a value x, we find the area under the curve to the left of x.
Finding Values from Probabilities (Working Backwards):
Sometimes we know the percentile or proportion and need to find the corresponding data value (x). This is the inverse of the process above.
- We use a calculator function ( $in v N or m$ ) which takes a percentile (the area to the left) and returns the corresponding value on the distribution.
- The process is:
  1. Determine the area to the left of the desired value.
  2. Use $in v N or m$ with this area, the mean, and the standard deviation to find the value x.

Key Vocabulary

Normal Distribution: A continuous, symmetric, bell-shaped probability distribution described by its mean (μ) and standard deviation (σ).
Empirical Rule (68-95-99.7 Rule): A rule stating the approximate percentage of data that falls within 1, 2, and 3 standard deviations of the mean in a bell-shaped distribution.
Z-score: A standardized value that indicates how many standard deviations an observation is from the mean of its distribution.
Standard Normal Distribution: A specific Normal distribution with a mean of 0 and a standard deviation of 1.
Percentile: The value at or below which a given percentage of observations in a group of observations falls. For example, the 80th percentile is the value greater than or equal to 80% of the other values.

Calculator Tech (TI-84)

Calculations for the Normal distribution are performed using the DISTR (Distributions) menu.

$n or ma l c df ()$ : Finds Probability/Area
This function calculates the area (proportion or probability) under the Normal curve between two boundaries.
- Keystrokes:2nd -> $V A RS [D I STR]$ -> 2: normalcdf()
- Syntax: $n or ma l c df (l o w er, u pp er, μ, σ)$
- Inputs:
  - $l o w er$ : The lower boundary of the area you want to find. If you want all the area to the left of a value, use a very small number like -1E99 (type $- 1$ , then 2nd, then $,$ for EE, then $99$ ).
  - $u pp er$ : The upper boundary. If you want all the area to the right of a value, use a very large number like 1E99.
  - $μ$ : The mean of the distribution.
  - $σ$ : The standard deviation of the distribution.
- Example: To find P(X < 85) for a distribution N(100, 15): $n or ma l c df (- 1 E 99, 85, 100, 15)$
$in v N or m ()$ : Finds Value from Percentile/Area
This function calculates the boundary value (x-value or z-score) corresponding to a given percentile (area to the left).
- Keystrokes:2nd -> $V A RS [D I STR]$ -> 3: invNorm()
- Syntax: $in v N or m (a re a, μ, σ)$
- Inputs:
  - $a re a$ : The area to the left of the value you want to find. This must be a value between 0 and 1.
  - $μ$ : The mean of the distribution.
  - $σ$ : The standard deviation of the distribution.
- Example: To find the value for the 90th percentile (top 10%) for N(100, 15): $in v N or m (0.90, 100, 15)$

How to Show Work on the FRQ

To earn full credit on Free Response Questions involving Normal distribution calculations, you must clearly communicate your process. Use the following 4-step structure.

Template for Normal Calculations:

State the Distribution and Parameters: Define the variable of interest and state that it follows a Normal distribution with a specific mean and standard deviation.
- Example: "Let X = the score on the AP exam. The scores are Normally distributed with a mean μ = 100 and standard deviation σ = 15, or N(100, 15)."
State the Probability of Interest: Write the probability you are trying to find in proper notation.
- Example: "We want to find the probability that a randomly selected student scores above 120, P(X > 120)."
Show the Calculation: Demonstrate your understanding by showing the standardization (z-score calculation). This is crucial even if you use the calculator's μ and σ inputs directly.
- Example: $z = (120 - 100) /15 = 1.33$ . Then, restate the probability in terms of Z: $P (Z > 1.33)$ .
- Alternative (also acceptable): Label your calculator inputs clearly: normalcdf(lower: 120, upper: 1E99, μ: 100, σ: 15). The z-score method is generally preferred as it shows more statistical understanding.
Answer in Context: State your final numerical answer and interpret it in the context of the problem.
- Example: "The probability that a randomly selected student scores above 120 is 0.0918. There is about a 9.18% chance of a student scoring above 120."

Practice Problems

Problem 1:

The scores on a standardized test are approximately Normally distributed with a mean of 500 and a standard deviation of 100. What proportion of students score between 450 and 620?

Solution:

State the Distribution and Parameters: Let X = the score on the standardized test. The scores follow a Normal distribution with μ = 500 and σ = 100, or N(500, 100).
State the Probability of Interest: We want to find the proportion of students scoring between 450 and 620, which is P(450 < X < 620).
Show the Calculation: We calculate the z-scores for both boundaries.
- For x = 450: $z = (450 - 500) /100 = - 0.5$
- For x = 620: $z = (620 - 500) /100 = 1.2$
- The probability is P(-0.5 < Z < 1.2).
- Using the calculator: normalcdf(lower: 450, upper: 620, μ: 500, σ: 100) = 0.5763.
Answer in Context: The proportion of students who score between 450 and 620 on this standardized test is approximately 0.5763.

Problem 2:

The heights of adult women in the United States are approximately Normally distributed with a mean of 64.5 inches and a standard deviation of 2.5 inches. To be a member of a "Tall Club," a woman must be in the top 5% of heights. What is the minimum height required to join the club?