Chapter Notes: Normal Distributions
In the real world, many things vary in predictable patterns. Heights of adults, test scores, measurement errors, and even the weights of apples from an orchard tend to cluster around an average value, with fewer and fewer observations as you move farther away from that average. The normal distribution, also called the Gaussian distribution or bell curve, is a mathematical model that describes this pattern precisely. Understanding normal distributions allows you to make predictions, calculate probabilities, and interpret data in fields ranging from biology and psychology to quality control and economics.
What Is a Normal Distribution?
A normal distribution is a continuous probability distribution that is symmetric and bell-shaped. It is completely determined by two parameters: the mean (denoted by the Greek letter μ, pronounced "mu") and the standard deviation (denoted by the Greek letter σ, pronounced "sigma").
The mean μ tells you the center of the distribution-the value around which data tends to cluster. The standard deviation σ measures how spread out the data is. A small σ means the data is tightly packed around the mean; a large σ means the data is more spread out.
We write that a variable \( X \) follows a normal distribution as:
\[ X \sim N(\mu, \sigma^2) \]The symbol \( \sim \) means "is distributed as," and \( N(\mu, \sigma^2) \) indicates a normal distribution with mean μ and variance σ².
Key Characteristics of the Normal Distribution
- Symmetric: The left and right sides of the curve are mirror images of each other.
- Bell-shaped: The curve rises to a single peak at the mean and falls off smoothly in both directions.
- Mean = Median = Mode: All three measures of center coincide at the peak of the curve.
- Total area under the curve = 1: This represents 100% probability.
- Asymptotic: The tails of the curve approach, but never touch, the horizontal axis. This means extreme values are possible but increasingly rare.
The Probability Density Function
The shape of a normal distribution is defined mathematically by its probability density function (PDF):
\[ f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2\sigma^2}} \]In this formula:
- \( x \) is the value of the variable
- \( \mu \) is the mean
- \( \sigma \) is the standard deviation
- \( \pi \approx 3.14159 \)
- \( e \approx 2.71828 \) (the base of natural logarithms)
You do not need to compute this formula by hand in practice. Instead, you will use tables, calculators, or software to find probabilities. However, understanding that the formula exists helps you appreciate why the distribution has its characteristic shape.
The Standard Normal Distribution
A special case of the normal distribution is the standard normal distribution, which has mean μ = 0 and standard deviation σ = 1. We write this as:
\[ Z \sim N(0, 1) \]The variable \( Z \) is often called a z-score or standard score. The standard normal distribution is extremely useful because any normal distribution can be converted to it using a simple transformation.
Converting to Z-Scores
To convert a value \( x \) from a normal distribution with mean μ and standard deviation σ to a z-score, use the formula:
\[ z = \frac{x - \mu}{\sigma} \]This formula tells you how many standard deviations \( x \) is above or below the mean. A positive z-score means \( x \) is above the mean; a negative z-score means \( x \) is below the mean.
Example: The heights of adult women in a population are normally distributed with a mean of 65 inches and a standard deviation of 3 inches.
What is the z-score for a woman who is 71 inches tall?
Solution:
Given: μ = 65 inches, σ = 3 inches, x = 71 inches
Use the z-score formula:
\( z = \frac{x - \mu}{\sigma} = \frac{71 - 65}{3} = \frac{6}{3} = 2 \)
The z-score is 2, meaning this woman's height is 2 standard deviations above the mean.
The Empirical Rule (68-95-99.7 Rule)
For any normal distribution, a simple rule describes how data is spread around the mean. This is called the Empirical Rule or the 68-95-99.7 Rule:
- Approximately 68% of the data falls within 1 standard deviation of the mean (between μ - σ and μ + σ).
- Approximately 95% of the data falls within 2 standard deviations of the mean (between μ - 2σ and μ + 2σ).
- Approximately 99.7% of the data falls within 3 standard deviations of the mean (between μ - 3σ and μ + 3σ).
This rule is a powerful tool for quickly estimating probabilities without using tables or technology.
Example: SAT math scores are normally distributed with a mean of 500 and a standard deviation of 100.
What percentage of students score between 400 and 600?
Solution:
400 is one standard deviation below the mean: 500 - 100 = 400
600 is one standard deviation above the mean: 500 + 100 = 600
By the Empirical Rule, approximately 68% of students score between 400 and 600.
Example: IQ scores are normally distributed with a mean of 100 and a standard deviation of 15.
What percentage of people have IQ scores between 70 and 130?
