SAT Data Analysis — Advanced Practice Questions
SAT Data Analysis — Advanced Practice
Eight original SAT-style problems built around the kinds of statistical reasoning students are expected to recognize, interpret, and solve under test conditions.
How to use this practice set
Each problem begins with the kind of visual information you may encounter on the SAT: a histogram, a table, a distribution, or a frequency display. Before calculating anything, first ask: What is the question really asking me to compare or determine?
Try each question on your own first. Then use the solution to study the reasoning, not just the answer. The goal is to recognize the underlying idea quickly when the SAT presents it in a different context.
Read a Histogram → Estimate the Mean
Use the distribution to estimate a measure of center.
- A) 4.5
- B) 6.5
- C) 7.5
- D) 9.5
Step-by-step solution
Find the midpoint of each interval. The representative values are 1, 4, 7, 10, and 13.
Multiply each representative value by its frequency:
Add the products:
There are 20 students, so divide the approximate total by 20:
The estimate is about 7 books. Among the choices, 7.5 is the closest.
Outliers → Mean, Median & Range
Understand how an unusually large value changes statistics.
18, 20, 21, 22, 23, 24, 25, 26, 27, 28, 91
The value 91 is considered an outlier. If the value 91 is removed, which statement is true?
- A) The mean decreases, the median decreases, and the range decreases.
- B) The mean decreases, the median stays the same, and the range decreases.
- C) The mean stays the same, the median decreases, and the range decreases.
- D) The mean decreases, the median stays the same, and the range stays the same.
Step-by-step solution
Start with the mean. The value 91 is much larger than the other values, so it pulls the mean upward. Removing 91 therefore makes the mean decrease.
Find the original median. There are 11 values, so the 6th value is the median:
After removing 91, there are 10 values. The two middle values are 23 and 24:
Compare the ranges. Originally:
After removing 91:
Rate of Change from a Data Table
Distinguish total change from average change per interval.
| Year | Annual spending |
|---|---|
| 2021 | $2,400,000 |
| 2022 | $2,550,000 |
| 2023 | $2,760,000 |
| 2024 | $2,970,000 |
- A) $150,000
- B) $190,000
- C) $200,000
- D) $570,000
Step-by-step solution
Find the total increase from 2021 to 2024:
From 2021 to 2024 there are three one-year intervals: 2021→2022, 2022→2023, and 2023→2024.
Divide the total increase by 3:
Target Average → Find the Missing Value
Work backward from a required mean.
| Quiz 1 | Quiz 2 | Quiz 3 | Quiz 4 | Quiz 5 | Quiz 6 |
|---|---|---|---|---|---|
| 72 | 81 | 84 | 88 | 90 | ? |
72, 81, 84, 88, 90
What is the minimum score the student must earn on the sixth quiz to have an average score of at least 85?
- A) 92
- B) 93
- C) 94
- D) 95
Step-by-step solution
Add the five existing scores:
Let the sixth score be x. The desired average is at least 85:
Multiply both sides by 6:
Subtract 415:
Therefore, the smallest possible score is 95.
Combined Mean from Two Groups
Combine group means using group sizes.
Group A
80 birds
Mean wingspan: 31 cm
Group B
120 birds
Mean wingspan: 34 cm
- A) 32.2 cm
- B) 32.8 cm
- C) 33.0 cm
- D) 33.5 cm
Step-by-step solution
Convert Group A’s mean into its total wingspan:
Convert Group B’s mean into its total wingspan:
Add the two totals:
Divide by the total number of birds:
Add a Value → What Happens to Mean and Median?
Separate the effects on center measures.
Which statement must be true?
- A) The new mean is greater than 49, and the new median is greater than 52.
- B) The new mean is greater than 49, but the new median is 52.
- C) The new mean is 49, and the new median is greater than 52.
- D) Both the new mean and the new median are greater than their original values.
Step-by-step solution
Consider the mean first. The original mean is 49, while the added value is 90.
Adding a value larger than the current mean pulls the mean upward. Therefore, the new mean must be greater than 49.
Now consider the median. There were 50 values, so the original median is determined by the 25th and 26th values. After adding 90, there are 51 values, and the median is the 26th value.
Because 90 is added at the high end of the ordered data, the central position remains at the original median value. Thus the new median is 52.
Sample Data → Population Interpretation
Use a random sample to estimate a larger population.
| Number of siblings | School A Sample |
School B Sample |
|---|---|---|
| 0 | 82 | 110 |
| 1 | 66 | 52 |
| 2 | 32 | 24 |
| 3 | 14 | 10 |
| 4 | 6 | 4 |
A random sample of 200 students was selected from each school. School A has 2,000 students and School B has 3,000 students.
- A) Exactly 820 students at School A have no siblings.
- B) Approximately 820 students at School A have no siblings.
- C) Exactly 1,100 students at School A have no siblings.
- D) Approximately 1,100 students at School A have no siblings.
Step-by-step solution
In the sample, 82 of the 200 students reported having zero siblings.
So approximately 41% of the students in the school would be expected to have zero siblings.
School A has 2,000 students:
Because the 200 students were a sample rather than every student in the school, the result is an estimate.
Median from a Frequency Table
Locate the middle observations using cumulative frequency.
| Number of siblings | Number of students |
|---|---|
| 0 | 12 |
| 1 | 18 |
| 2 | 16 |
| 3 | 9 |
| 4 | 5 |
- A) 1
- B) 1.5
- C) 2
- D) 2.5
Step-by-step solution
There are 60 students. Because 60 is even, the median is the average of the 30th and 31st values.
Build cumulative frequencies:
| Siblings | Frequency | Cumulative frequency |
|---|---|---|
| 0 | 12 | 12 |
| 1 | 18 | 30 |
| 2 | 16 | 46 |
| 3 | 9 | 55 |
| 4 | 5 | 60 |
The cumulative frequency reaches 30 at a sibling count of 1. Therefore, the 30th value is 1.
The next observation, the 31st value, belongs to the next row. Therefore, the 31st value is 2.
