8.4 Standard Deviation & The Normal Distribution (Empirical Rule)
Key Takeaways
- Sample variance ($s^2 = \frac{\sum (x - \bar{x})^2}{n-1}$) and sample standard deviation ($s = \sqrt{s^2}$) use Bessel's correction ($n-1$) to provide an unbiased estimate of population dispersion, measuring the typical distance of data points from the mean.
- The standard normal distribution is a continuous, unimodal, bell-shaped density curve perfectly symmetric about mean $\mu$ where $\text{Mean} = \text{Median} = \text{Mode}$ and the total area under the curve equals $1.0$ ($100\%$).
- The Empirical Rule (68–95–99.7% Rule) states that for bell-shaped distributions: approximately $68\%$ of data falls within $\mu \pm 1\sigma$, $95\%$ falls within $\mu \pm 2\sigma$, and $99.7\%$ falls within $\mu \pm 3\sigma$.
- Exact regional tail slice percentages partitioned by standard deviations are: $34.0\%$ ($\mu$ to $\mu + 1\sigma$), $13.5\%$ ($\mu + 1\sigma$ to $\mu + 2\sigma$), $2.35\%$ ($\mu + 2\sigma$ to $\mu + 3\sigma$), and $0.15\%$ ($> \mu + 3\sigma$).
- The standardized z-score ($z = \frac{x - \mu}{\sigma}$) measures the exact number of standard deviations a data value lies above ($z > 0$) or below ($z < 0$) the mean, enabling objective comparison across different distributions.
8.4 Standard Deviation & The Normal Distribution (Empirical Rule)
In statistical analysis, standard deviation is the universal benchmark metric for quantifying numerical dispersion. When data values follow a symmetric, bell-shaped distribution, the standard deviation pairs with the mean to completely define the Normal Distribution. On the CLEP College Mathematics exam, questions extensively test sample vs. population standard deviation, the Empirical Rule (68–95–99.7% Rule), regional slice calculations, and standardized z-scores.
1. Variance & Standard Deviation
Standard deviation measures the typical or average distance that data points deviate from their arithmetic mean. Because simple deviations $\sum (x - \bar{x})$ always sum to zero, deviations are squared before averaging.
Sample vs. Population Formulas
| Statistical Parameter | Population ($N$) | Sample ($n$) |
|---|---|---|
| Variance | ||
| Standard Deviation | ||
| Units of Measure | Squared original units (e.g., $\text{dollars}^2$) | Identical to original data units (e.g., $\text{dollars}$) |
Why Divide by $(n - 1)$? (Bessel's Correction)
When computing sample variance from a subset of a larger population, using the sample mean $\bar{x}$ instead of the true population mean $\mu$ tends to underestimate the true dispersion. Dividing by the degrees of freedom $n - 1$ (Bessel's correction) slightly inflates the variance, providing an unbiased estimator of the population variance $\sigma^2$.
Step-by-Step Manual Calculation Protocol
- Calculate the sample mean: $\bar{x} = \frac{\sum x}{n}$.
- Compute the deviation of each value from the mean: $(x_i - \bar{x})$.
- Square each deviation: $(x_i - \bar{x})^2$.
- Sum all squared deviations: $\sum (x_i - \bar{x})^2$.
- Divide by $n - 1$ to obtain the sample variance $s^2$.
- Take the positive square root $\sqrt{s^2}$ to obtain the sample standard deviation $s$.
2. Properties of the Normal Distribution
The Normal Distribution (Gaussian distribution) is a continuous probability distribution defined by its mean ($\mu$) and standard deviation ($\sigma$).
