SAT Math Quiz: Center Shape And Spread Of Data

What this quiz covers

This quiz focuses on Center Shape And Spread Of Data, giving you a quick way to practice the rules, question types, and explanations that matter most for SAT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 20.5
2. 15
3. 23
4. 22 (correct answer)

Explanation: This question asks for the median, which is the middle value when data is arranged in order. The data set 8, 12, 15, 22, 22, 23, 30 is already arranged from least to greatest and contains 7 values. For an odd number of values, the median is the value in the middle position, which is the 4th position. Counting from the left: 1st=8, 2nd=12, 3rd=15, 4th=22, so the median is 22.

Question 2

1. Class A, because its scores are higher
2. Class A, because its range is larger
3. Class B, because it has more extreme scores (correct answer)
4. They have the same standard deviation

Explanation: This question compares the standard deviations (spread around the mean) of two classes. Class A: 6,7,7,8,8,8,9,9 has values clustered near the center. Class B: 2,6,7,8,8,8,9,10 has more extreme values (2 and 10) farther from the center. The standard deviation measures how far values typically are from the mean - Class B's extreme values (2 and 10) create larger deviations from the mean than Class A's values. Even with the same median, the presence of extreme scores increases standard deviation. When comparing spreads visually, look for values far from the center of the distribution.

Question 3

1. X has greater standard deviation because all values match the mean.
2. Y has greater standard deviation because its values vary around the mean. (correct answer)
3. X and Y have the same standard deviation because both have mean 70.
4. Y has standard deviation 0 because it is symmetric around 70.

Explanation: This question compares the standard deviation (spread) of two data sets with the same mean of 70. Data set X has all values equal to 70, so every value is exactly at the mean with zero deviation, giving a standard deviation of 0. Data set Y has values 60, 65, 70, 75, 80 that vary around the mean of 70, with deviations of -10, -5, 0, +5, +10, resulting in a positive standard deviation. Standard deviation measures how spread out values are from the mean - when all values equal the mean, standard deviation is 0; when values vary, it's positive. The key insight is that identical values have no spread, while varied values create spread even if they're symmetric around the mean.

Question 4

1. Range 32; outlier 90 (correct answer)
2. Range 32; outlier 58
3. Range 90; outlier 90
4. Range 7; outlier 65

Explanation: This question asks for the range and identification of an outlier in temperature data: 58, 60, 61, 63, 65, 90. The range is the difference between the maximum and minimum values: 90 - 58 = 32°F. The value 90 stands out as an outlier because it's much higher than the other temperatures, which cluster between 58-65. The other values increase gradually by 1-3 degrees, while 90 jumps 25 degrees from the next highest value (65). A common error is calculating range as the largest value (90) instead of the difference, or identifying the minimum value as the outlier. When identifying outliers, look for values that are unusually far from the rest of the data cluster.

Question 5

1. 15
2. 16
3. 14 (correct answer)
4. 17

Explanation: The question asks for the mean of the data set: 8, 10, 11, 13, 15, 17, 19, 22. To calculate the mean, add all values and divide by the number of values. Sum = 8 + 10 + 11 + 13 + 15 + 17 + 19 + 22 = 115. There are 8 values, so mean = 115 ÷ 8 = 14.375, which rounds to 14. A common error is making arithmetic mistakes in addition or division, or confusing the mean with the median. Always double-check your addition and ensure you're dividing by the correct count of values.

Question 6

1. The mean increases noticeably because it uses all values in the sum. (correct answer)
2. The median increases noticeably because it is the largest value.
3. The mode increases noticeably because an outlier changes frequency most.
4. The range is unchanged because only the middle values matter.

Explanation: This question asks which measure of center is most affected by the outlier value of 46 in the push-up data. To find the mean with the outlier: (18+20+21+21+22+23+24+25+46)÷9 = 220÷9 ≈ 24.4; without the outlier: (18+20+21+21+22+23+24+25)÷8 = 174÷8 = 21.75, showing a significant increase. The median with the outlier is 22 (the 5th value when ordered), and without it's 21.5 (average of 21 and 22), showing minimal change. The mode remains 21 in both cases, and the range changes from 7 to 28, but range isn't a measure of center. The mean is most sensitive to outliers because it uses every value in its calculation, while the median only depends on the middle value(s).

