SHSAT Math Quiz: Comparing Data Sets
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Comparing Data SetsQuestion 1 of 19

A teacher collected test scores from two classes. Class 1 has 15 students with a mean score of 82. Class 2 has 25 students with a mean score of 78. If 5 students transfer from Class 2 to Class 1, and these 5 students have a combined score of 390, what is the new mean for Class 1?

The new mean for Class 1 will be exactly 80.5 points
The new mean for Class 1 will be exactly 81.0 points
The new mean for Class 1 will be exactly 81.5 points
The new mean for Class 1 will be exactly 82.0 points
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SHSAT Math Quiz

SHSAT Math Quiz: Comparing Data Sets

Practice Comparing Data Sets in SHSAT Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Comparing Data Sets, giving you a quick way to practice the rules, question types, and explanations that matter most for SHSAT 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.

All questions

Question 1

A teacher collected test scores from two classes. Class 1 has 15 students with a mean score of 82. Class 2 has 25 students with a mean score of 78. If 5 students transfer from Class 2 to Class 1, and these 5 students have a combined score of 390, what is the new mean for Class 1?

  1. The new mean for Class 1 will be exactly 80.5 points
  2. The new mean for Class 1 will be exactly 81.0 points (correct answer)
  3. The new mean for Class 1 will be exactly 81.5 points
  4. The new mean for Class 1 will be exactly 82.0 points
Explanation: Class 1 originally has 15 students with mean 82, so total points = 15 × 82 = 1230. The 5 transferring students have combined score of 390. After transfer, Class 1 has 20 students with total points = 1230 + 390 = 1620. New mean = 1620/20 = 81.0. Choice A (80.5) results from calculation errors. Choice C (81.5) might result from incorrectly averaging the means. Choice D (82.0) incorrectly assumes the mean stays the same.

Question 2

The heights in centimeters of the plants in two separate garden beds are recorded below.

Bed A: 42, 45, 47, 49, 5042,\ 45,\ 47,\ 49,\ 50
Bed B: 38, 41, 46, 52, 5538,\ 41,\ 46,\ 52,\ 55

Which bed has the greater median height?

  1. Bed A has the greater median. (correct answer)
  2. Bed B has the greater median.
  3. The two beds have the same median.
  4. The median cannot be determined from the data given.
Explanation: When you encounter questions about median, remember that the median is the middle value when data is arranged in order. Since both datasets are already sorted from smallest to largest, you can find the median directly. For Bed A: 42,45,47,49,5042, 45, 47, 49, 50 With 5 values, the median is the 3rd value (middle position): 4747 cm. For Bed B: 38,41,46,52,5538, 41, 46, 52, 55 With 5 values, the median is also the 3rd value: 4646 cm. Since 47>4647 > 46, Bed A has the greater median height. Looking at the answer choices: Choice A correctly identifies that Bed A has the greater median. Choice B is wrong because Bed B's median (46 cm) is actually smaller than Bed A's median (47 cm). Choice C is incorrect since the medians are different values—47 cm versus 46 cm. Choice D is wrong because the median can absolutely be determined; both datasets provide complete information with clearly ordered values. Strategy tip: When finding the median with an odd number of values, always look for the middle position: for 5 values, it's the 3rd; for 7 values, it's the 4th; and so on. The formula is n+12\frac{n+1}{2} where nn is the number of values. Don't get distracted by the range or spread of the data—focus only on that middle value's position.

Question 3

A runner recorded her daily mile times (in minutes) for two separate weeks.

Week 1: 9.2,9.4,9.5,9.5,9.6,9.8,10.09.2, 9.4, 9.5, 9.5, 9.6, 9.8, 10.0
Week 2: 8.8,9.0,9.1,9.3,11.2,11.4,11.58.8, 9.0, 9.1, 9.3, 11.2, 11.4, 11.5

Considering the median as the measure of center, which week shows faster typical performance (lower median time)?

