Digital SAT Math · Problem-Solving & Data Analysis
One-variable data (center & spread)
A whole column of the test hands you one variable drawn as a picture — a dot plot, a histogram, a box plot, a frequency table — and asks a question you answer by counting, not by calculating. Almost nothing here needs a formula. What it needs is the discipline to find the middle observation instead of the middle of the axis, and to weight a table by its frequencies instead of averaging the labels.
On the test
| Domain | Problem-Solving and Data Analysis (score report) |
| CB skill | Distributions and measures of center and spread |
| What it looks like | One display — dot plot, histogram, box plot or frequency table — plus a short question about center, spread, or what changes |
| Often asked | “What is the median…?”, “Which statement must be true?”, “How does the mean change if…?”, “Which data set has the greater standard deviation?” |
| Format | Multiple choice and student-produced response |
| Calculator | Allowed throughout; useful for a weighted mean, useless for everything else on this page |
Recognition cues: dot plot, histogram, frequency, box plot, median, mean, range, standard deviation, must be true, greater spread, after the value was corrected.
Pattern recognition
Name the display before you read the question. The display decides the method, and there are only four of them.
- List — values printed in a sentence. Sort first, always. The median is the middle observation, not the middle of the printed order.
- Dot plot — one dot per observation stacked over its value. The dots are the frequencies; count them.
- Frequency table — a value column and a count column. Every statistic must be weighted by the count column.
- Box plot — the five-number summary and nothing else. It shows min, Q1, median, Q3, max. It does not show the mean, the number of observations, or any individual value.
Method
- Name the display and the total. Write down — the number of observations, not the number of bars, dots-per-stack, or table rows. On a dot plot, is the total number of dots; on a frequency table, is the sum of the count column.
- Median by cumulative count. The middle observation sits at position when is odd, and is the average of positions and when is even. Run a running total up the value column until you pass that position, then read the value you landed on.
- Mean by weighting. From a frequency table or a dot plot,
The denominator is , never the number of distinct values. 4. Spread by looking. Range is . Standard deviation is not the range — it measures typical distance from the mean, so judge it by clustering: the set whose values hug the center has the smaller standard deviation. You will never need to compute one. 5. Answer the asked statistic. Median, mean, range, interquartile range and standard deviation are five different answers to five different questions, and the choice list usually contains at least three of them.
| Words in the stem | What to do |
|---|---|
| median | count to the middle observation |
| mean / average | weight by frequency, then divide by |
| range | — two numbers only |
| interquartile range | — the width of the box |
| standard deviation | compare clustering; do not calculate |
| must be true | test each statement against what the display actually shows |
| how does … change | ask which statistic an extreme value can move |
Worked example 1 — median from a frequency display
Stem. Nineteen students reported how many languages they speak.
| Languages spoken | Number of students |
|---|---|
| 1 | 4 |
| 2 | 7 |
| 3 | 5 |
| 4 | 2 |
| 5 | 1 |
What is the median number of languages spoken?
Step 1 — find . students. Nineteen observations, not five.
Step 2 — locate the middle position. is odd, so the median is the observation at position .
Step 3 — cumulative count up the value column. Through 1 language: 4 students. Through 2 languages: students. The running total passes 10 while it is still on the value 2, so observations 5 through 11 all report 2 languages — and observation 10 is one of them.
Check. Nine students sit below the tenth and nine sit above it, which is what a median of 19 values requires. Answer: 2 languages.
Trap watch. The value column runs 1 to 5, so the middle of the column is 3 — a wrong answer that feels right because it is symmetric. And 10 is the median’s position, not its value. Both errors disappear the moment you insist on ending your count on a number that a student actually reported.
Worked example 2 — mean from a frequency table
Stem. Twenty planting crews recorded how many trees each crew lost over the winter.
| Trees lost | Number of crews |
|---|---|
| 0 | 5 |
| 1 | 9 |
| 2 | 4 |
| 3 | 2 |
What is the mean number of trees lost per crew?
Step 1 — total the observations. crews, so .
Step 2 — total the values, weighted.
Twenty crews lost 23 trees between them.
Step 3 — divide.
Check. Most crews lost 0 or 1 tree, so a mean just above 1 is the right neighbourhood. Answer: 1.15 trees per crew.
Trap watch. Averaging the left-hand column gives — the classic unweighted mean, and it is wrong because it treats the 2 crews that lost 3 trees as though they carried the same weight as the 9 crews that lost 1. Averaging the right-hand column gives , which is a mean number of crews per row and answers nothing anyone asked.
