• Bell-shaped and symmetric around mean
• Mean = median = mode
• Defined by two parameters: mean μ and standard deviation σ
• Total area under curve = 1 (100% of probability)

Visual Summary:

• Center: mean μ
• Width: controlled by σ (larger σ → wider, flatter)
• Asymptotic: tails approach x-axis but never touch

The 68-95-99.7 Rule

For normally distributed data:

• 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
• (remaining 0.3% split equally in two tails)

Example: SAT scores normally distributed with μ = 1000, σ = 200

• 68% score between 800 and 1200
• 95% score between 600 and 1400
• 99.7% score between 400 and 1600

Z-Scores (Standardization)

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

Interpretation:

• z-score = number of standard deviations away from mean
• z > 0: above mean
• z < 0: below mean
• z = 0: at mean

Standardization benefit: transforms any normal distribution into standard normal (μ = 0, σ = 1)

Example: Score x = 1200 on SAT (μ = 1000, σ = 200): $z = \frac{1200 - 1000}{200} = \frac{200}{200} = 1$

Score is 1 standard deviation above mean.

Finding Normal Probabilities

Using normalcdf(lower, upper, μ, σ):

• Calculates \(P(a < X < b)\) for normal distribution
• Returns area under curve between a and b

Example: SAT scores X ~ N(1000, 200)

• \(P(X < 1200) = P(Z < 1) =\) normalcdf(−999, 1, 0, 1) ≈ 0.8413

Finding Percentiles with invNorm

Using invNorm(percentile, μ, σ):

• Returns value corresponding to given percentile
• Inverse of normalcdf

Example: 90th percentile of SAT scores:

• invNorm(0.90, 1000, 200) ≈ 1256
• 90% of test-takers score below 1256

Worked Example

Scenario: Adult female heights normally distributed with μ = 64 inches, σ = 2.5 inches.

Question 1: What percent of women are taller than 69 inches?

Step 1: Standardize $z = \frac{69 - 64}{2.5} = \frac{5}{2.5} = 2$

Step 2: Find probability \(P(X > 69) = P(Z > 2) =\) 1 − normalcdf(−999, 2, 0, 1) ≈ 1 − 0.9772 = 0.0228 or 2.28%

Question 2: What height separates the shortest 25% from the rest?

Step 1: Use invNorm(0.25, 64, 2.5) ≈ 62.32 inches

Interpretation: 25% of women are shorter than 62.32 inches.

Common Student Mistakes

1. Wrong z-score direction: \(z = \frac{x - \mu}{\sigma}\), not \(\frac{\mu - x}{\sigma}\)
2. Confusing normalcdf and invNorm: normalcdf(value) → probability; invNorm(probability) → value
3. Forgetting to standardize: must convert to z-score before using standard normal table
4. Area mistakes: P(Z > 2) ≠ P(Z < 2); use complement if needed
5. Ignoring context: answer "0.0228" instead of "2.28% of women"

When NOT to Use Normal Model

• Data is clearly skewed (check histogram/boxplot)
• Sample size too small (rule of thumb: n ≥ 30, or visually normal)
• Data has multiple peaks
• Outliers present

AP Exam Tip

On calculator problems:

• Clearly state the distribution: "Let X ~ N(μ, σ)"
• Show your z-score: \(z = \frac{x - \mu}{\sigma}\) = ...
• State calculator function: "Using normalcdf(lower, upper, μ, σ)..."
• Interpret result in context: "Therefore, approximately _____% of SAT scores fall between ___ and ___."

Common FRQ mistake: Not showing work on calculator commands. Examiners want to see setup even if you use calc.

For $x = 600$: $z = \frac{600 - 500}{100} = +1$

Step 2: Use the Empirical Rule (68-95-99.7)

The Empirical Rule states that in a normal distribution:

• 68% of data falls within 1 SD of the mean (between $\mu \pm 1\sigma$)
• 95% within 2 SDs (between $\mu \pm 2\sigma$)
• 99.7% within 3 SDs

Since scores between 400 and 600 represent the range from $\mu - 1\sigma$ to $\mu + 1\sigma$:

Answer: 68% of test-takers score between 400 and 600.

Upper boundary ($x = 10.2$): $z = \frac{10.2 - 10}{0.1} = +2$

Upper z-score answer: z = +2

Step 2: Find the proportion using Empirical Rule

The range 9.8 to 10.2 is $\mu \pm 2\sigma$ (from -2 to +2 SD).

By the Empirical Rule, 95% of bolts fall within 2 standard deviations.

Step 3: Find proportion outside acceptable range

Proportion defective = 100% − 95% = 5%

This means 2.5% are too small (below 9.8) and 2.5% are too large (above 10.2).

Interpretation: The process produces about 95 acceptable bolts per 100, leaving a 5% defect rate. The manufacturer might adjust the process to center it more tightly or reduce SD.