  1. Week 1, because its median is lower.
  2. Week 2, because its median is lower. (correct answer)
  3. Both weeks have the same median.
  4. The median cannot be found because the data values are not integers.
Explanation: When you encounter questions about comparing datasets, focus on what measure of center or spread is being asked for—here it's the median, which represents the middle value when data is arranged in order. To find each week's median, you need to identify the middle value in each ordered dataset. Since both weeks have 7 values, the median will be the 4th value (the middle position). For Week 1: 9.2,9.4,9.5,9.5,9.6,9.8,10.09.2, 9.4, 9.5, \mathbf{9.5}, 9.6, 9.8, 10.0 The median is 9.59.5 minutes. For Week 2: 8.8,9.0,9.1,9.3,11.2,11.4,11.58.8, 9.0, 9.1, \mathbf{9.3}, 11.2, 11.4, 11.5
The median is 9.39.3 minutes. Since 9.3<9.59.3 < 9.5, Week 2 has the lower (faster) median time. Looking at the wrong answers: Choice A incorrectly states that Week 1 has the lower median—this reverses the comparison. Choice C claims both weeks have the same median, but 9.5≠9.39.5 \neq 9.3. Choice D suggests you can't find the median because values aren't integers, but the median can absolutely be calculated from decimal values—you simply arrange them in order and find the middle position. Study tip: For the SHSAT, remember that the median is always the middle value when data is ordered, regardless of whether the numbers are integers or decimals. With an odd number of values, it's simply the middle position; don't overthink it based on the type of numbers involved.

Question 4

Two soccer teams played 66 games each. Team Jets scored {0,1,1,2,3,5}\{0,1,1,2,3,5\} goals per game, and Team Comets scored {2,2,2,3,3,4}\{2,2,2,3,3,4\} goals per game. Which team's distribution has the higher median?

  1. Team Jets has the higher median.
  2. Team Comets has the higher median. (correct answer)
  3. Both teams have equal medians.
  4. Neither distribution has a single defined median because each has an even number of games.
Explanation: When you encounter a question about medians, remember that the median is the middle value when data is arranged in order. For datasets with an even number of values, the median is the average of the two middle values. Let's find each team's median. Team Jets scored {0,1,1,2,3,5}\{0,1,1,2,3,5\} goals per game. Since this data is already ordered and contains 6 values, the median is the average of the 3rd and 4th values: 1+22=1.5\frac{1+2}{2} = 1.5. Team Comets scored {2,2,2,3,3,4}\{2,2,2,3,3,4\} goals per game. Again with 6 values, the median is the average of the 3rd and 4th values: 2+32=2.5\frac{2+3}{2} = 2.5. Since 2.5>1.52.5 > 1.5, Team Comets has the higher median. Choice A is wrong because Team Jets' median (1.5) is lower than Team Comets' median (2.5). Choice C is incorrect since the medians are clearly different values. Choice D reveals a fundamental misunderstanding—having an even number of data points doesn't prevent you from finding a median. You simply average the two middle values, which gives you a perfectly valid median. The key strategy here is remembering that with even-numbered datasets, you always take the average of the two middle values. Don't let the even number of data points confuse you into thinking no median exists—this is a common trap on standardized tests.

Question 5

The math club kept track of the number of practice problems solved by each of its two study groups last week. Group X solved 8,10,12,14,168, 10, 12, 14, 16 problems, and Group Y solved 9,11,11,13,159, 11, 11, 13, 15 problems. Which study group had the greater mean number of problems solved?

  1. Group X had the greater mean. (correct answer)
  2. Group Y had the greater mean.
  3. The two groups had equal means.
  4. There is not enough information to determine the means.
Explanation: When you encounter a problem asking you to compare means (averages), you need to calculate the mean for each data set and then compare the results. To find the mean, add all values and divide by the number of values. For Group X: 8+10+12+14+16=608 + 10 + 12 + 14 + 16 = 60, so the mean is 60÷5=1260 ÷ 5 = 12 problems. For Group Y: 9+11+11+13+15=599 + 11 + 11 + 13 + 15 = 59, so the mean is 59÷5=11.859 ÷ 5 = 11.8 problems. Since 12 > 11.8, Group X had the greater mean, making choice A correct. Choice B is wrong because Group Y's mean (11.8) is actually smaller than Group X's mean (12). Choice C is incorrect because the means are different—12 and 11.8 are not equal. Choice D is wrong because we have complete data sets for both groups, giving us all the information needed to calculate and compare the means. Notice that even though some individual values in Group Y are higher than some in Group X, what matters for the mean is the total sum. Group X's sum (60) is higher than Group Y's sum (59), and since both groups solved the same number of problems (5 each), Group X must have the higher average. Strategy tip: When comparing means, always calculate both completely before deciding. Don't be misled by individual high or low values—the mean depends on the total sum divided by the count.

