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 10 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

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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 10 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

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 when data is arranged in order.

❓ Question:

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

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.

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

❓ Question:

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

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.

❓ 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?

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.

❓ 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?

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.

❓ 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?

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.

❓ 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?

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.

❓ 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.

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.

❓ 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?

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.

❓ 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?

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.

❓ Question:

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.

Statistics and Data Interpretation

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Statistics and Data Interpretation on the SAT

Central Tendency: Mean, Median, Mode

Mean (Average)

Median

📚 Practice Problems

1Problem 1easy

❓ Question:

2Problem 2easy

Practice Lab

Practice with Flashcards

Browse All Topics

📌 Related Topics in Problem Solving and Data Analysis

❓ Frequently Asked Questions

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

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6Problem 6medium

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7Problem 7hard

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8Problem 8hard

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9Problem 9expert

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10Problem 10expert

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