+-----------------------------------------------------------------------------+
| PROPERTIES OF THE NORMAL DISTRIBUTION CURVE |
| |
| 1. BELL-SHAPED & CONTINUOUS: Smooth, symmetric curve over (-inf, +inf). |
| 2. PERFECT SYMMETRY: Mean = Median = Mode at the exact central peak (μ). |
| 3. TOTAL PROBABILITY AREA = 1.0 (100%): Exactly 50% lies below μ and |
| 50% lies above μ. |
| 4. ASYMPTOTIC: The tails extend infinitely in both directions, approaching|
| the horizontal axis closely but never touching or crossing it. |
| 5. INFLECTION POINTS: The curve transitions from concave downward to |
| concave upward at exactly μ - σ and μ + σ. |
+-----------------------------------------------------------------------------+
3. The Empirical Rule (68–95–99.7% Rule)
For any dataset that is normally distributed (or approximately bell-shaped), the Empirical Rule provides exact percentage approximations based on standard deviation intervals away from the mean:
+-----------------------------------------------------------------------------+
| THE EMPIRICAL RULE (68 - 95 - 99.7%) |
| |
| [------ 68.2% ------] |
| [------------- 95.4% -------------] |
| [-------------------- 99.7% --------------------] |
| |
| 34.0% 34.0% |
| 13.5% 13.5% |
| 2.35% 2.35% |
| 0.15% 0.15% |
| --+-------+-------+-------+-------+-------+-------+-------+-- |
| μ-3σ μ-2σ μ-1σ μ μ+1σ μ+2σ μ+3σ |
+-----------------------------------------------------------------------------+
Regional Slice Breakdown Table
Dividing the normal curve into standardized one-standard-deviation slices yields the universal values tested on the CLEP:
| Interval from Mean | Standard Deviation Range | Slice Percentage | Cumulative Area from $-\infty$ |
|---|---|---|---|
| Extreme Lower Tail | Below $\mu - 3\sigma$ | $0.15%$ | $0.15%$ |
| Outer Lower Slice | $\mu - 3\sigma$ to $\mu - 2\sigma$ | $2.35%$ | $2.50%$ (Lower $2.5%$ Cutoff) |
| Inner Lower Slice | $\mu - 2\sigma$ to $\mu - 1\sigma$ | $13.50%$ | $16.00%$ (Lower $16%$ Cutoff) |
| Central Lower Slice | $\mu - 1\sigma$ to $\mu$ | $34.00%$ | $50.00%$ (Median) |
| Central Upper Slice | $\mu$ to $\mu + 1\sigma$ | $34.00%$ | $84.00%$ (Upper $16%$ Tail) |
| Inner Upper Slice | $\mu + 1\sigma$ to $\mu + 2\sigma$ | $13.50%$ | $97.50%$ (Upper $2.5%$ Tail) |
| Outer Upper Slice | $\mu + 2\sigma$ to $\mu + 3\sigma$ | $2.35%$ | $99.85%$ |
| Extreme Upper Tail | Above $\mu + 3\sigma$ | $0.15%$ | $100.00%$ |
4. Standardized Z-Scores
A standardized z-score indicates how many standard deviations a particular observation ($x$) lies above or below the mean.
Z-Score Formula
Interpretation of Z-Scores
- $z = 0$: The data value is exactly equal to the mean.
- $z > 0$ (Positive): The data value lies above the mean (e.g., $z = +1.5$ is $1.5$ standard deviations above average).
- $z < 0$ (Negative): The data value lies below the mean (e.g., $z = -2.0$ is $2.0$ standard deviations below average).
- Standard Normal Distribution: Converting all raw scores $x$ to $z$-scores transforms any normal distribution into the standard normal distribution with mean $\mu_z = 0$ and standard deviation $\sigma_z = 1$.
Comparing Relative Standing Across Different Distributions
Z-scores allow direct, objective comparisons between individuals scored on completely different scales (e.g., comparing an ACT score of $30$ with an SAT score of $1380$).
5. Step-by-Step Worked Examples
Worked Example 1: Calculating Sample Variance and Standard Deviation
Problem: A quality control technician measures the weight (in grams) of $5$ sample metal components: ${10, ; 12, ; 14, ; 15, ; 19}$. Calculate the sample mean $\bar{x}$, sample variance $s^2$, and sample standard deviation $s$.
Solution:
- Sample mean: $\bar{x} = \frac{10 + 12 + 14 + 15 + 19}{5} = \frac{70}{5} = 14.0\text{ g}$.
