Normal Distributions

Properties and calculations with normal curves

Normal Distributions

Introduction

The normal distribution (also called Gaussian distribution or bell curve) is the most important probability distribution in statistics. Many natural phenomena approximately follow a normal distribution, and it forms the foundation for much of statistical inference.

Characteristics of Normal Distributions

Shape Properties

1. Bell-shaped curve:

  • Symmetric around the center
  • Single peak at the mean
  • Tails extend infinitely in both directions (approaching but never touching the x-axis)

2. Symmetric:

  • Left side mirrors right side
  • Mean = Median = Mode
  • If folded at center, both halves match perfectly

3. Unimodal:

  • Single peak (at the mean)
  • Highest point at center
  • Decreases smoothly on both sides

4. Asymptotic:

  • Tails get closer and closer to x-axis
  • Never actually reach zero
  • Theoretically extends to -\infty and ++\infty

The Normal Curve Equation

Probability density function:

f(x)=1σ2πe12(xμσ)2f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}

Don't memorize this! Just know:

  • Defined by two parameters: μ\mu (mean) and σ\sigma (standard deviation)
  • Shape determined entirely by μ\mu and σ\sigma

Parameters: Mean and Standard Deviation

Mean (μ\mu)

Controls location:

  • Center of distribution
  • Peak of curve
  • Balance point

Changing μ\mu:

  • Shifts distribution left or right
  • Doesn't change shape
  • Doesn't change spread

Example:

  • Distribution A: μ=50\mu = 50, centered at 50
  • Distribution B: μ=70\mu = 70, centered at 70
  • B is shifted 20 units right from A

Standard Deviation (σ\sigma)

Controls spread:

  • How spread out distribution is
  • Width of bell curve
  • Distance from mean to inflection points

Changing σ\sigma:

  • Larger σ\sigma → wider, flatter curve
  • Smaller σ\sigma → narrower, taller curve
  • Doesn't change center
  • Total area under curve stays 1.0

Example:

  • Distribution A: σ=5\sigma = 5, narrow and tall
  • Distribution B: σ=15\sigma = 15, wide and flat
  • Both centered at same μ\mu, but B more spread out

The Empirical Rule (68-95-99.7 Rule)

For normal distributions:

68% of data within 1 standard deviation of mean
μ1σ\mu - 1\sigma to μ+1σ\mu + 1\sigma

95% of data within 2 standard deviations of mean
μ2σ\mu - 2\sigma to μ+2σ\mu + 2\sigma

99.7% of data within 3 standard deviations of mean
μ3σ\mu - 3\sigma to μ+3σ\mu + 3\sigma

Example Application

IQ scores: μ=100\mu = 100, σ=15\sigma = 15

68% of people have IQ between: 10015=85100 - 15 = 85 and 100+15=115100 + 15 = 115

95% of people have IQ between: 10030=70100 - 30 = 70 and 100+30=130100 + 30 = 130

99.7% of people have IQ between: 10045=55100 - 45 = 55 and 100+45=145100 + 45 = 145

Using the Empirical Rule

Quick mental calculations:

Example: Heights of adult males: μ=70\mu = 70 inches, σ=3\sigma = 3 inches

Q: What percentage between 67 and 73 inches?
A: 67 to 73 is μ±1σ\mu \pm 1\sigma68%

Q: What percentage above 76 inches?
A: 76 is μ+2σ\mu + 2\sigma, so 95% are between 64 and 76
Above 76 = (100% - 95%) / 2 = 2.5%

Q: What percentage below 64 inches?
A: 64 is μ2σ\mu - 2\sigma
Below 64 = (100% - 95%) / 2 = 2.5%

The Standard Normal Distribution (Z-distribution)

Definition

Standard Normal: Normal distribution with μ=0\mu = 0 and σ=1\sigma = 1

Denoted: N(0,1)N(0, 1) or Z-distribution

Why it matters:

  • Reference distribution
  • All normal distributions can be standardized to this
  • Tables and calculators use standard normal

Z-Scores

Z-score (standardized score): Number of standard deviations from the mean

Formula: z=xμσz = \frac{x - \mu}{\sigma}

Where:

  • xx = observed value
  • μ\mu = mean
  • σ\sigma = standard deviation

Interpretation:

  • z=0z = 0: At the mean
  • z=1z = 1: One SD above mean
  • z=1z = -1: One SD below mean
  • z=2.5z = 2.5: 2.5 SD above mean

Calculating Z-Scores

Example: Test scores with μ=75\mu = 75, σ=8\sigma = 8

Score of 83: z=83758=88=1z = \frac{83 - 75}{8} = \frac{8}{8} = 1

Score is 1 SD above mean

Score of 67: z=67758=88=1z = \frac{67 - 75}{8} = \frac{-8}{8} = -1

Score is 1 SD below mean

Score of 91: z=91758=168=2z = \frac{91 - 75}{8} = \frac{16}{8} = 2

Score is 2 SD above mean

Using Z-Scores

Purposes:

  1. Standardize different distributions for comparison
  2. Find probabilities using standard normal table
  3. Identify unusual values (typically |z| > 2 or 3)
  4. Compare across different scales

Example comparison:

Student A: Math score 85 (class μ=75\mu = 75, σ=8\sigma = 8)
z=85758=1.25z = \frac{85-75}{8} = 1.25

Student B: English score 88 (class μ=80\mu = 80, σ=5\sigma = 5)
z=88805=1.6z = \frac{88-80}{5} = 1.6

Student B performed better relative to their class (higher z-score)!

