One-variable data: center & spread
These questions ask you to describe a single set of numbers: where its center sits (mean, median, mode) and how spread out it is (range, standard deviation). The most-tested idea is how an outlier pulls the mean but barely moves the median.
What College Board tests
Computing mean and median (including from a frequency table), reasoning about how a changed or added value affects each, comparing the spread of two data sets, and knowing which measure of center is more resistant to outliers.
Center and spread
An outlier drags the mean toward it but leaves the median almost untouched. When a question adds an extreme value, expect the mean to move and the median to hold.
Four worked examples in SAT format. Read the approach, try it yourself, then tap Show the full solution.
1 · Find the median
A data set is shown: . What is the median of the data set?
Approach The data is already sorted. With six values (an even count), the median is the average of the two middle ones — the 3rd and 4th.
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Answer: B
Why the other choices are wrong
A Picks only the 3rd value instead of averaging the middle two.
C Picks only the 4th value.
D Computes the mean, not the median.
2 · How an outlier affects center
A data set is . If the value 15 is replaced by 95, how do the mean and median change?
- Both the mean and the median increase.
- The mean increases; the median stays the same.
- The median increases; the mean stays the same.
- Neither changes.
Approach Think about what each measure depends on. The mean uses every value, so a huge new number pulls it up. The median depends only on the middle position — check whether replacing the largest value changes what sits in the middle.
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Answer: B
Replacing the top value with a far larger one leaves the middle position (13) untouched, so the median holds while the mean jumps.
Why the other choices are wrong
A The median doesn't increase — 13 is still the middle value.
C Reverses the two — it's the mean that moves, not the median.
D The mean clearly changes (12.8 → 28.8).
3 · Mean from a frequency table
The table shows how many households reported each number of pets.
| Pets | Households |
|---|---|
| 1 | 2 |
| 2 | 3 |
| 3 | 5 |
What is the mean number of pets per household?
Approach Don't average 1, 2, 3. Each value is repeated by its frequency. Multiply each pet-count by its number of households, total those, then divide by the total number of households.
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Answer: B
Why the other choices are wrong
A Averages 1, 2, 3 directly, ignoring the frequencies.
C Divides by the number of distinct values (3) instead of total households.
D Takes the most common value (the mode), not the mean.
4 · Compare spread
Two data sets each have five values with the same mean of 22.
Set P: Set Q:
Which statement correctly compares the standard deviations?
- Set P has the larger standard deviation.
- Set Q has the larger standard deviation.
- They have equal standard deviations.
- Standard deviation can't be compared without computing it exactly.
Approach Standard deviation measures how far values sit from the mean. You don't need to compute it — just see which set's values are bunched near the mean and which are flung far from it.
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Answer: B
Both sets center on 22, but Set Q's values are far from that center while Set P's hug it. Greater distance from the mean means greater standard deviation.
Why the other choices are wrong
A Reverses it — Set P is the tightly bunched one.
C Equal means would not force equal spread; their spreads clearly differ.
D You can compare spread by inspection when one set is plainly more dispersed.