Worked example 3 — spread, and what one extreme value does
Stem. Two batches of seedlings were measured in centimetres.
- Batch R: 68, 70, 71, 72, 74
- Batch S: 60, 66, 71, 76, 82
(a) Both batches have the same mean. Which batch has the greater standard deviation? (b) A sixth seedling of 140 cm is added to batch R. What happens to R’s mean and to R’s median?
(a) Compare, do not compute. Each batch sums to 355, so each has mean . Batch R’s values sit within 3 cm of 71; batch S’s reach 11 cm away on both sides. Distances from the mean are larger in S, so batch S has the greater standard deviation. No formula, no calculator — and notice that the ranges (6 versus 22) point the same way here only by luck, not by rule.
(b) Mean versus median. Before: mean 71, median 71 (the third of five values). After adding 140:
and with six values the median is the average of the third and fourth, .
Check. The mean climbed 11.5 cm; the median climbed 0.5 cm. Every value pulls on a balance point, but only a crossing changes a middle position — exactly the asymmetry the callout promised. Answer: (a) batch S; (b) the mean rises sharply to 82.5 while the median barely moves, to 71.5.
Trap watch. “The batches have the same mean, so they have the same spread” confuses center with spread. “The ranges differ, so the standard deviations differ” gets the right answer here for the wrong reason and will fail you the moment a stem gives two sets the same range on purpose. And assuming the median jumps along with the mean in part (b) is the outlier error that this skill is built around.
Practice
Answer before you open the explanation. Two items are student-produced response (type the number, no choices), matching the real test, and five are built on a display — a dot plot, a histogram, a box plot or a frequency table — because that is how the test actually presents this skill. Every wrong choice below is a specific error with a name.
Question 1 Warm-up
A commuter recorded how many minutes late a train arrived on each of 7 mornings: 9, 4, 12, 2, 19, 6, 4. What is the median number of minutes late?
Show the answer Choice B
Why it is right
Sort the seven values first: 2, 4, 4, 6, 9, 12, 19. With an odd count of 7 observations the median is the value in position (7 + 1) / 2 = 4, which is 6. Three values lie below it and three lie above it, which is exactly what a median of seven observations must do.
Why each other choice fails
- Choice A
- 4 is the value that appears most often, so this reports the mode instead of the median. The most common value and the middle value are different statistics and here they differ by 2 minutes.
- Choice C
- 8 is the mean: (2 + 4 + 4 + 6 + 9 + 12 + 19) / 7 = 56 / 7 = 8. The stem asked for the median, and the 19-minute delay pulls the mean above the middle observation.
- Choice D
- 10.5 is the midrange, the halfway point between the smallest and largest values: (2 + 19) / 2 = 10.5. That averages only the two extremes and ignores the five observations in between.
Question 2 Standard
The dot plot shows the number of days absent last term for each of the 25 students in a seminar. What is the median number of days absent?
Show the answer Choice A
Why it is right
There are 25 dots, so the median is the observation at position (25 + 1) / 2 = 13. Run a cumulative count up the axis: 3 dots at 0 days, 3 + 6 = 9 through 1 day, and 9 + 7 = 16 through 2 days. The running total first passes 13 while it is still on the value 2, so observation 13 is a student with 2 days absent.
Why each other choice fails
- Choice B
- 2.24 is the mean, not the median: (0·3 + 1·6 + 2·7 + 3·4 + 4·2 + 5·2 + 6·1) / 25 = 56 / 25 = 2.24. The two students at 5 days and the one at 6 pull the mean above the middle observation.
- Choice C
- 3 is the midpoint of the horizontal axis, which runs from 0 to 6. The median is the middle observation, not the middle of the axis, and most of the dots are stacked to the left of 3.
- Choice D
- 13 is the position of the median, not its value. Cumulative counting locates observation number 13; the answer is the number of days that observation records, which is 2.
Question 3 Standard
A zine library holds 6 issues whose mean length is 82 pages. A seventh issue, 96 pages long, is added to the collection. What is the mean length, in pages, of the 7 issues?
Show the answer Choice B
Why it is right
A mean is a total divided by a count, so recover the total first: the 6 original issues hold 6 × 82 = 492 pages. Adding the seventh gives 492 + 96 = 588 pages spread over 7 issues, so the new mean is 588 / 7 = 84 pages. The mean rose by 2 because the new issue is 14 pages above the old mean and that surplus is shared among all 7 issues.
Why each other choice fails
- Choice A
- 82 assumes the mean is unchanged. Adding a value above the current mean must raise the mean; only a value exactly equal to 82 would leave it alone.
