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Comparing Data Distributions

SAT Math · Problem-Solving & Data Analysis · Updated June 2026

Comparing two data sets means looking at shape, center, and spread together — and picking the center measure that fits the shape.

Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.

What the SAT tests here

For roughly symmetric data, the mean is a good center; for skewed data or data with outliers, the median is more representative. Spread is described by range or standard deviation.

How to solve one, step by step

Example: a strongly right-skewed income data set.

  1. A few very large values pull the mean upward.
  2. The median better represents the typical value.

The mistakes that cost points

Practice questions

Try these the way you would on test day, then open the solution to check your method.

Easy
The dot plots show quiz scores for Class A and Class B. Which class has the greater median score?
  • AClass A
  • BCannot be determined from dot plots
  • CClass B
  • DBoth classes have the same median
Show solution
Answer: A, Class A. Class A: $10$ values, median $= $ avg of 5th and 6th $= \frac{78 + 80}{2} = 79$. Class B: median $= $ avg of 5th and 6th $= \frac{72 + 74}{2} = 73$. Class A has the greater median.
Medium
The dot plots show the quiz scores for students in Class A and Class B. Which of the following statements must be true?
  • AThe medians are equal and the standard deviation of Class A is less than that of Class B.
  • BThe mean of Class A is greater than the mean of Class B.
  • CThe standard deviation of Class A is greater than that of Class B.
  • DThe median of Class A is greater than the median of Class B.
Show solution
Answer: A, The medians are equal and the standard deviation of Class A is less than that of Class B.. Class A median: 10 values, average of 5th and 6th = (80+80)/2 = 80. Class B median: average of 5th and 6th = (80+80)/2 = 80. Medians are equal. Class A is more tightly clustered around 80; Class B has more spread (60 to 100), so Class A has a smaller standard deviation.
Hard
The dot plot shows the number of hours per week spent exercising for 12 adults. If one adult who exercises 20 hours per week is added to the data, which of the following best describes the effect on the mean and median?
  • AThe mean increases more than the median
  • BThe median increases more than the mean
  • CBoth the mean and median increase by the same amount
  • DNeither the mean nor the median changes significantly
Show solution
Answer: A, The mean increases more than the median. The new value of $20$ is far above the existing data (max $\approx 8$). It will pull the mean up significantly. The median, being a middle value, shifts only slightly.

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