Finding Areas Under the Normal Curve

Methods

1. Empirical Rule (for z = ±1, ±2, ±3)

2. Standard Normal Table (z-table)

  • Gives area to LEFT of z-score
  • Also called cumulative probability

3. Calculator

  • normalcdf function
  • More accurate, easier

Using the Table

Area to the left of z:

  • Look up z in table directly
  • Example: z = 1.23 → area = 0.8907
  • Meaning: 89.07% of data below z = 1.23

Area to the right of z:

  • Area to right = 1 - area to left
  • Example: z = 1.23 → area to left = 0.8907
  • Area to right = 1 - 0.8907 = 0.1093 (10.93%)

Area between two z-scores:

  • Find area to left of each
  • Subtract smaller from larger
  • Example: Between z = -1 and z = 1
    • Area left of 1: 0.8413
    • Area left of -1: 0.1587
    • Between: 0.8413 - 0.1587 = 0.6826 (68.26%)

Calculator Method

TI-83/84:

normalcdf(lower, upper, mean, SD)

Examples:

Area between 65 and 75 (μ=70\mu = 70, σ=5\sigma = 5):
normalcdf(65, 75, 70, 5) → 0.6827

Area above 80:
normalcdf(80, 999999, 70, 5) → 0.0228

Area below 60:
normalcdf(-999999, 60, 70, 5) → 0.0228

Finding Values from Areas (Inverse Normal)

The Inverse Problem

Given: Probability (area)
Find: Corresponding x-value or z-score

Example: Find the score such that 90% of students score below it

Calculator Method

TI-83/84:

invNorm(area to left, mean, SD)

Examples:

90th percentile (μ=70\mu = 70, σ=5\sigma = 5):
invNorm(0.90, 70, 5) → 76.4

Meaning: 90% score below 76.4

25th percentile (Q1):
invNorm(0.25, 70, 5) → 66.6

75th percentile (Q3):
invNorm(0.75, 70, 5) → 73.4

Assessing Normality

Why It Matters

Many statistical methods assume normality. We need to check if data is approximately normal before applying these methods.

Methods to Assess Normality

1. Histogram/Dotplot:

  • Look for bell shape
  • Check for symmetry
  • Quick visual check

2. Normal Probability Plot (Q-Q Plot):

  • Plot observed values vs. expected normal values
  • If roughly linear → approximately normal
  • If curved or non-linear → not normal

3. Numerical Checks:

  • Mean ≈ Median (symmetry)
  • Few outliers by 1.5 × IQR rule
  • Most data within μ±2σ\mu \pm 2\sigma

What to Look For

Approximately normal: ✓ Bell-shaped histogram
✓ Linear normal probability plot
✓ Mean ≈ Median
✓ About 68% within 1 SD, 95% within 2 SD

Not normal: ❌ Skewed histogram
❌ Curved normal probability plot
❌ Mean ≠ Median
❌ Many outliers
❌ Gaps or multiple peaks

Common Mistakes

Assuming all data is normal
Many distributions are NOT normal!

Confusing z-scores with original values
z-scores are standardized, no units

Using empirical rule for non-normal data
Only valid for normal distributions

Forgetting to standardize before using table
Must convert to z-scores first

Reading wrong side of table
Most tables give area to LEFT

Not checking normality assumption
Methods based on normality won't work if data isn't normal

Quick Reference

Normal Distribution:

  • Parameters: μ\mu (mean), σ\sigma (SD)
  • Notation: N(μ,σ)N(\mu, \sigma)
  • Properties: Symmetric, bell-shaped, unimodal

Empirical Rule (68-95-99.7):

  • 68% within μ±1σ\mu \pm 1\sigma
  • 95% within μ±2σ\mu \pm 2\sigma
  • 99.7% within μ±3σ\mu \pm 3\sigma

Z-Score:

  • Formula: z=xμσz = \frac{x - \mu}{\sigma}
  • Interpretation: # of SDs from mean
  • Standard normal: μ=0\mu = 0, σ=1\sigma = 1

Calculator Commands:

  • normalcdf(lower, upper, μ, σ) for area/probability
  • invNorm(area, μ, σ) for x-value

Assessing Normality:

  • Histogram: bell-shaped?
  • Normal plot: linear?
  • Mean ≈ Median?

Remember: The normal distribution is powerful but not universal. Always check if the normality assumption is reasonable before using methods that require it!

📚 Practice Problems

1Problem 1easy

Question:

What percentage of data in a normal distribution falls within: a) 1 standard deviation of the mean? b) 2 standard deviations of the mean? c) 3 standard deviations of the mean?