- Compute squared deviations:
| Value ($x$) | Deviation ($x - \bar{x}$) | Squared Deviation $(x - \bar{x})^2$ |
|---|---|---|
| $10$ | $10 - 14 = -4$ | $(-4)^2 = 16$ |
| $12$ | $12 - 14 = -2$ | $(-2)^2 = 4$ |
| $14$ | $14 - 14 = 0$ | $(0)^2 = 0$ |
| $15$ | $15 - 14 = +1$ | $(+1)^2 = 1$ |
| $19$ | $19 - 14 = +5$ | $(+5)^2 = 25$ |
| Total | $\sum (x - \bar{x}) = 0$ | $\sum (x - \bar{x})^2 = 46$ |
- Sample variance ($n - 1 = 5 - 1 = 4$):
- Sample standard deviation:
Worked Example 2: Empirical Rule Analysis
Problem: Scores on a national mathematics exam are normally distributed with a mean of $\mu = 500$ and a standard deviation of $\sigma = 100$. Out of $10{,}000$ test takers:
- What percentage of students scored between $400$ and $700$?
- Approximately how many students scored above $700$?
- What score corresponds to the $84\text{th}$ percentile?
Solution:
- Identify standard deviation milestones:
- $400 = 500 - 1(100) = \mu - 1\sigma$
- $700 = 500 + 2(100) = \mu + 2\sigma$
- Sum the corresponding regional slices:
- Score above $700$ ($> \mu + 2\sigma$):
- Tail percentage $= 2.35% + 0.15% = 2.5%$
- Number of students $= 10{,}000 \times 0.025 = 250\text{ students}$
- $84\text{th}$ percentile:
- Cumulative area up to $\mu + 1\sigma = 50.0% + 34.0% = 84.0%$
- Score $= \mu + 1\sigma = 500 + 100 = 600$
Worked Example 3: Comparative Z-Score Analysis
Problem: Jessica scored $88$ on an English literature exam (class mean $\mu_E = 76$, standard deviation $\sigma_E = 8$). On her Chemistry exam, she scored $81$ (class mean $\mu_C = 69$, standard deviation $\sigma_C = 6$). On which exam did Jessica perform better relative to her peers?
Solution:
- Calculate Jessica's English z-score:
- Calculate Jessica's Chemistry z-score:
- Comparison: Jessica scored $1.5$ standard deviations above average in English, but $2.0$ standard deviations above average in Chemistry. Because $+2.00 > +1.50$, Jessica performed better relative to her peers in Chemistry.
6. Common CLEP Traps & Strategic Checkpoints
- Trap 1: Applying the Empirical Rule to Non-Normal Distributions: The 68–95–99.7% Rule applies only to bell-shaped, symmetric normal distributions. It cannot be used for uniform or skewed data.
- Trap 2: Dividing by $n$ Instead of $n - 1$ for Samples: For sample standard deviation ($s$), always divide the sum of squared deviations by $n - 1$.
- Trap 3: Forgetting Negative Signs on Z-Scores: If a raw score is below the mean ($x < \mu$), the z-score MUST be negative (e.g., $z = -1.25$). Omitting the negative sign reverses the percentile ranking.
- Trap 4: Confusing Variance with Standard Deviation: Variance is in squared units ($s^2$). To find the standard deviation, you must take the square root ($s = \sqrt{s^2}$).
The distribution of heights of adult males in a region is normally distributed with a mean of μ = 70 inches and a standard deviation of σ = 3 inches. Using the Empirical Rule (68–95–99.7% Rule), what percentage of adult males in this region have heights between 64 inches and 76 inches?
A student scores 84 on a biology exam where the class mean is 76 with a standard deviation of 4. On a chemistry exam, the student scores 88 where the class mean is 80 with a standard deviation of 5. On which exam did the student perform better relative to the class, and what are the corresponding z-scores?
When calculating the sample standard deviation s from a data set of n = 6 observations, why is the sum of squared deviations divided by (n - 1) = 5 instead of n = 6?