The Empirical Rule: 68–95–99.7

For any normally distributed dataset with mean μ and standard deviation σ:

Empirical rule: 68% of data lies within 1σ of the mean, 95% within 2σ, 99.7% within 3σ. The remaining percentages outside each band: 32%, 5%, and 0.3%. Memorise as 68-95-99.7.

Pause, copy the empirical rule table: 1σ → 68% inside, 32% outside; 2σ → 95% inside, 5% outside; 3σ → 99.7% inside, 0.3% outside, memorise as 68-95-99.7 into your book.

Applying the Empirical Rule

The empirical rule says 68% of data lies within 1σ of the mean, 95% within 2σ, and 99.7% within 3σ, with 32%, 5%, and 0.3% lying outside those bands. Applying the rule to a real dataset means first identifying how many standard deviations each boundary is from the mean: subtract the boundary from μ then divide by σ, and match the resulting number to the 68/95/99.7 table.

To apply the empirical rule, first identify how many standard deviations each boundary is from the mean.

Worked example: IQ scores are normally distributed with μ = 100, σ = 15. What percentage of people have IQ between 70 and 130?

1. Find the distance from the mean: 100 − 70 = 30 = 2σ and 130 − 100 = 30 = 2σ.
2. So the interval (70, 130) is within 2 standard deviations of the mean.
3. By the empirical rule: approximately 95% of people have IQ between 70 and 130.

Worked example 2: What percentage have IQ between 85 and 115?

1. 100 − 85 = 15 = 1σ and 115 − 100 = 15 = 1σ.
2. Interval (85, 115) is within 1 standard deviation.
3. Approximately 68%.

To apply the empirical rule: (1) subtract each boundary from μ, (2) divide by σ to get the number of standard deviations, (3) apply 68/95/99.7. If the boundaries are at ±kσ from the mean, use the matching percentage.

Pause, copy the three-step application process: (1) subtract each boundary from μ; (2) divide by σ to get the number of standard deviations; (3) match to 68/95/99.7 and read off the corresponding percentage into your book.

One-Sided Intervals Using Symmetry

Applying the empirical rule gives a two-sided percentage: 68% lies within ±1σ, meaning 32% lies outside. But many questions ask about one tail only, for example, "what percentage scores above μ + σ?" Because the normal distribution is perfectly symmetric about its mean, the 32% outside ±1σ splits equally: 16% in each tail.

The empirical rule gives two-sided (symmetric) percentages. Use symmetry to find one-sided percentages.

Key facts derived from symmetry:

• Since 68% is within 1σ, the remaining 32% is outside 1σ. By symmetry, 16% is below μ − σ and 16% is above μ + σ.
• Since 95% is within 2σ, 5% is outside 2σ. By symmetry, 2.5% is below μ − 2σ and 2.5% is above μ + 2σ.

Worked example: μ = 100, σ = 15. What percentage have IQ above 130?

1. 130 = μ + 2σ, so we want the percentage above μ + 2σ.
2. 2.5% is above μ + 2σ (from 5% ÷ 2).
3. Answer: approximately 2.5%.

One-sided intervals use symmetry: since 68% lies within 1σ, 16% lies below μ − σ and 16% above μ + σ. For 2σ: 2.5% in each tail. For 3σ: 0.15% in each tail. Divide the total outside percentage by 2 for each tail.

Pause, copy the symmetry halving rule (outside % ÷ 2 = each tail) and the three one-sided tail values: beyond ±1σ: 16% each tail; beyond ±2σ: 2.5% each tail; beyond ±3σ: 0.15% each tail into your book.

Combining Intervals

Using symmetry to split the outside percentages, 16% in each tail beyond ±1σ, 2.5% beyond ±2σ, 0.15% beyond ±3σ, lets you find one-sided probabilities for any standard deviation boundary. For values that fall between different σ boundaries, subtract the inner percentage from the outer to find the percentage in that band, then use symmetry to split it if needed.

More complex questions ask for intervals between different multiples of σ. Use the empirical rule plus subtraction.

Example: μ = 100, σ = 15. What percentage lies between 85 and 115 but NOT between 70 and 130?

This asks: what is in the 95% band but not in the 68% band?

• 95% − 68% = 27% lies between 1σ and 2σ from the mean (on both sides combined).
• By symmetry: 13.5% is between μ + 1σ and μ + 2σ, and 13.5% between μ − 2σ and μ − 1σ.

Similarly, between 2σ and 3σ on each side: (99.7 − 95) ÷ 2 = 2.35%.

Identifying unusual values: a value is considered unusual if it lies more than 2 standard deviations from the mean (i.e., |z| > 2), placing it in the outer 5% of the distribution. Values beyond 3σ are very rare (0.3%).

Pause, copy the two classification thresholds: |z| > 2 places a value in the outer 5% (unusual), and |z| > 3 places it in the outer 0.3% (very rare), and note a value is considered statistically unusual when it falls more than 2 standard deviations from the mean into your book.

Identifying Unusual Values

A value is unusual if it lies beyond 2σ from the mean, placing it in the outer 5%, or very rare if beyond 3σ (outer 0.3%). These thresholds classify values as typical or unusual, but to measure exactly how unusual a specific value is, the z-score formula gives the precise number of standard deviations: z = (x − μ) / σ.

A value is considered unusual if it falls more than 2 standard deviations from the mean. This is because only about 5% of data lies outside the 2σ band, making such values relatively rare.

Worked example: A factory produces bolts with length μ = 10 mm, σ = 0.2 mm. A bolt measures 10.5 mm. Is this unusual?

1. Distance from mean: 10.5 − 10 = 0.5 mm
2. In terms of σ: 0.5 ÷ 0.2 = 2.5 standard deviations above the mean
3. Since 2.5 > 2, this value is more than 2σ from the mean, it is unusual.

z-score: z = (x − μ) / σ. A positive z means x is above the mean; negative means below. The z-score tells you exactly how many standard deviations a value is from the mean, a standardised measure allowing comparisons across datasets.

Pause, copy z = (x − μ) / σ with definitions, the sign convention (z > 0 means x is above mean; z < 0 means below), and the purpose: z-scores standardise values so they can be compared across different distributions into your book.