Normal Distribution Intro
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Normal Distribution
The normal distribution (also called the bell curve or Gaussian distribution) is a symmetric, bell-shaped distribution where most values cluster around the mean, with fewer values as you move away from the center. It is one of the most important distributions in statistics.
Many natural phenomena follow a normal distribution: heights, weights, test scores, measurement errors, and many biological characteristics. Understanding the normal distribution helps us make predictions and interpret data.
Key Characteristics of Normal Distributions
- Symmetric: The left and right halves are mirror images
- Bell-shaped: Highest in the middle, tapering on both sides
- Mean = Median = Mode: All three measures of center are equal
- Defined by two parameters: Mean (center) and standard deviation (spread)
- Asymptotic: The tails approach but never touch the x-axis
The Empirical Rule (68-95-99.7 Rule)
For any normal distribution:
| Within this range | Approximate % of data |
|---|---|
| Mean +/- 1 standard deviation | 68% |
| Mean +/- 2 standard deviations | 95% |
| Mean +/- 3 standard deviations | 99.7% |
Understanding Z-Scores
Z-Score
A z-score tells you how many standard deviations a value is from the mean.
Formula: z = (x - mean) / standard deviation
A z-score of 0 means the value equals the mean. Positive z-scores are above the mean; negative z-scores are below.
Interpreting Z-Scores
| Z-Score | Interpretation | % Below (approximately) |
|---|---|---|
| -2 | 2 standard deviations below mean | 2.5% |
| -1 | 1 standard deviation below mean | 16% |
| 0 | At the mean | 50% |
| +1 | 1 standard deviation above mean | 84% |
| +2 | 2 standard deviations above mean | 97.5% |
SAT/ACT Connection
The SAT and ACT frequently test the Empirical Rule and z-scores. You should be able to: apply the 68-95-99.7 rule to find percentages, calculate z-scores, and compare values from different distributions using z-scores. These concepts appear in the Problem Solving and Data Analysis section.
Examples
Work through these examples to understand normal distributions.
Example 1: Using the Empirical Rule
Problem: Test scores are normally distributed with mean 75 and standard deviation 8. What percentage of students scored between 67 and 83?
Step 1: Identify how many standard deviations from the mean
67 = 75 - 8 = mean - 1 SD
83 = 75 + 8 = mean + 1 SD
Step 2: Apply the Empirical Rule
Within 1 standard deviation of the mean contains approximately 68% of data.
Answer: Approximately 68% of students scored between 67 and 83.
Example 2: Finding Range for 95%
Problem: Heights of adult men are normally distributed with mean 70 inches and standard deviation 3 inches. Between what heights do 95% of men fall?
Step 1: Apply the Empirical Rule for 95%
95% falls within 2 standard deviations of the mean.
Step 2: Calculate the range
Lower bound: 70 - 2(3) = 70 - 6 = 64 inches
Upper bound: 70 + 2(3) = 70 + 6 = 76 inches
Answer: 95% of adult men are between 64 and 76 inches tall.
Example 3: Calculating a Z-Score
Problem: In a class where the mean score is 80 and standard deviation is 5, a student scored 92. What is their z-score?
Step 1: Apply the z-score formula
z = (x - mean) / standard deviation
z = (92 - 80) / 5
z = 12 / 5 = 2.4
Answer: The z-score is 2.4, meaning the score is 2.4 standard deviations above the mean.
Example 4: Comparing with Z-Scores
Problem: Anna scored 85 on a test with mean 75 and SD 5. Ben scored 90 on a test with mean 82 and SD 4. Who performed better relative to their class?
Step 1: Calculate Anna's z-score
z = (85 - 75) / 5 = 10/5 = 2
Step 2: Calculate Ben's z-score
z = (90 - 82) / 4 = 8/4 = 2
Answer: Both have z-scores of 2, so they performed equally well relative to their respective classes. Both are 2 standard deviations above their class means.
Example 5: Finding Percentile from Z-Score
Problem: A data value has z-score of -1.5. Using the Empirical Rule, approximately what percentile is this?
Step 1: Interpret the z-score
Z = -1.5 means 1.5 standard deviations below the mean
Step 2: Estimate using Empirical Rule
16% is below z = -1, and 2.5% is below z = -2
Z = -1.5 is between these, closer to 7% (roughly)
Answer: Approximately the 7th percentile (about 7% of values are below this point).