Question 7

1. 22 (correct answer)
2. 25
3. 27
4. 30

Explanation: This question asks for the mode, which is the value that appears most frequently in a dataset. To find the mode, count how many times each value appears in the dataset: 22, 25, 27, 22, 30, 26, 31, 29, 22. Counting the frequencies: 22 appears 3 times, 25 appears 1 time, 27 appears 1 time, 30 appears 1 time, 26 appears 1 time, 31 appears 1 time, and 29 appears 1 time. Since 22 appears most frequently (3 times), it is the mode. A common error is confusing mode with median or mean, so remember that mode specifically refers to the most frequent value.

Question 8

1. 40
2. 50
3. 30 (correct answer)
4. 20

Explanation: When you're given the definition of a function as you are here, $f(x)=x^2 - x$, your job to calculate a function is to take the value in parentheses and plug that in for $x$ wherever it appears in the definition. Here, qualitatively, you're being told "whatever $x$ is, square it and then subtract $x$ from that square." That means that: $f(9)=9^2 - 9$ So: $f(9) = 81 - 9 = 72$ And for f(7) you'd have: $f(7)=7^2 - 7$ So: $f(7) = 49 - 7 = 42$ This means that $f(9)-f(7) = 72 - 42 = 30$ so the correct answer is 30.

Question 9

1. 3
2. 4 (correct answer)
3. 6
4. 5

Explanation: Range = largest minus smallest = 6 − 2 = 4. The distractors come from subtracting the wrong values or misreading the data.

Question 10

1. Median
2. Mean (correct answer)
3. Mode
4. All change equally

Explanation: This question asks which measure of center changes most when removing the outlier value 30. Let's calculate each measure with and without the outlier. With all 8 values: mean = 91/8 = 11.375, median = (8+9)/2 = 8.5, mode = 7 and 10 (both appear twice). Without the outlier (7 values): mean = 61/7 ≈ 8.71, median = 8, mode = 7 and 10. The mean changes by about 2.66, while the median changes by only 0.5, and the mode doesn't change at all. The mean is most sensitive to outliers because it uses all values in its calculation, while the median only depends on the middle value(s). When comparing changes in measures of center, calculate the actual differences to avoid guessing.

Question 11

1. 14
2. 13 (correct answer)
3. 12
4. 13.5

Explanation: This question asks for the median of 9 practice session scores. To find the median, we first verify the data is already ordered: 11, 12, 12, 13, 13, 13, 14, 15, 18. With 9 values (odd number), the median is the middle value, which is the 5th value in the ordered list: 13. A common error would be averaging the two middle values (13 and 13), but this is only necessary when we have an even number of values. When finding the median, always count carefully to identify the exact middle position in odd-numbered data sets.

Question 12

1. Set A has greater standard deviation.
2. Set B has greater standard deviation. (correct answer)
3. They have equal standard deviation.
4. Standard deviation cannot be compared without the median.

Explanation: This question compares standard deviations of two datasets with the same mean (50). Set A: 49, 50, 50, 51 has values very close to the mean. Set B: 20, 50, 50, 80 has values much farther from the mean. Standard deviation measures spread around the mean - the farther values are from the mean, the larger the standard deviation. Set B clearly has greater spread (values differ from mean by up to 30) compared to Set A (values differ by at most 1). Therefore, Set B has the greater standard deviation. When comparing standard deviations visually, look at how far values spread from the center.

Question 13

1. 23
2. 27
3. 8 (correct answer)
4. 12

Explanation: This question asks for the IQR from a box plot's five-number summary. Given: min = 48, Q1 = 52, median = 56, Q3 = 60, max = 75. The interquartile range is calculated as IQR = Q3 - Q1 = 60 - 52 = 8. Don't confuse IQR with the full range (75 - 48 = 27) or use the median in the calculation. The IQR represents the spread of the middle 50% of the data. When reading box plots, remember that IQR only involves the first and third quartiles, not the minimum, maximum, or median.

