A data set can be summarized by its center and spread. On the SAT, the main challenge is often not arithmetic but deciding which measure the question describes and preserving the correct number of data values.
Four essential measures
| Measure | What it describes | How to find it |
|---|---|---|
| Mean | Arithmetic balance point | Add all values and divide by the count |
| Median | Middle of an ordered list | Sort first; average the two middle values when the count is even |
| Mode | Most frequent value or values | Compare frequencies; a set can have one mode, several modes, or no mode |
| Range | Total spread from least to greatest | Subtract the minimum from the maximum |
Here \(\sum x_i\) is the sum of all data values and \(n\) is the number of values.
Only the two endpoint values determine the range.
Summarize one data set
Find the mean, median, mode, and range of \(4,7,7,9,13\).
- Mean
The sum is \(40\), so the mean is \(40/5=8\).
- Median
The list is ordered and the middle value is \(7\).
- Mode
The value \(7\) occurs twice; every other value occurs once.
- Range
Subtract the endpoints: \(13-4=9\).
Median and mode details
Mode outcomes
One mode
One value has the unique greatest frequency, as in \(2,2,5,8\).
Multiple modes
Two or more values tie for the greatest frequency, as in \(1,1,4,4,7\).
No mode
No value occurs more often than another, as in \(3,6,9\).
Reverse the mean relationship
Mean times count gives the total. This is the fastest route to missing-value and combined-group questions.
Find a missing value
The mean of \(6,10,12,x\) is \(11\). Find \(x\).
- Required total
Four values with mean \(11\) must total \(4(11)=44\).
- Known total
The known values total \(6+10+12=28\).
- Subtract
Therefore \(x=44-28=16\).
Frequency tables and weighted means
| Score \(x\) | Frequency \(f\) | Contribution \(xf\) |
|---|---|---|
| \(2\) | \(1\) | \(2\) |
| \(5\) | \(3\) | \(15\) |
| \(8\) | \(2\) | \(16\) |
| Total | \(6\) | \(33\) |
Multiply each value by its frequency, add those contributions, then divide by the total frequency.
Combine groups with different sizes
A group of \(12\) students has mean \(74\), and a group of \(18\) students has mean \(84\). Find the combined mean.
- Recover totals
The group totals are \(12(74)=888\) and \(18(84)=1512\).
- Combine
The total score is \(2400\) for \(30\) students.
- Divide
The combined mean is \(2400/30=80\).
How changes affect the measures
- Adding \(k\) to every value adds \(k\) to the mean and median, leaves frequencies attached to shifted values, and leaves the range unchanged.
- Multiplying every value by a positive factor \(k\) multiplies the mean, median, mode values, and range by \(k\).
- One extreme outlier can move the mean and range substantially; the median is generally more resistant.
- Adding one value changes both the total and the count, so update both before recomputing the mean.
Common mistakes and traps
- Finding the median before sorting the data.
- Dividing by the wrong count after adding or removing a value.
- Treating a frequency-table row as one observation instead of \(f\) observations.
- Averaging group means without weighting them by group size.
- Reporting no mode when two values tie for the greatest frequency.
- Computing range as maximum plus minimum instead of maximum minus minimum.
Check your understanding
Five values have mean \(18\). A sixth value, \(30\), is added. What is the new mean?
- \(18\)
- \(20\)
- \(21\)
- \(24\)
Show answer and explanation
Answer: \(20\)
The original total is \(5(18)=90\). The new total is \(120\), and \(120/6=20\).
What to remember
- Mean is total divided by count; reverse it with \(\text{total}=n\bar{x}\).
- Sort before finding a median, and average the two middle values for an even count.
- Mode is determined by frequency; range uses only the minimum and maximum.
- Frequency and combined-group means are weighted by their counts.
- Outliers usually affect the mean and range more than the median.
Put these notes into practice
Apply the ideas with SAT-style questions, then reinforce key details with flashcards.