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
📚 Practice Problems
1Problem 1easy
❓ Question:
The heights of adult men follow a normal distribution with mean 70 inches and standard deviation 2.5 inches. What is the z-score for a man who is 75 inches tall?
Use the Normal distribution, z-scores, and the empirical rule to find probabilities.
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Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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What course covers Normal Distributions?▾
Normal Distributions is part of the AP Statistics course on Study Mondo, specifically in the Unit 1: Exploring One-Variable Data section. You can explore the full course for more related topics and practice resources.
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Z-Scores (Standardization)
Formula:z=σx−μ
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=2001200−1000=200200=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
Confusing normalcdf and invNorm: normalcdf(value) → probability; invNorm(probability) → value
Forgetting to standardize: must convert to z-score before using standard normal table
Area mistakes: P(Z > 2) ≠ P(Z < 2); use complement if needed
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.
σ
x−μ
where x is the value, μ is the mean, and σ is the standard deviation.
z=2.575−70=2.55=2
Interpretation: A height of 75 inches is 2 standard deviations above the mean. This is quite tall but not extremely rare (about 2.3% of men are taller).
2Problem 2medium
❓ Question:
SAT scores are normally distributed with mean 500 and standard deviation 100. What percentage of test-takers score between 400 and 600?
💡 Show Solution
Step 1: Find z-scores
For x=400: z=100400−500=−1
For x=600: z=100600−500=
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 μ±1σ)
95% within 2 SDs (between μ±2σ)
99.7% within 3 SDs
Since scores between 400 and 600 represent the range from μ−1σ to μ+1σ:
Answer: 68% of test-takers score between 400 and 600.
3Problem 3hard
❓ Question:
A manufacturing process produces bolts with diameter normally distributed: mean = 10 mm, SD = 0.1 mm. The acceptable range is 9.8 to 10.2 mm. What proportion of bolts are acceptable? What z-score defines the upper boundary?
💡 Show Solution
Step 1: Find z-scores for the boundaries
Lower boundary (x=9.8): z=0.19.8−10=−2
Upper boundary (x=10.2): z=0.110.2−10=
Upper z-score answer: z = +2
Step 2: Find the proportion using Empirical Rule
The range 9.8 to 10.2 is μ±2σ (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.
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Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.