- Choice C
- 89 is (82 + 96) / 2, averaging the old mean with the new value as if each carried equal weight. The old mean already stands for 6 issues, so it must be weighted 6 times as heavily as the single new one.
- Choice D
- 98 is 588 / 6, the correct new total divided by the old count. Once the seventh issue joins, the divisor has to be 7.
Question 4 Standard
The table shows how many nights each of the 30 guests who checked out of a hostel last week had stayed. What is the mean number of nights stayed per guest?
| Nights stayed | Number of guests |
|---|---|
| 1 | 7 |
| 2 | 8 |
| 3 | 6 |
| 4 | 5 |
| 5 | 4 |
Show the answer Choice B
Why it is right
Weight every value by its frequency. The 30 guests stayed 1·7 + 2·8 + 3·6 + 4·5 + 5·4 = 7 + 16 + 18 + 20 + 20 = 81 nights in total, and the count column sums to 7 + 8 + 6 + 5 + 4 = 30 guests. The mean is therefore 81 / 30 = 2.7 nights per guest.
Why each other choice fails
- Choice A
- 2.5 is the median, not the mean. Guests 15 and 16 of the 30 stayed 2 and 3 nights, so the median is (2 + 3) / 2 = 2.5; the stem asked for the mean.
- Choice C
- 3 averages the left-hand column only: (1 + 2 + 3 + 4 + 5) / 5 = 3. That is the unweighted mean, and it wrongly treats the 4 guests who stayed 5 nights as equal in weight to the 8 guests who stayed 2.
- Choice D
- 6 is 30 / 5, the number of guests divided by the number of table rows. That averages the wrong column and produces guests per row, not nights per guest.
Question 5 Standard
A studio recorded the number of pieces fired in each of its last 8 kiln loads: 44, 31, 38, 29, 33, 27, 41, 35. What is the median number of pieces per load?
Show the answer Choice C
Why it is right
Sort the eight values: 27, 29, 31, 33, 35, 38, 41, 44. With an even count there is no single middle observation, so the median is the average of the two central ones, in positions 4 and 5: (33 + 35) / 2 = 34. Four loads fired fewer than 34 pieces and four fired more, which is the balance a median of eight values requires.
Why each other choice fails
- Choice A
- 29 is the fourth value in the order the stem prints them, so this skips the sorting step entirely. The printed order is chronological, not numerical, and 29 is only the second-smallest load.
- Choice B
- 33 is the lower of the two central values. With eight observations you cannot stop at one of them; the median of an even-sized set is the average of both middle values.
- Choice D
- 34.75 is the mean: (27 + 29 + 31 + 33 + 35 + 38 + 41 + 44) / 8 = 278 / 8 = 34.75. The stem asked for the median, which comes from position, not from a total.
Question 6 Harder
The histogram shows the number of hours worked last season by each of a park's 40 registered volunteers. How many of these volunteers worked fewer than 20 hours?
Show the answer Choice C
Why it is right
Every volunteer who worked fewer than 20 hours falls into one of the first two intervals, 0 to under 10 hours and 10 to under 20 hours. Their bar heights are 18 and 6, so the count is 18 + 6 = 24 volunteers. As a check, the five bars total 18 + 6 + 5 + 6 + 5 = 40, which matches the 40 volunteers named in the stem.
Why each other choice fails
- Choice A
- 16 is the number who worked 20 hours or more (5 + 6 + 5), the complement of what was asked. It is the right count of the wrong group.
- Choice B
- 18 is the height of the first bar alone. That counts only volunteers under 10 hours and drops the 6 who worked between 10 and 20, all of whom also worked fewer than 20.
- Choice D
- 29 is 18 + 6 + 5, which sweeps in the 20-to-under-30 bar. Those 5 volunteers worked at least 20 hours, so the boundary was read one interval too far to the right.
Question 7 Harder Student-produced response
A test garden has 20 plots. At harvest, 6 plots had produced 3 melons each, 9 plots had produced 4 melons each, and 5 plots had produced 5 melons each. What is the mean number of melons per plot?
Show the answer 3.95
Why it is right
Weight each yield by the number of plots that produced it: 6 × 3 = 18 melons, 9 × 4 = 36 melons, and 5 × 5 = 25 melons, for a total of 18 + 36 + 25 = 79 melons. The plot counts sum to 6 + 9 + 5 = 20, matching the 20 plots in the stem, so the mean is 79 / 20 = 3.95 melons per plot.
