Statistics and Data Interpretation

Analyze data using mean, median, mode, standard deviation, and data displays.

Written and reviewed by the Study Mondo Education TeamLast updated April 18, 2026

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Statistics and Data Interpretation on the SAT

Central Tendency: Mean, Median, Mode

Mean (Average)

$\text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}}$

Key property: The mean is sensitive to outliers.

Median

The middle value when data is arranged in order.

Odd number of values: the middle one
Even number of values: average of the two middle ones

Key property: The median is resistant to outliers.

Mode

The most frequently occurring value. A data set can have no mode, one mode, or multiple modes.

When to Use Each Measure

Situation	Best Measure	Why
Symmetric data, no outliers	Mean	Representative of all values
Skewed data or outliers	Median	Not pulled by extremes
Categorical data	Mode	Most common category

SAT Tip: The SAT often asks "which measure best represents the data" — choose median when there are outliers.

Spread: Range, Standard Deviation, IQR

Range

$\text{Range} = \text{Maximum} - \text{Minimum}$

Interquartile Range (IQR)

$\text{IQR} = Q_3 - Q_1$ Measures the spread of the middle 50% of data.

Standard Deviation

Measures how spread out data is from the mean.

Small SD: Data is clustered near the mean
Large SD: Data is spread out from the mean

You do NOT need to calculate standard deviation on the SAT — just understand what it means!

Box Plots (Box-and-Whisker)

A box plot displays the 5-number summary:

Minimum
$Q_1$ (25th percentile)
Median (50th percentile)
$Q_3$ (75th percentile)
Maximum

The box spans from $Q_1$ to $Q_3$ ; the line inside is the median.

Reading Tables and Graphs

Histograms

$x$ -axis: intervals (bins)
$y$ -axis: frequency
To find total: add all bar heights
To find median: count from left until you reach the middle value

Scatterplots

Look for positive, negative, or no association
A line of best fit (regression line) approximates the trend
Correlation coefficient $r$ : close to +1 (strong positive), close to -1 (strong negative), close to 0 (no correlation)

Dot Plots

Each dot = one data point
Easy to find mode, range, and count

Margin of Error and Confidence Intervals

$\text{Confidence Interval} = \text{Estimate} \pm \text{Margin of Error}$

Example: A survey estimates 45% support with a margin of error of 3%. This means the true value is likely between 42% and 48%.

To decrease margin of error: Increase sample size.

SAT Question Types

Type 1: Calculate Mean/Median

"Find the mean of 12, 15, 18, 18, 22" → $\frac{85}{5} = 17$

Type 2: Effect of Adding/Removing Values

"If value 100 is added to the set above, how does it affect the mean vs. median?" Mean changes significantly (sensitive to outliers), median barely changes.

Type 3: Interpret Graphs

Read values from bar charts, histograms, line graphs. Pay attention to axis labels and scales!

Type 4: Compare Distributions

"Set A has mean 50 and SD 5. Set B has mean 50 and SD 10. Which is more spread out?" Set B (larger SD).

Common SAT Mistakes

Confusing mean and median — the SAT specifically tests whether you know which is affected by outliers
Misreading graph scales — check if axes start at 0
Confusing correlation with causation — correlation does NOT prove one variable causes another
Forgetting to order data before finding the median
Misinterpreting standard deviation — it's about spread, not about the mean itself

📚 Practice Problems

1Problem 1easy

❓ Question:

The ages of 5 students are: 14, 16, 15, 14, 17. What is the median age?

💡 Show Solution

Step 1: Arrange in order: 14, 14, 15, 16, 17

Step 2: Find the middle value. With 5 values, the median is the 3rd value.

Answer: Median = 15

Note: Don't forget to sort first! The original order doesn't matter for median.

2Problem 2easy

❓ Question:

The ages of 5 students are: 14, 16, 15, 14, 17. What is the median age?

💡 Show Solution

Step 1: Arrange in order: 14, 14, 15, 16, 17

Step 2: Find the middle value. With 5 values, the median is the 3rd value.

Answer: Median = 15

Note: Don't forget to sort first! The original order doesn't matter for median.

