A measure of center describes a typical or representative value in a distribution. The two most common are the mean and the median.
Mean ($\bar{x}$): The arithmetic average. Add all values and divide by $n$.
$$\bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n} = \frac{\sum x_i}{n}$$
Median ($M$): The middle value when data are ordered. If $n$ is even, average the two middle values.
The choice depends on the shape of the distribution and the presence of outliers:
AP Exam Tip: Outliers and skewness pull the mean toward the tail but leave the median relatively unchanged. In right-skewed distributions, mean > median. In left-skewed distributions, mean < median.
Seven students' test scores: 72, 75, 78, 80, 82, 85, 43.
Step 1 — Order the data: 43, 72, 75, 78, 80, 82, 85
Step 2 — Median: Middle value (4th of 7) = 78
Step 3 — Mean: $\bar{x} = \frac{43+72+75+78+80+82+85}{7} = \frac{515}{7} \approx \mathbf{73.6}$
The outlier 43 dragged the mean down to 73.6, far below six of the seven scores. The median 78 better represents a typical score.
Daily high temperatures (°F) for a week: 68, 71, 70, 73, 69, 95, 72. Find the mean and median. Which better describes a typical day?
A measure of spread describes how variable the data are. More spread means more variability. We study three: range, IQR, and standard deviation.
Range = Maximum − Minimum. Simple but not resistant to outliers.
Quartiles: $Q_1$ = median of lower half; $Q_3$ = median of upper half.
IQR = $Q_3 - Q_1$. The spread of the middle 50% of data. Resistant to outliers.
Exam scores: 55, 62, 70, 74, 78, 82, 85, 88, 91, 96
n = 10, so the median = average of 5th and 6th values = (78+82)/2 = 80
Lower half: 55, 62, 70, 74, 78 → $Q_1$ = 70
Upper half: 82, 85, 88, 91, 96 → $Q_3$ = 88
IQR = 88 − 70 = 18
Outlier check: Any value below $Q_1 - 1.5 \cdot \text{IQR} = 70 - 27 = 43$ or above $Q_3 + 1.5 \cdot \text{IQR} = 88 + 27 = 115$ is an outlier. Score of 55 is not an outlier.
The sample standard deviation $s_x$ measures the typical distance from each data value to the mean.
$$s_x = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$$
The variance is $s_x^2$. Divide by $n-1$ (not $n$) for an unbiased estimate of the population variance.
Key properties of standard deviation:
Data: 3, 7, 7, 6, 5. Find $s_x$.
Step 1 — Mean: $\bar{x} = (3+7+7+6+5)/5 = 28/5 = 5.6$
Step 2 — Deviations squared:
| $x_i$ | $x_i - \bar{x}$ | $(x_i - \bar{x})^2$ |
|---|---|---|
| 3 | −2.6 | 6.76 |
| 7 | 1.4 | 1.96 |
| 7 | 1.4 | 1.96 |
| 6 | 0.4 | 0.16 |
| 5 | −0.6 | 0.36 |
| Sum | 11.20 |
Step 3: $s_x = \sqrt{11.20/(5-1)} = \sqrt{2.80} \approx \mathbf{1.67}$
Data: 2, 4, 4, 4, 5, 5, 7, 9. Find the IQR and determine whether any outliers exist.
| Situation | Center | Spread |
|---|---|---|
| Symmetric, no outliers | Mean ($\bar{x}$) | Standard deviation ($s_x$) |
| Skewed or outliers present | Median ($M$) | IQR |
This pairing matters: always report mean with $s_x$ and median with IQR. Mixing them (e.g., median with $s_x$) is penalized on the AP exam.
AP Exam Tip: When comparing distributions, always address shape, center, spread, and outliers (SOCS) and use comparative language in context: "The IQR for Group A (18 pts) is larger than for Group B (12 pts), indicating Group A has more variability in scores."
A z-score measures how many standard deviations a value $x$ is from the mean:
$$z = \frac{x - \bar{x}}{s_x}$$
A positive z-score means the value is above the mean; negative means below. Z-scores have no units.
Z-scores allow comparison across different scales. A student scoring 82 on Test A ($\bar{x}=75, s=8$) and 76 on Test B ($\bar{x}=70, s=4$) performed better on Test B ($z_A = 0.875$, $z_B = 1.5$).