Solution:
70 is two standard deviations below the mean: 100 - 2(15) = 100 - 30 = 70
130 is two standard deviations above the mean: 100 + 2(15) = 100 + 30 = 130
By the Empirical Rule, approximately 95% of people have IQ scores between 70 and 130.
Finding Probabilities Using the Standard Normal Table
The standard normal table (also called the z-table) provides the area under the standard normal curve to the left of a given z-score. This area represents the probability that \( Z \) is less than or equal to that z-score.
Most z-tables give \( P(Z \leq z) \), the cumulative probability from the far left tail up to \( z \).
Reading the Z-Table
To find \( P(Z \leq z) \):
- Round your z-score to two decimal places.
- Find the row corresponding to the first two digits (including the sign and first decimal place).
- Find the column corresponding to the second decimal place.
- The intersection gives the cumulative probability.
Example: Find the probability that a standard normal variable \( Z \) is less than 1.25.
What is \( P(Z \leq 1.25) \)?
Solution:
Locate z = 1.25 in the standard normal table.
Row for 1.2, column for 0.05.
The table gives approximately 0.8944.
Therefore, \( P(Z \leq 1.25) = \) 0.8944, or about 89.44%.
Finding Probabilities for Z Greater Than a Value
Because the total area under the curve is 1, we have:
\[ P(Z > z) = 1 - P(Z \leq z) \]Example: Find the probability that \( Z \) is greater than 0.75.
What is \( P(Z > 0.75) \)?
Solution:
From the z-table, \( P(Z \leq 0.75) \approx 0.7734 \).
Use the complement rule:
\( P(Z > 0.75) = 1 - 0.7734 = 0.2266 \)
Therefore, \( P(Z > 0.75) = \) 0.2266, or about 22.66%.
Finding Probabilities Between Two Values
To find \( P(a < z="">< b)="" \),="">
\[ P(a < z="">< b)="P(Z" \leq="" b)="" -="" p(z="" \leq="" a)="" \]="">Example: Find the probability that \( Z \) is between -1.0 and 1.5.
What is \( P(-1.0 < z="">< 1.5)="">
Solution:
From the z-table, \( P(Z \leq 1.5) \approx 0.9332 \).
From the z-table, \( P(Z \leq -1.0) \approx 0.1587 \).
\( P(-1.0 < z="">< 1.5)="0.9332" -="" 0.1587="0.7745">
Therefore, \( P(-1.0 < z="">< 1.5)="\)">0.7745, or about 77.45%.
Finding Probabilities for Any Normal Distribution
When working with a normal distribution that is not standard (μ ≠ 0 or σ ≠ 1), convert the values to z-scores first, then use the z-table.
Example: The amount of soda dispensed by a machine is normally distributed with a mean of 12 ounces and a standard deviation of 0.5 ounces.
What is the probability that a randomly selected cup contains more than 13 ounces?
Solution:
Given: μ = 12, σ = 0.5, x = 13
Convert to a z-score: \( z = \frac{13 - 12}{0.5} = \frac{1}{0.5} = 2.0 \)
From the z-table, \( P(Z \leq 2.0) \approx 0.9772 \).
\( P(Z > 2.0) = 1 - 0.9772 = 0.0228 \)
The probability that a cup contains more than 13 ounces is 0.0228, or about 2.28%.
Finding Values Given Probabilities (Inverse Normal Problems)
Sometimes you know the probability and need to find the corresponding value or z-score. This is called an inverse normal problem or finding a percentile.
Finding a Z-Score from a Percentile
To find the z-score corresponding to a cumulative probability \( P(Z \leq z) = p \):
- Locate the probability \( p \) inside the body of the z-table.
- Read the corresponding z-score from the row and column headers.
Example: Find the z-score such that 90% of the standard normal distribution lies to the left of it.
What is the value of \( z \) such that \( P(Z \leq z) = 0.90 \)?
Solution:
Look in the body of the z-table for the probability closest to 0.90.
The closest value is 0.8997, corresponding to z = 1.28.
Alternatively, 0.9015 corresponds to z = 1.29.
Using linear interpolation or a calculator, the z-score is approximately 1.28.
Finding Values in Any Normal Distribution
To find a value \( x \) given a probability when \( X \sim N(\mu, \sigma^2) \):
- Find the corresponding z-score using the z-table.
- Convert the z-score back to the original scale using \( x = \mu + z \sigma \).
Example: Test scores are normally distributed with a mean of 75 and a standard deviation of 10.
What score represents the 85th percentile?Find the score such that 85% of students score below it.
Solution:
Find the z-score such that \( P(Z \leq z) = 0.85 \).
