SAT Math · Problem-Solving and Data Analysis

Mean, median and spread on the SAT

One-variable data questions ask about center and spread. The mean is the sum divided by the count, and the median is the middle value once the data is in order. Range is the largest value minus the smallest. Standard deviation measures how far values typically sit from the mean. SAT questions mostly compare data sets and ask how an outlier or a changed value moves each measure.

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The key idea

mean=sum of valuesnumber of valuessum=mean×count\text{mean} = \frac{\text{sum of values}}{\text{number of values}} \qquad \text{sum} = \text{mean} \times \text{count}

The second formula is the most useful one on the test. When a question gives an average and changes the data, work with the total, not the average.

For the median, sort the values first. With an odd count, it is the middle value. With an even count, it is the mean of the two middle values.

What a very large outlier does
MeasureEffect of adding a very large value
MeanGoes up a lot: the outlier is in the sum
MedianMoves a little or not at all: it only depends on the middle
RangeGoes up: the maximum changes
Standard deviationGoes up: values are more spread out

You will not need to compute a standard deviation by hand on the SAT. You need to compare: the data set whose values sit farther from its mean has the larger standard deviation.

Worked examples

Example 1: mean and median

Problem Find the mean and median of 4, 9, 3, 12, 7, 9.

  1. Sort the values: 3, 4, 7, 9, 9, 12. There are 6 values, so the median is the mean of the 3rd and 4th.
    7+92=8\frac{7 + 9}{2} = 8
  2. Add all values and divide by 6.
    3+4+7+9+9+126=446=223\frac{3 + 4 + 7 + 9 + 9 + 12}{6} = \frac{44}{6} = \frac{22}{3}

Answer Median 8, mean 223\frac{22}{3}, about 7.33

Example 2: the total trick

Problem The mean of a student's 5 test scores is 84. What score on a 6th test would make the mean of all 6 tests exactly 86?

  1. Total of the first 5 scores.
    5×84=4205 \times 84 = 420
  2. Total needed for 6 tests with mean 86.
    6×86=5166 \times 86 = 516
  3. The 6th score is the difference.
    516−420=96516 - 420 = 96

Answer 96

Example 3: a frequency table

Problem In a survey, 3 students have 1 pet, 5 have 2 pets, 8 have 3 pets and 4 have 4 pets. Find the median and mean number of pets.

  1. Count the students.
    3+5+8+4=203 + 5 + 8 + 4 = 20
  2. With 20 values, the median is the mean of the 10th and 11th. In order, values 1 to 3 are 1 pet, 4 to 8 are 2 pets, and 9 to 16 are 3 pets. So the 10th and 11th are both 3.
  3. For the mean, multiply each value by how often it appears.
    1(3)+2(5)+3(8)+4(4)20=5320=2.65\frac{1(3) + 2(5) + 3(8) + 4(4)}{20} = \frac{53}{20} = 2.65

Answer Median 3, mean 2.65

Example 4 (SAT-hard): remove an outlier

Problem A data set is 12, 15, 15, 18, 20, 64. If 64 is removed, how much do the mean and the median each decrease?

  1. Mean before.
    12+15+15+18+20+646=24\frac{12 + 15 + 15 + 18 + 20 + 64}{6} = 24
  2. Mean after.
    12+15+15+18+205=16\frac{12 + 15 + 15 + 18 + 20}{5} = 16
  3. Median before: the 3rd and 4th values are 15 and 18.
    15+182=16.5\frac{15 + 18}{2} = 16.5
  4. Median after: 5 values, so the middle one, 15. The median drops by 1.5.

Answer The mean decreases by 8 and the median by 1.5.

Common mistakes

  • Taking the median without sorting. The middle of 4, 9, 3, 12, 7, 9 as written is not the median. Fix: always sort first.
  • Averaging averages with different group sizes. A class of 20 averaging 78 and a class of 30 averaging 88 do not average to 83. Fix: combine totals: 20(78)+30(88)50=84\frac{20(78) + 30(88)}{50} = 84.
  • Treating a frequency table as a short list. The values 1, 2, 3, 4 with frequencies 3, 5, 8, 4 are 20 data points, not 4. Fix: multiply each value by its frequency.
  • Thinking standard deviation is about size. 101, 102, 103 has the same spread as 1, 2, 3. Fix: look at distances from the mean, not the values.
  • Guessing that an outlier moves the median a lot. It usually moves it a little or not at all. Fix: find the new middle and compare.

Quick method

Practice

5 SAT-style questions

  1. What is the median of 14, 3, 8, 21, 9, 5, 17?

    1. 8
    2. 9
    3. 11
    4. 21
    Show answer

    Answer: 9

    Sorted: 3, 5, 8, 9, 14, 17, 21. The 4th of 7 values is 9. 11 is the mean, and 21 is the middle of the unsorted list.

  2. The mean of 7 numbers is 12. When one number, 30, is removed, what is the mean of the remaining 6 numbers?

    1. 6
    2. 9
    3. 10
    4. 12
    Show answer

    Answer: 9

    Total is 7×12=847 \times 12 = 84. Remove 30 to get 54, then 54÷6=954 \div 6 = 9. Removing a value above the mean always lowers the mean, so 12 cannot be right.

  3. Which data set has the largest standard deviation?

    1. 10, 10, 10, 10, 10
    2. 6, 8, 10, 12, 14
    3. 2, 10, 10, 10, 18
    4. 9, 10, 10, 10, 11
    Show answer

    Answer: 2, 10, 10, 10, 18

    All four sets have mean 10. In 2, 10, 10, 10, 18, two values sit 8 away from the mean, which is more total spread than 6, 8, 10, 12, 14 (distances 4, 2, 0, 2, 4). The all-10 set has standard deviation 0.

  4. A data set is 20, 22, 23, 25, 26, 27, 29. The 29 is replaced by 90. Which statement is true?

    1. The mean and the median both stay the same.
    2. The mean increases and the median stays the same.
    3. The mean stays the same and the median increases.
    4. The mean and the median both increase.
    Show answer

    Answer: The mean increases and the median stays the same.

    The sum goes up by 61, so the mean goes up. The data is still in the same order with 25 in the middle, so the median stays 25.

  5. Student-produced response: a class of 20 students has a mean test score of 78. Another class of 30 students has a mean of 88. What is the mean score of all 50 students?

    Show answer

    Answer: 84

    Totals: 20×78=156020 \times 78 = 1560 and 30×88=264030 \times 88 = 2640. Combined: 420050=84\frac{4200}{50} = 84. 83 is the trap from averaging the two means.

Frequently asked questions

When should I use the median instead of the mean?

The median is better when the data has outliers or is skewed, like home prices or incomes, because a few huge values pull the mean up. The mean is fine for data without extreme values. SAT questions often ask which one better describes a typical value.

Do I need to calculate standard deviation on the SAT?

Almost never by hand. You need to compare spreads: the set whose values sit farther from its mean has the larger standard deviation. Desmos can compute it with stdev\text{stdev} or stdevp\text{stdevp} if a question needs a value.

What is the difference between range and standard deviation?

Range uses only the largest and smallest values. Standard deviation uses every value's distance from the mean. Two data sets can have the same range but very different standard deviations.

Sources

  1. College Board: SAT Math, Problem-Solving and Data Analysis skills (one-variable data), accessed October 1, 2026

Try asking Ducky

  • “Why doesn't the median change when I replace 29 with 90?”
  • “How do I find the median from a frequency table?”
  • “Which of these two sets has the bigger standard deviation, and why?”

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