AP Statistics scores: $\bar{x} = 3.1$, $s_x = 0.9$. A student scored 5. Find and interpret the z-score.
$$z = \frac{5 - 3.1}{0.9} = \frac{1.9}{0.9} \approx \mathbf{2.11}$$
Interpretation: The student's score of 5 is about 2.11 standard deviations above the mean AP Statistics score.
Interactive: Explore how mean and standard deviation shape a Normal distribution
Figure 2.1 — Normal Distribution with Adjustable Mean and Standard Deviation
For data that follow a Normal distribution, the Empirical Rule gives the percentage of values within 1, 2, and 3 standard deviations of the mean:
$\approx \mathbf{68\%}$ of values fall within $\bar{x} \pm 1s_x$
$\approx \mathbf{95\%}$ of values fall within $\bar{x} \pm 2s_x$
$\approx \mathbf{99.7\%}$ of values fall within $\bar{x} \pm 3s_x$
Adult male heights are approximately Normal with $\bar{x} = 70$ in and $s_x = 3$ in.
(a) What percent of men are between 67 and 73 inches?
67 = 70 − 3 = $\bar{x} − s_x$ and 73 = $\bar{x} + s_x$ → approximately 68%
(b) What percent are taller than 76 inches?
76 = 70 + 2(3) = $\bar{x} + 2s_x$. 5% are outside $\bar{x} \pm 2s_x$; half (2.5%) are above 76 inches. → approximately 2.5%
(c) Find the z-score for a man who is 74 inches tall.
$$z = \frac{74-70}{3} = \frac{4}{3} \approx 1.33$$
Interactive: 68-95-99.7 Rule — shaded regions show the empirical rule regions
Figure 2.2 — The Empirical Rule for Normal Distributions
SAT Math scores are approximately Normal with $\mu = 530$ and $\sigma = 115$. (a) What percent of students score between 415 and 645? (b) What is the z-score for a student who scored 760?
Interactive: Z-score visualizer — see where a value falls relative to the distribution
Figure 2.3 — Z-Score Position on the Normal Curve
Find the mean and median of: 4, 8, 6, 5, 3, 2, 8, 9, 3, 7. Which is larger? What does this suggest about the distribution's shape?
Home prices in a neighborhood: $280K, $305K, $290K, $315K, $1,200K. Which measure of center better represents a typical home price? Explain.
Data: 12, 15, 18, 20, 22, 25, 28. Find (a) Q₁, (b) Q₃, (c) IQR, (d) any outliers using the 1.5×IQR rule.
A distribution has $\bar{x}=50$ and $s_x=10$. Find the z-scores for values 35, 50, and 68. Interpret each.
IQ scores are Normal with μ=100 and σ=15. Using the Empirical Rule: (a) What percent score between 70 and 130? (b) What percent score above 145?
Class A scores: mean=78, s=12. Class B scores: mean=78, s=4. Both classes have the same mean. What does the difference in standard deviation tell you?
(AP Free Response Style) Two running groups recorded their weekly mileage. Group A: mean=42, median=38, s=18. Group B: mean=35, median=36, s=6. Compare the distributions of weekly mileage for the two groups.
A student scored 88 on a test where $\bar{x}=80$ and $s=6$. On a second test, she scored 74 where $\bar{x}=65$ and $s=9$. On which test did she perform better relative to her class?
$\bar{x} = \dfrac{\sum x_i}{n}$ — the arithmetic average. Sensitive to outliers and skewness. Use when distribution is roughly symmetric.
The middle value when data is sorted. Resistant to outliers. Use when distribution is skewed or has outliers. For skewed data: median is preferred.
$s = \sqrt{\dfrac{\sum(x_i - \bar{x})^2}{n-1}}$ — typical distance from the mean. Use with mean for symmetric distributions.
$\text{IQR} = Q_3 - Q_1$ — spread of the middle 50% of data. Resistant to outliers. Use with median.
$\text{Range} = \max - \min$ — simplest spread measure. Not resistant to outliers.
A value is an outlier if it falls below $Q_1 - 1.5\times\text{IQR}$ or above $Q_3 + 1.5\times\text{IQR}$.