3Problem 3easy

❓ Question:

The ages of 5 students are: 14, 16, 15, 14, 17. What is the median age?

💡 Show Solution

Step 1: Arrange in order: 14, 14, 15, 16, 17

Step 2: Find the middle value. With 5 values, the median is the 3rd value.

Answer: Median = 15

Note: Don't forget to sort first! The original order doesn't matter for median.

4Problem 4medium

❓ Question:

A data set has a mean of 72 and contains 8 values. If a 9th value of 90 is added, what is the new mean?

💡 Show Solution

Step 1: Find the current sum. $\text{Mean} = \frac{\text{Sum}}{n} \implies \text{Sum} = \text{Mean} \times n = 72 \times 8 = 576$

Step 2: Add the new value. $\text{New Sum} = 576 + 90 = 666$

Step 3: Calculate the new mean. $\text{New Mean} = \frac{666}{9} = 74$

Answer: New mean = 74

SAT Tip: To find the sum from a mean, multiply: Sum = Mean × Count.

5Problem 5medium

❓ Question:

A data set has a mean of 72 and contains 8 values. If a 9th value of 90 is added, what is the new mean?

💡 Show Solution

Step 1: Find the current sum. $\text{Mean} = \frac{\text{Sum}}{n} \implies \text{Sum} = \text{Mean} \times n = 72 \times 8 = 576$

Step 2: Add the new value. $\text{New Sum} = 576 + 90 = 666$

Step 3: Calculate the new mean. $\text{New Mean} = \frac{666}{9} = 74$

Answer: New mean = 74

SAT Tip: To find the sum from a mean, multiply: Sum = Mean × Count.

6Problem 6medium

❓ Question:

A data set has a mean of 72 and contains 8 values. If a 9th value of 90 is added, what is the new mean?

💡 Show Solution

Step 1: Find the current sum. $\text{Mean} = \frac{\text{Sum}}{n} \implies \text{Sum} = \text{Mean} \times n = 72 \times 8 = 576$

Step 2: Add the new value. $\text{New Sum} = 576 + 90 = 666$

Step 3: Calculate the new mean. $\text{New Mean} = \frac{666}{9} = 74$

Answer: New mean = 74

SAT Tip: To find the sum from a mean, multiply: Sum = Mean × Count.

7Problem 7medium

❓ Question:

A class of 20 students has a mean test score of 78. A class of 30 students has a mean test score of 84. What is the combined mean for all 50 students?

💡 Show Solution

Step 1: Find each class's total points. Class 1: $78 \times 20 = 1{,}560$ Class 2: $84 \times 30 = 2{,}520$

Step 2: Find the combined mean. $\text{Combined Mean} = \frac{1{,}560 + 2{,}520}{20 + 30} = \frac{4{,}080}{50} = 81.6$

Answer: 81.6

Key: You cannot just average the two means (that would give 81). The combined mean is a weighted average because the groups have different sizes.

8Problem 8medium

❓ Question:

A class of 20 students has a mean test score of 78. A class of 30 students has a mean test score of 84. What is the combined mean for all 50 students?

💡 Show Solution

Step 1: Find each class's total points. Class 1: $78 \times 20 = 1{,}560$ Class 2: $84 \times 30 = 2{,}520$

Step 2: Find the combined mean. $\text{Combined Mean} = \frac{1{,}560 + 2{,}520}{20 + 30} = \frac{4{,}080}{50} = 81.6$

Answer: 81.6

Key: You cannot just average the two means (that would give 81). The combined mean is a weighted average because the groups have different sizes.

9Problem 9medium

❓ Question:

A class of 20 students has a mean test score of 78. A class of 30 students has a mean test score of 84. What is the combined mean for all 50 students?

💡 Show Solution

Step 1: Find each class's total points. Class 1: $78 \times 20 = 1{,}560$ Class 2: $84 \times 30 = 2{,}520$

Step 2: Find the combined mean. $\text{Combined Mean} = \frac{1{,}560 + 2{,}520}{20 + 30} = \frac{4{,}080}{50} = 81.6$

Answer: 81.6

Key: You cannot just average the two means (that would give 81). The combined mean is a weighted average because the groups have different sizes.