💡 Show Solution

The Empirical Rule (68-95-99.7 Rule):

Step 1: Within 1 standard deviation (μ ± 1σ) Approximately 68% of data From (mean - SD) to (mean + SD) About 2/3 of all data

Step 2: Within 2 standard deviations (μ ± 2σ) Approximately 95% of data From (mean - 2SD) to (mean + 2SD) Nearly all data (only 5% outside)

Step 3: Within 3 standard deviations (μ ± 3σ) Approximately 99.7% of data From (mean - 3SD) to (mean + 3SD) Almost everything (only 0.3% outside)

Visual representation: |---68%---| |-----95%-----| |-------99.7%-------| μ-3σ μ-2σ μ-1σ μ μ+1σ μ+2σ μ+3σ

Answer: a) 68% b) 95% c) 99.7%

2Problem 2easy

Question:

IQ scores are normally distributed with mean 100 and standard deviation 15. What IQ scores represent the middle 68% of the population?

💡 Show Solution

Step 1: Identify given information μ (mean) = 100 σ (standard deviation) = 15 Need: Middle 68%

Step 2: Apply empirical rule 68% of data falls within 1 standard deviation of the mean Range: μ ± 1σ

Step 3: Calculate lower bound Lower bound = μ - 1σ = 100 - 15 = 85

Step 4: Calculate upper bound Upper bound = μ + 1σ = 100 + 15 = 115

Step 5: Interpret 68% of people have IQs between 85 and 115 About 16% have IQs below 85 About 16% have IQs above 115

Answer: IQ scores from 85 to 115 represent the middle 68% of the population.

3Problem 3medium

Question:

Heights of adult women are normally distributed with mean 64 inches and standard deviation 2.5 inches. A woman is 69 inches tall. What is her z-score, and what does it mean?

💡 Show Solution

Step 1: Identify given information μ = 64 inches σ = 2.5 inches x = 69 inches

Step 2: Calculate z-score z = (x - μ) / σ z = (69 - 64) / 2.5 z = 5 / 2.5 z = 2

Step 3: Interpret the z-score z = 2 means:

  • This woman's height is 2 standard deviations above the mean
  • She is taller than the average woman
  • Her height is 5 inches above the mean

Step 4: Find percentile using empirical rule 95% of women are within 2 SD of mean (between 59" and 69") This means 2.5% are above 69" (upper tail) She is taller than about 97.5% of women

Step 5: Context A z-score of 2 is quite high (unusual) Values beyond z = ±2 are sometimes considered "unusual" She's quite tall compared to most women

Answer: z = 2. This woman's height is 2 standard deviations above the mean, making her taller than approximately 97.5% of women.

4Problem 4medium

Question:

SAT scores are normally distributed with mean 1050 and standard deviation 200. What percentage of students score above 1450? Use the empirical rule.

💡 Show Solution

Step 1: Identify given information μ = 1050 σ = 200 x = 1450 Find: P(X > 1450)

Step 2: Calculate z-score z = (x - μ) / σ z = (1450 - 1050) / 200 z = 400 / 200 z = 2

Step 3: Apply empirical rule z = 2 means 1450 is 2 SD above mean 95% of data is within ±2 SD (between 650 and 1450) This leaves 5% outside this range

Step 4: Find upper tail 5% total is split between two tails Upper tail (above 1450): 5% / 2 = 2.5%

Step 5: Visualize |-------95%--------| 650 1050 1450 μ-2σ μ μ+2σ 2.5% | 2.5%

Answer: 2.5% of students score above 1450.

5Problem 5hard

Question:

Test scores are normally distributed with mean 72 and standard deviation 8. Two students scored 84 and 60 respectively. Compare their performances using z-scores. Which score is more unusual?

💡 Show Solution

Step 1: Calculate z-score for student 1 (score = 84) z₁ = (x - μ) / σ z₁ = (84 - 72) / 8 z₁ = 12 / 8 z₁ = 1.5

Interpretation:

  • 1.5 standard deviations ABOVE mean
  • Positive z-score = above average
  • Score is 12 points above mean

Step 2: Calculate z-score for student 2 (score = 60) z₂ = (60 - 72) / 8 z₂ = -12 / 8 z₂ = -1.5

Interpretation:

  • 1.5 standard deviations BELOW mean
  • Negative z-score = below average
  • Score is 12 points below mean

Step 3: Compare absolute values |z₁| = 1.5 |z₂| = 1.5 Same distance from mean!

Step 4: Determine which is more unusual Both scores are EQUALLY unusual Both are exactly 1.5 SD from the mean One is high, one is low Both within the typical range (68% rule: within ±1 SD, 95% rule: within ±2 SD)

Step 5: Context using empirical rule About 68% are within ±1 SD About 95% are within ±2 SD Both scores are between 1 and 2 SD away Not extremely unusual, but somewhat uncommon Roughly 13-14% of students score this far from mean in each direction

Step 6: Percentiles (approximate) Student 1 (z = 1.5): ~93rd percentile (better than ~93%) Student 2 (z = -1.5): ~7th percentile (better than only ~7%)

Answer: Both scores are equally unusual - both have |z| = 1.5, meaning they're the same distance from the mean (1.5 standard deviations). However, student 1 performed much better (top 7%) while student 2 performed poorly (bottom 7%).