Question 14

1. They do not affect the mean or the median.
2. They decrease both the mean and the median.
3. They increase the median more than the mean.
4. They increase the mean more than the median. (correct answer)

Explanation: This question examines how high scores (90 and 100) affect the mean and median of quiz scores. The median of 11 scores is the 6th value: 72. The mean is 825/11 = 75. Without the two high scores, the median would be the 5th of 9 values: 71, and the mean would be 635/9 ≈ 70.6. The high scores pull the mean up more (from 70.6 to 75, a change of 4.4) than they affect the median (from 71 to 72, a change of 1). This demonstrates that the mean is more sensitive to extreme values than the median. When analyzing the effect of outliers, consider that the mean uses all values while the median only depends on the middle value(s).

Question 15

1. Mean; median $=5$ (correct answer)
2. Median; median $=5$
3. Mode; median $=4$
4. Mean; median $=6$

Explanation: This question asks which measure of center is most affected by the outlier (20) and what the median is. To find the median of 9 values, we order them: 3, 4, 4, 5, 5, 5, 6, 6, 20, then take the 5th value (middle position), which is 5. The mean is (3+4+4+5+5+5+6+6+20)÷9 = 58÷9 ≈ 6.4, while without the outlier it would be 38÷8 = 4.75, showing the mean changes dramatically. The median remains 5 whether we include the outlier or not, demonstrating that the mean is most affected by outliers. When dealing with outliers, always check how each measure of center responds to extreme values.

Question 16

1. Median is better because it is less affected by outliers (correct answer)
2. Median is worse because it uses all data values
3. Mean is better because it ignores extreme values
4. Mean and median are equally affected by outliers

Explanation: This question asks which measure of center better represents typical customer traffic when there's an outlier (50 customers). The mean is (12+15+15+16+16+18+20+20+50)/9 = 182/9 ≈ 20.2 customers. The median is the 5th value when ordered: 12, 15, 15, 16, 16, 18, 20, 20, 50, so median = 16 customers. The outlier of 50 pulls the mean up significantly (to 20.2) while the median remains at 16, which better represents the typical hour. The median is less affected by outliers because it only depends on the middle position(s), not the actual values. When data contains outliers, the median often provides a better measure of central tendency than the mean.

Question 17

1. 7.0
2. 7.1 (correct answer)
3. 7.15
4. 7.2

Explanation: This question asks for the mode of mile times: 7.2, 7.1, 7.0, 7.3, 7.1, 7.0, 7.1. The mode is the value that appears most frequently. Counting occurrences: 7.0 appears 2 times, 7.1 appears 3 times, 7.2 appears 1 time, 7.3 appears 1 time. Therefore, the mode is 7.1 minutes since it appears most often (3 times). A common error is confusing mode with median (middle value) or mean (average). When finding the mode, always count how many times each distinct value appears. If multiple values tie for the highest frequency, the data set can have multiple modes.

Question 18

1. Median stays 14 (correct answer)
2. Median decreases
3. Median becomes 15
4. Median increases

Explanation: This question asks what happens to the median when we replace 25 with 19 in the ordered data: 6, 9, 12, 13, 15, 16, 18, 25. With 8 values, the median is the average of the 4th and 5th values: (13 + 15)/2 = 14. After replacing 25 with 19, the data becomes: 6, 9, 12, 13, 15, 16, 18, 19 (still ordered). The median is still the average of the 4th and 5th values: (13 + 15)/2 = 14. The median stays the same because we only changed a value that was not in the middle positions. This illustrates that the median is resistant to changes in extreme values - only changes to values at or near the median positions affect it.

Question 19

1. 22
2. 24 (correct answer)
3. 26
4. 28

Explanation: Combined total needed is 13×18 = 234; known totals are 5×14 = 70 and 7×20 = 140, so $x = 234 − (70 + 140) = 24$. Other choices result from arithmetic slips.

Question 20

1. 9
2. 10
3. 12
4. 11 (correct answer)

Explanation: To make the 4th value 10, one inserted value must be 10 and the other must be just above 10; taking $x = 10$ gives the least $y = 11$. Smaller choices place the median below 10 or violate $x < y$.