10Problem 10hard

❓ Question:

A survey of 500 residents found that 62% support a new park, with a margin of error of 4%. Which of the following is a valid conclusion?

A) Exactly 62% of all residents support the park. B) Between 58% and 66% of all residents likely support the park. C) At least 58% of all residents definitely support the park. D) The survey is unreliable because of the margin of error.

💡 Show Solution

Analysis of each option:

A) "Exactly 62%" — No. The 62% is an estimate, not an exact figure. ✗

B) "Between 58% and 66% likely support" — Yes! The confidence interval is $62\% \pm 4\% = [58\%, 66\%]$ . "Likely" is the right word because it's a probability statement. ✓

C) "At least 58% definitely" — No. "Definitely" is too strong. There's a small chance the true value is outside the interval. ✗

D) "Survey is unreliable" — No. All surveys have margins of error; this doesn't make them unreliable. ✗

Answer: B

SAT Tip: Confidence intervals give a RANGE of plausible values, not a guarantee. Watch for words like "definitely" or "exactly" — they're usually wrong.

11Problem 11hard

❓ Question:

A survey of 500 residents found that 62% support a new park, with a margin of error of 4%. Which of the following is a valid conclusion?

💡 Show Solution

Analysis of each option:

A) "Exactly 62%" — No. The 62% is an estimate, not an exact figure. ✗

B) "Between 58% and 66% likely support" — Yes! The confidence interval is $62\% \pm 4\% = [58\%, 66\%]$ . "Likely" is the right word because it's a probability statement. ✓

C) "At least 58% definitely" — No. "Definitely" is too strong. There's a small chance the true value is outside the interval. ✗

D) "Survey is unreliable" — No. All surveys have margins of error; this doesn't make them unreliable. ✗

Answer: B

SAT Tip: Confidence intervals give a RANGE of plausible values, not a guarantee. Watch for words like "definitely" or "exactly" — they're usually wrong.

12Problem 12hard

❓ Question:

A survey of 500 residents found that 62% support a new park, with a margin of error of 4%. Which of the following is a valid conclusion?

💡 Show Solution

Analysis of each option:

A) "Exactly 62%" — No. The 62% is an estimate, not an exact figure. ✗

B) "Between 58% and 66% likely support" — Yes! The confidence interval is $62\% \pm 4\% = [58\%, 66\%]$ . "Likely" is the right word because it's a probability statement. ✓

C) "At least 58% definitely" — No. "Definitely" is too strong. There's a small chance the true value is outside the interval. ✗

D) "Survey is unreliable" — No. All surveys have margins of error; this doesn't make them unreliable. ✗

Answer: B

SAT Tip: Confidence intervals give a RANGE of plausible values, not a guarantee. Watch for words like "definitely" or "exactly" — they're usually wrong.

13Problem 13expert

❓ Question:

Two data sets each have 10 values. Set A: {2, 3, 4, 5, 5, 5, 6, 7, 8, 15}. Set B: {4, 4, 5, 5, 5, 5, 6, 6, 7, 8}. Which set has the greater standard deviation, and which measure of center (mean or median) would differ more between the sets?

💡 Show Solution

Step 1: Compare the spreads.

Set A has values ranging from 2 to 15, with the outlier 15 pulling the data wide. Set B has values from 4 to 8, tightly clustered.

Set A has the greater standard deviation because it is more spread out, especially due to the outlier 15.

Step 2: Compare the means. Set A mean: $\frac{2+3+4+5+5+5+6+7+8+15}{10} = \frac{60}{10} = 6.0$ Set B mean: $\frac{4+4+5+5+5+5+6+6+7+8}{10} = \frac{55}{10} = 5.5$

Step 3: Compare the medians. Set A median: average of 5th and 6th values = $\frac{5+5}{2} = 5$ Set B median: average of 5th and 6th values = $\frac{5+5}{2} = 5$

Step 4: The means differ by 0.5, but the medians are identical. So the mean differs more between the sets.

