Normal Distribution and Z-Scores

What is Normal Distribution?

Normal distribution: Bell-shaped probability distribution

Also called: Gaussian distribution, bell curve

Properties:

• Symmetric around mean
• Mean = median = mode
• Total area under curve = 1
• Defined by mean (μ) and standard deviation (σ)

Notation: N(μ, σ²) or N(μ, σ)

Example: Heights

Adult male heights in US:

• Mean: μ ≈ 70 inches
• Standard deviation: σ ≈ 3 inches
• Distribution: approximately normal

Standard Deviation Review

Standard deviation (σ): Measures spread of data

Small σ: Data clustered near mean

Large σ: Data spread out

Formula: σ = √[Σ(x - μ)² / N]

Example: Compare Spreads

Dataset A: Mean = 50, σ = 2 (tightly clustered)

Dataset B: Mean = 50, σ = 10 (widely spread)

Both centered at 50, but B more variable

Empirical Rule (68-95-99.7)

For normal distributions:

68% of data within 1σ of mean (μ ± σ)

95% of data within 2σ of mean (μ ± 2σ)

99.7% of data within 3σ of mean (μ ± 3σ)

Powerful tool for estimation!

Example: Apply Empirical Rule

Test scores: μ = 75, σ = 10

Find percentage between 65 and 85:

• 65 = 75 - 10 (one σ below)
• 85 = 75 + 10 (one σ above)

Answer: ≈68% of scores between 65 and 85

Find percentage between 55 and 95:

• 55 = 75 - 20 (two σ below)
• 95 = 75 + 20 (two σ above)

Answer: ≈95% between 55 and 95

Z-Score

Z-score: Number of standard deviations from mean

Formula: z = (x - μ) / σ

Meaning:

• z = 0: at the mean
• z > 0: above mean
• z < 0: below mean
• z = 2: two standard deviations above mean

Standardizes different datasets for comparison

Example 1: Calculate Z-Score

Test score: x = 85 Mean: μ = 75 Std dev: σ = 10

Calculate:

z = (85 - 75) / 10
  = 10 / 10
  = 1

Score is 1 standard deviation above mean

Example 2: Negative Z-Score

Height: x = 64 inches Mean: μ = 70 inches Std dev: σ = 3 inches

Calculate:

z = (64 - 70) / 3
  = -6 / 3
  = -2

Height is 2 standard deviations below mean

Standard Normal Distribution

Standard normal: N(0, 1)

• Mean = 0
• Standard deviation = 1

Any normal distribution can be standardized using z-scores

Z-table: Shows probabilities for standard normal

Using Z-Tables

Z-table gives P(Z < z): Area to left of z

To find probabilities:

1. Calculate z-score
2. Look up z in table
3. Interpret based on question

Example: Find Probability

IQ scores: μ = 100, σ = 15

Find P(IQ < 115):

Step 1: Calculate z

z = (115 - 100) / 15
  = 1

Step 2: Look up z = 1 From table: P(Z < 1) ≈ 0.8413

Answer: 84.13% have IQ below 115

Finding Upper Tail Probability

P(X > x) = 1 - P(X < x)

Use complement

Example: Upper Tail

Test scores: μ = 70, σ = 8

Find P(score > 78):

Calculate z:

z = (78 - 70) / 8 = 1

From table: P(Z < 1) = 0.8413

Upper tail:

P(Z > 1) = 1 - 0.8413
         = 0.1587

Answer: 15.87% score above 78

Finding Probability Between Two Values

P(a < X < b) = P(X < b) - P(X < a)

Example: Between Two Values

Heights: μ = 65, σ = 2.5

Find P(63 < height < 68):

Calculate z-scores:

z₁ = (63 - 65) / 2.5 = -0.8
z₂ = (68 - 65) / 2.5 = 1.2

From table:

• P(Z < -0.8) ≈ 0.2119
• P(Z < 1.2) ≈ 0.8849

Calculate:

P(63 < X < 68) = 0.8849 - 0.2119
                = 0.6730

Answer: 67.30% have heights between 63 and 68 inches

Finding Value from Percentile

Given probability, find x value

Steps:

1. Find z from table (or calculator)
2. Use x = μ + zσ

Example: Find Cutoff

Scores: μ = 500, σ = 100

Top 10% get scholarship. What's minimum score?

Top 10% means P(X > x) = 0.10 So P(X < x) = 0.90

From table: z ≈ 1.28 for P(Z < z) = 0.90

Calculate x:

x = 500 + 1.28(100)
  = 500 + 128
  = 628

Answer: Need score ≥ 628

Applications: Quality Control

Manufacturing: Check if products within specifications

Defect rate: Use normal distribution to estimate

Example: Light Bulb Lifetime

Lifetime: μ = 1000 hours, σ = 50 hours

Reject if < 900 hours. What % rejected?

Calculate z:

z = (900 - 1000) / 50 = -2

From table: P(Z < -2) ≈ 0.0228

Answer: 2.28% rejected

Applications: Standardized Tests

SAT, ACT, IQ tests: Designed to be normally distributed

Percentiles: Easily calculated using z-scores

Example: SAT Score

SAT Math: μ = 500, σ = 100

Score of 650 is what percentile?

Calculate z:

z = (650 - 500) / 100 = 1.5

From table: P(Z < 1.5) ≈ 0.9332

Answer: 93rd percentile

When to Use Normal Distribution

Appropriate when:

• Data symmetric and bell-shaped
• Large sample size (Central Limit Theorem)
• Natural phenomena (heights, measurements)

Not appropriate when:

• Data highly skewed
• Small sample with outliers
• Discrete counts (use binomial/Poisson instead)

Central Limit Theorem

Key result: Sample means tend toward normal distribution

Even if population not normal!

Requirements: Large enough sample (usually n ≥ 30)

Implication: Normal distribution very powerful in statistics

Comparing Across Different Scales

Z-scores allow comparison across different tests/measurements

Example: Compare Scores

Student A:

• Math: 85 (μ = 75, σ = 10)
• English: 90 (μ = 82, σ = 8)

Which is relatively better?

Math z-score: (85 - 75) / 10 = 1.0

English z-score: (90 - 82) / 8 = 1.0

Same relative performance!

Technology Tools

Calculators: normalcdf, invNorm functions

Software: Excel (NORM.DIST, NORM.INV), R, Python

Online calculators: Many z-score calculators available

Practice