Answers students type instead
- 4
- Averages the three distinct yields, (3 + 4 + 5) / 3 = 4, as if each yield described the same number of plots. The 9 plots at 4 melons must count nine times, not once.
- 79
- The total number of melons harvested. That is the numerator of the mean, not the mean; it still has to be divided by the 20 plots.
- 6.67
- Averages the plot counts instead of the yields, (6 + 9 + 5) / 3 = 20 / 3, which is a mean number of plots per row of the data and answers nothing the stem asked.
Question 8 Harder
A technician discovers that a probe under-reported every reading, so each of the 24 soil-moisture readings in a data set is increased by 5 units. Which of the following describes the effect on the mean and on the standard deviation of the data set?
Show the answer Choice D
Why it is right
Adding 5 to every reading adds 5·24 = 120 to the total, so the mean rises by 120 / 24 = 5. The standard deviation measures how far the readings sit from their mean, and both the readings and the mean have moved 5 units in the same direction, so every distance is exactly what it was. Shifting a whole data set relocates its center and leaves its spread untouched.
Why each other choice fails
- Choice A
- Treats the standard deviation as though it shifts along with the center. A standard deviation is a distance, not a location, and a shift that moves every reading and the mean together changes no distance.
- Choice B
- Says the mean is unchanged, which would only be true if the increases cancelled out. Every reading rose, so the total rose and the mean must rise with it.
- Choice C
- Applies the correct conclusion about the standard deviation to the mean as well. The set really did move, and the mean is exactly the statistic that records where it moved to.
Question 9 Harder
The box plots summarize the patient wait times, in minutes, recorded at Clinic P and at Clinic Q on the same afternoon. Based on the box plots, which of the following statements must be true?
Show the answer Choice C
Why it is right
The median is the line drawn inside each box, and both boxes have that line at 24 minutes, so the medians are equal. Every other comparison here fails: the boxes run 18 to 30 and 22 to 26, so the interquartile ranges are 12 and 4 minutes; the whiskers run 12 to 36 and 20 to 44, so both ranges are 24 minutes; and a box plot never reports a mean.
Why each other choice fails
- Choice A
- The interquartile range is the width of the box, Q3 minus Q1. Clinic P's box spans 30 - 18 = 12 minutes and Clinic Q's spans 26 - 22 = 4 minutes, so they are far from equal.
- Choice B
- Range is maximum minus minimum. Clinic Q runs 44 - 20 = 24 minutes and Clinic P runs 36 - 12 = 24 minutes, so the ranges are equal. Clinic Q's long right whisker makes it look wider, but a single stretched whisker is not the range.
- Choice D
- A box plot shows only five numbers - minimum, Q1, median, Q3 and maximum - and the mean is not one of them. Without the individual wait times, no statement about the means can be forced to be true.
Question 10 Hardest
Two fundraisers each collected 7 pledges, in dollars. Data set X: 10, 30, 30, 30, 30, 30, 50. Data set Y: 10, 20, 25, 30, 35, 40, 50. Which of the following statements about the two data sets is true?
Show the answer Choice D
Why it is right
Both sets run from 10 to 50, so both have range 50 - 10 = 40, and both total 210 over seven pledges, so both have mean 30. Standard deviation measures typical distance from the mean, and there the sets part company: five of X's seven pledges sit exactly on 30, while Y's pledges are spaced out at 10, 5, 0, 5, 10 and 20 dollars from the mean. Y's values are farther from the center on average, so Y has the greater standard deviation.
Why each other choice fails
- Choice A
- Equal means say where two data sets are centered, not how tightly they cluster. These two sets share a mean of 30 and still differ in spread, which is precisely why center and spread are reported separately.
- Choice B
- Range and standard deviation are different measures. The range looks only at the two extreme pledges, which are 10 and 50 in both sets, so it cannot see that X packs five pledges onto the mean while Y scatters them.
- Choice C
- Gets the direction backwards. Repeating the middle value pulls values toward the mean, which shrinks typical distance rather than growing it, so those five pledges at 30 make X's standard deviation the smaller of the two.
Question 11 Hardest
The dot plot shows the number of overtime shifts worked last month by each of 20 employees. A payroll audit then found that the two employees plotted at 10 shifts had each worked 4 shifts, and the plot was corrected. Which of the following describes how the mean and the median of the 20 values change?
Show the answer Choice A
Why it is right
Before the correction the 20 shifts total 0·3 + 1·5 + 2·6 + 3·2 + 4·2 + 10·2 = 51, so the mean is 51 / 20 = 2.55. Replacing the two 10s with 4s removes 12 shifts, giving 39 and a mean of 39 / 20 = 1.95. The median is the average of observations 10 and 11; the cumulative count reaches 3 at 0 shifts, 8 at 1 shift and 14 at 2 shifts both before and after, so both observations still read 2 and the median stays 2.
