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Analyze data using mean, median, mode, standard deviation, and data displays.
Learn step-by-step with practice exercises built right in.
Key property: The mean is sensitive to outliers.
The when data is arranged in order.
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.
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Key property: The median is resistant to outliers.
The most frequently occurring value. A data set can have no mode, one mode, or multiple modes.
| 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.
Measures the spread of the middle 50% of data.
Measures how spread out data is from the mean.
You do NOT need to calculate standard deviation on the SAT — just understand what it means!
A box plot displays the 5-number summary:
The box spans from to ; the line inside is the median.
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.
"Find the mean of 12, 15, 18, 18, 22" →
"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.
Read values from bar charts, histograms, line graphs. Pay attention to axis labels and scales!
"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).
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.
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.
Step 2: Add the new value.
Step 3: Calculate the new mean.
Answer: New mean = 74
SAT Tip: To find the sum from a mean, multiply: Sum = Mean × Count.
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.
Step 2: Add the new value.
Step 3: Calculate the new mean.
Answer: New mean = 74
SAT Tip: To find the sum from a mean, multiply: Sum = Mean × Count.
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: Class 2:
Step 2: Find the combined mean.
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.
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: Class 2:
Step 2: Find the combined mean.
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.
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 . "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.
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 . "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.
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: Set B mean:
Step 3: Compare the medians. Set A median: average of 5th and 6th values = Set B median: average of 5th and 6th values =
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.
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: Set B mean:
Step 3: Compare the medians. Set A median: average of 5th and 6th values = Set B median: average of 5th and 6th values =
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.