Answer: Set A has the greater standard deviation. The mean differs more between the sets because it is affected by the outlier (15) in Set A, while the median is resistant to outliers.

14Problem 14expert

❓ Question:

💡 Show Solution

Step 1: Compare the spreads.

Set A has values ranging from 2 to 15, with the outlier 15 pulling the data wide. Set B has values from 4 to 8, tightly clustered.

Set A has the greater standard deviation because it is more spread out, especially due to the outlier 15.

Step 2: Compare the means. Set A mean: $\frac{2+3+4+5+5+5+6+7+8+15}{10} = \frac{60}{10} = 6.0$ Set B mean: $\frac{4+4+5+5+5+5+6+6+7+8}{10} = \frac{55}{10} = 5.5$

Step 3: Compare the medians. Set A median: average of 5th and 6th values = $\frac{5+5}{2} = 5$ Set B median: average of 5th and 6th values = $\frac{5+5}{2} = 5$

Step 4: The means differ by 0.5, but the medians are identical. So the mean differs more between the sets.

Answer: Set A has the greater standard deviation. The mean differs more between the sets because it is affected by the outlier (15) in Set A, while the median is resistant to outliers.

15Problem 15expert

❓ Question:

💡 Show Solution

Step 1: Compare the spreads.

Set A has values ranging from 2 to 15, with the outlier 15 pulling the data wide. Set B has values from 4 to 8, tightly clustered.

Set A has the greater standard deviation because it is more spread out, especially due to the outlier 15.

Step 2: Compare the means. Set A mean: $\frac{2+3+4+5+5+5+6+7+8+15}{10} = \frac{60}{10} = 6.0$ Set B mean: $\frac{4+4+5+5+5+5+6+6+7+8}{10} = \frac{55}{10} = 5.5$

Step 3: Compare the medians. Set A median: average of 5th and 6th values = $\frac{5+5}{2} = 5$ Set B median: average of 5th and 6th values = $\frac{5+5}{2} = 5$

Step 4: The means differ by 0.5, but the medians are identical. So the mean differs more between the sets.

Answer: Set A has the greater standard deviation. The mean differs more between the sets because it is affected by the outlier (15) in Set A, while the median is resistant to outliers.

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❓ Frequently Asked Questions

What is Statistics and Data Interpretation?▾

Analyze data using mean, median, mode, standard deviation, and data displays.

How can I study Statistics and Data Interpretation effectively?▾

Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 15 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

Is this Statistics and Data Interpretation study guide free?▾

Yes — all study notes, flashcards, and practice problems for Statistics and Data Interpretation on Study Mondo are free to access. No account is needed.

What course covers Statistics and Data Interpretation?▾

Statistics and Data Interpretation is part of the SAT Prep course on Study Mondo, specifically in the Problem-Solving and Data Analysis section. You can explore the full course for more related topics and practice resources.

Are there practice problems for Statistics and Data Interpretation?▾

Yes, this page includes 15 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Statistics and Data Interpretation

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Statistics and Data Interpretation on the SAT

Central Tendency: Mean, Median, Mode

Mean (Average)

Median

Mode

When to Use Each Measure

Spread: Range, Standard Deviation, IQR

Range

Interquartile Range (IQR)

Standard Deviation

Box Plots (Box-and-Whisker)

Reading Tables and Graphs

Histograms

Scatterplots

Dot Plots

Margin of Error and Confidence Intervals

SAT Question Types

Type 1: Calculate Mean/Median

Type 2: Effect of Adding/Removing Values

Type 3: Interpret Graphs

Type 4: Compare Distributions

Common SAT Mistakes

📚 Practice Problems

1Problem 1easy

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2Problem 2easy

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3Problem 3easy

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4Problem 4medium

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5Problem 5medium

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6Problem 6medium

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7Problem 7medium

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8Problem 8medium

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9Problem 9medium

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10Problem 10hard

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11Problem 11hard

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12Problem 12hard

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13Problem 13expert

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14Problem 14expert

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15Problem 15expert

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❓ Frequently Asked Questions