Why each other choice fails
- Choice B
- Assumes the median follows the mean. Only the two largest values changed, and they stayed above the middle of the ordered list, so observations 10 and 11 are the same values they were.
- Choice C
- Reverses both effects. The total fell by 12 shifts, so the mean must fall, while the median depends on position and nothing crossed the middle.
- Choice D
- Confuses the count of observations with their values. Keeping 20 values fixes the divisor of the mean, not the numerator, and the numerator dropped from 51 to 39.
Question 12 Hardest Student-produced response
At a warehouse, the 12 pickers on the morning shift filled a mean of 78 orders each, and the 18 pickers on the afternoon shift filled a mean of 88 orders each. No picker worked both shifts. What is the mean number of orders filled per picker across all 30 pickers?
Show the answer 84
Why it is right
Turn each mean back into a total before combining. The morning shift filled 12 × 78 = 936 orders and the afternoon shift filled 18 × 88 = 1584 orders, so all 30 pickers filled 936 + 1584 = 2520 orders. The combined mean is 2520 / 30 = 84 orders per picker. It sits closer to 88 than to 78 because the larger afternoon group carries more weight.
Answers students type instead
- 82
- Pairs each mean with the other shift's headcount: (18 × 78 + 12 × 88) / 30 = 2460 / 30 = 82. The weights are swapped, which tilts the answer toward the wrong shift.
- 83
- Averages the two shift means, (78 + 88) / 2 = 83, as if the shifts were the same size. The 18 afternoon pickers must count 18 times and the 12 morning pickers 12 times.
- 2520
- The combined number of orders filled. That is the total, not the mean, and it still has to be divided by the 30 pickers.
Common mistakes
- Median read off the middle of the axis — the midpoint of the number line under a dot plot, or the middle bar of a histogram, has nothing to do with where the observations pile up. Count.
- Median taken from the printed order — a list in a stem is almost never sorted, and the value sitting fourth in the sentence is not the fourth-smallest.
- Position reported instead of value — with 25 observations the median is at position 13; the answer is the value that observation 13 holds, not 13.
- Mean computed without the frequencies — averaging the value column of a frequency table treats a row representing 9 crews and a row representing 2 crews as equals.
- The wrong column averaged — dividing the total number of observations by the number of table rows produces a number with no meaning, and it is always in the choice list.
- Range treated as standard deviation — the range sees only the two endpoints, so two sets can share a range and differ sharply in spread. Judge standard deviation by how tightly values cluster around the mean.
- Assuming an outlier drags the median as far as the mean — a far-away value moves a balance point a long way and a middle position hardly at all.
- Adding a constant to every value assumed to change the spread — shifting the whole set moves the mean and the median by that constant and leaves the standard deviation exactly where it was.
- Reading a mean off a box plot — the line inside the box is the median. A box plot shows five numbers, none of which is the mean and none of which is .
- Answering a different statistic — the median when the stem said mean, the interquartile range when it said range, or the modal value’s frequency when it asked for the value.
FAQ
Do I ever have to compute a standard deviation? No. The Digital SAT asks you to compare or interpret spread, never to calculate it, and no standard-deviation formula appears on the reference sheet. If a question seems to demand the number, you have misread it — re-read for the word greater, smaller or unchanged.
What is the difference between range and standard deviation? Range is a subtraction between the two most extreme values; standard deviation summarises how far a typical value sits from the mean. One outlier can inflate the range without much affecting how clustered the rest of the data is, which is why the two can disagree, and why the test enjoys building an item on exactly that disagreement.
A box plot is on screen and the question asks for the mean. Now what? It cannot be answered. A box plot reports min, Q1, median, Q3 and max; individual values and the count are invisible, so the mean is not recoverable. On a “must be true” item, any statement about the mean of a box-plotted set is a distractor unless the stem supplies the values separately.
How do I type these answers into a grid-in? Type the number the last line asks for and nothing else — no units, no percent sign. A weighted mean often lands on a decimal such as 3.95, and the grid accepts it; it also accepts an equivalent fraction. If your answer is a count of observations, it will be a whole number, which is a useful sanity check on the arithmetic.
Does the mode show up? Rarely, and never as the point of a question. The tallest stack on a dot plot is worth noticing because it tells you where the data is dense, but College Board’s skill is named for center and spread. Do not report the height of the tallest stack when a stem asks for a value.