From the z-table, the closest probability is 0.8508, corresponding to z ≈ 1.04.
Convert to the original scale: \( x = \mu + z\sigma = 75 + 1.04(10) = 75 + 10.4 = 85.4 \)
The 85th percentile score is approximately 85.4.
Applications of the Normal Distribution
Normal distributions appear throughout statistics and real-world phenomena. Here are some common applications:
Quality Control
Manufacturing processes often produce items whose measurements (length, weight, diameter) vary normally around a target value. Companies use normal distribution models to set acceptable tolerance ranges and identify defective products.
For example, if bolts produced by a machine have lengths normally distributed with a mean of 5.0 cm and a standard deviation of 0.1 cm, a quality control engineer can calculate the percentage of bolts that fall outside acceptable limits, say between 4.8 cm and 5.2 cm.
Standardized Testing
Many standardized tests (SAT, ACT, IQ tests) are designed so that scores follow a normal distribution. This allows educators and psychologists to compare individual performance to the population and identify percentiles.
Biological and Physical Measurements
Heights, weights, blood pressure readings, and many biological traits in large populations approximate normal distributions. This enables researchers to model populations and make predictions.
Error Analysis
Measurement errors in scientific experiments often follow a normal distribution centered at zero. Understanding this helps scientists quantify uncertainty and improve experimental design.
Important Properties and Theorems
Linear Transformations of Normal Variables
If \( X \sim N(\mu, \sigma^2) \) and you apply a linear transformation \( Y = aX + b \), where \( a \) and \( b \) are constants, then \( Y \) is also normally distributed:
\[ Y \sim N(a\mu + b, a^2\sigma^2) \]This property means that adding a constant shifts the mean, and multiplying by a constant scales both the mean and the standard deviation.
Sums of Independent Normal Variables
If \( X_1 \sim N(\mu_1, \sigma_1^2) \) and \( X_2 \sim N(\mu_2, \sigma_2^2) \) are independent, then their sum is also normal:
\[ X_1 + X_2 \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2) \]Notice that variances add, not standard deviations.
Example: Two independent machines produce parts.
Machine A produces parts with weights \( N(50, 4) \) grams.
Machine B produces parts with weights \( N(30, 9) \) grams.What is the distribution of the total weight of one part from each machine?
Solution:
Let \( X_A \sim N(50, 4) \) and \( X_B \sim N(30, 9) \).
Total weight \( W = X_A + X_B \).
Mean: \( \mu_W = 50 + 30 = 80 \) grams
Variance: \( \sigma_W^2 = 4 + 9 = 13 \), so \( \sigma_W = \sqrt{13} \approx 3.61 \) grams
The total weight is distributed as \( N(80, 13) \), with mean 80 grams and standard deviation approximately 3.61 grams.
Common Mistakes and How to Avoid Them
- Confusing standard deviation with variance: Remember that variance is σ², the square of the standard deviation. When adding independent normal variables, you add variances, not standard deviations.
- Using the wrong tail: Pay careful attention to whether a problem asks for "less than," "greater than," or "between." Sketch a normal curve and shade the area you're finding to help visualize.
- Forgetting to convert to z-scores: Always standardize values using \( z = \frac{x - \mu}{\sigma} \) before using the z-table, unless you're already working with the standard normal distribution.
- Rounding too early: Keep extra decimal places during calculations and round only your final answer to avoid accumulation of rounding errors.
Technology and the Normal Distribution
While z-tables are useful for understanding the mechanics of normal distribution calculations, modern practice relies heavily on technology. Graphing calculators, spreadsheet software (like Excel or Google Sheets), and statistical programs (such as R or Python) have built-in functions to find normal probabilities and inverse values quickly and accurately.
For example, many calculators have functions like normalcdf(lower, upper, μ, σ) to find probabilities and invNorm(probability, μ, σ) to find values corresponding to percentiles.
Learning to use these tools efficiently is an important skill, but understanding the underlying concepts ensures you can interpret results correctly and catch errors.
Summary of Key Ideas
The normal distribution is a fundamental model in statistics characterized by its symmetric, bell-shaped curve. It is completely determined by its mean μ and standard deviation σ. The standard normal distribution \( N(0, 1) \) serves as a reference, and any normal variable can be converted to a z-score for easier probability calculations.
The Empirical Rule provides quick estimates: about 68% of data lies within one standard deviation, 95% within two, and 99.7% within three. For more precise probabilities, we use z-tables or technology.
Understanding normal distributions allows you to model real-world variation, make informed predictions, and interpret data across countless applications in science, business, education, and beyond.