Practice
Apply your knowledge of normal distributions to these problems.
1. Scores are normally distributed with mean 100 and SD 15. What percent of scores fall between 85 and 115?
A) 34% B) 68% C) 95% D) 99.7%
2. Using the same distribution (mean 100, SD 15), what percent fall between 70 and 130?
A) 68% B) 95% C) 99.7% D) 50%
3. If mean = 50 and SD = 6, what value is 2 standard deviations above the mean?
A) 56 B) 62 C) 52 D) 68
4. Calculate the z-score: x = 78, mean = 70, SD = 4
A) 1 B) 2 C) 0.5 D) -2
5. A z-score of -1 means the value is:
A) 1 unit below the mean B) 1 standard deviation below the mean C) At the 1st percentile D) Below 1% of values
6. What z-score corresponds to a value equal to the mean?
A) 1 B) 0 C) -1 D) It depends on the SD
7. Heights have mean 64 inches, SD 2.5 inches. What percentage of heights are above 69 inches?
A) 2.5% B) 16% C) 50% D) 84%
8. A value with z = 1 is at approximately what percentile?
A) 16th B) 50th C) 84th D) 97.5th
9. In a normal distribution, which is true?
A) Mean > Median B) Median > Mode C) Mean = Median = Mode D) Mode > Mean
10. Test A: mean 75, SD 10. Test B: mean 80, SD 5. A student scores 85 on Test A and 87 on Test B. On which test did they perform better relative to others?
A) Test A B) Test B C) Equal performance D) Cannot determine
Click to reveal answers
- B) 68% - 85 to 115 is mean +/- 1 SD (100 +/- 15)
- B) 95% - 70 to 130 is mean +/- 2 SD (100 +/- 30)
- B) 62 - 50 + 2(6) = 50 + 12 = 62
- B) 2 - z = (78-70)/4 = 8/4 = 2
- B) 1 standard deviation below the mean - Z-scores measure SDs from mean
- B) 0 - When x = mean, z = (mean - mean)/SD = 0
- A) 2.5% - 69 = 64 + 2(2.5), so z = 2; about 2.5% above z = 2
- C) 84th - 68% between -1 and 1; 16% below -1, so 84% below +1
- C) Mean = Median = Mode - This is a key property of normal distributions
- A) Test A - Test A: z = (85-75)/10 = 1; Test B: z = (87-80)/5 = 1.4. Higher z = better relative performance
Check Your Understanding
Answer these reflection questions to deepen your understanding.
1. Why is the normal distribution sometimes called the "bell curve"?
Reveal Answer
The normal distribution is called a bell curve because its shape resembles a bell: it's highest in the middle (at the mean) and symmetrically tapers off on both sides. The curve is smooth and continuous, with most data clustered near the center and progressively fewer values as you move toward the tails. This bell shape visually represents how most values are "average" while extreme values are rare.
2. Explain why z-scores are useful for comparing values from different distributions.
Reveal Answer
Z-scores standardize data by expressing how many standard deviations a value is from its mean. This allows fair comparison across different scales or distributions. For example, scoring 80 on a test with mean 70 isn't directly comparable to scoring 85 on a test with mean 80 - the raw numbers don't tell the whole story. But z-scores (2.0 vs. 1.0, if SDs are 5) reveal that the first score is relatively better. Z-scores put everything on the same scale: "standard deviations from mean."
3. If a distribution is not symmetric (skewed), can you still use the Empirical Rule?
Reveal Answer
No, the Empirical Rule (68-95-99.7) only applies to normal distributions, which are symmetric. If data is skewed, these percentages will not be accurate. For skewed distributions, the mean and median differ, and the spread is not symmetric around the center. You would need different methods to describe skewed data, or you could transform the data to make it more normal before applying these rules.
4. A student has a z-score of 0.5 on an exam. What does this tell you about their performance?
Reveal Answer
A z-score of 0.5 means the student scored 0.5 standard deviations above the mean. This indicates above-average performance, but not dramatically so. Using the Empirical Rule approximations, about 69% of students scored below this level (50% below mean, plus about 19% between mean and z=0.5). The student performed better than roughly two-thirds of the class but is not in the top tier (which would require z > 1 or 2).
🚀 Next Steps
- Review any concepts that felt challenging
- Move on to the next lesson when ready
- Return to practice problems periodically for review