Question 6

Class A has 2020 students with an average test score of 7878. Class B has 2525 students with an average test score of 7474. When the two classes are combined, which statement about the overall mean test score is true?

  1. The overall mean is exactly 76.076.0.
  2. The overall mean is between 7474 and 7878 but closer to 7878.
  3. The overall mean is between 7474 and 7878 but closer to 7474. (correct answer)
  4. The overall mean is lower than 7474.
Explanation: When you encounter problems about combining groups with different averages, remember that the overall average will always fall between the two group averages, but it gets "pulled" toward the group with more members. To find the exact overall mean, you need the weighted average. Class A contributes 20×78=156020 \times 78 = 1560 total points, while Class B contributes 25×74=185025 \times 74 = 1850 total points. The combined total is 1560+1850=34101560 + 1850 = 3410 points across 20+25=4520 + 25 = 45 students. Therefore, the overall mean is 341045=75.78\frac{3410}{45} = 75.78. Since 75.7875.78 is between 7474 and 7878 but closer to 7474 (the difference from 7474 is 1.781.78, while the difference from 7878 is 2.222.22), answer choice C is correct. Choice A incorrectly assumes the overall mean is simply the arithmetic average of the two class means: 78+742=76\frac{78 + 74}{2} = 76. This ignores the fact that Class B has more students. Choice B makes the right prediction that the overall mean falls between the two averages, but incorrectly assumes it's closer to 7878. Since Class B (with the lower average) has more students, the overall mean gets pulled toward 7474. Choice D is impossible since weighted averages of positive numbers must fall between the original values. Strategy tip: When combining groups with different averages, the overall average always gets "pulled" toward the larger group. Don't just average the averages—weight them by group size.

Question 7

The table shows the daily temperatures (°F) recorded in two cities for one week. Use the table to answer the question.

If the temperature recorded as 95°F in City M is later discovered to be a typo and should be 59°F, how will the mean and median of City M change relative to City N?

  1. The new mean of City M will be less than the mean of City N, but the new median of City M will still be greater than the median of City N. (correct answer)
  2. The new mean and new median of City M will both be less than those of City N.
  3. The new mean of City M will equal the mean of City N, and the new median of City M will equal the median of City N.
  4. The new mean of City M will be less than the mean of City N, and the new median of City M will equal the median of City N.
Explanation: City M original: 70, 72, 75, 78, 80, 82, 95. Replace 95 with 59: 59, 70, 72, 75, 78, 80, 82. Mean = 516/7 ≈ 73.71. Median (4th value) = 75. City N: 68, 71, 73, 74, 76, 78, 80. Mean = 520/7 ≈ 74.29. Median = 74. Compare: new mean M (73.71) < mean N (74.29). New median M (75) > median N (74). Answer A.

Question 8

The table shows scores of two teams in a trivia contest. Refer to the table.

If each score in Team Alpha is doubled and each score in Team Beta is increased by 20, which statement about the resulting data sets is true?

  1. The new mean of Team Alpha equals the new mean of Team Beta. (correct answer)
  2. The new median of Team Alpha is greater than the new median of Team Beta.
  3. The new mean of Team Alpha is greater than the new mean of Team Beta, but the new median of Team Alpha is less than the new median of Team Beta.
  4. The new mean of Team Alpha is less than the new mean of Team Beta, but the new median of Team Alpha is greater than the new median of Team Beta.
Explanation: Team Alpha original: 20, 25, 30, 35, 40. Mean=30, median=30. Doubled: 40, 50, 60, 70, 80. New mean=60, new median=60. Team Beta original: 30, 35, 40, 45, 50. Mean=40, median=40. Add 20: 50, 55, 60, 65, 70. New mean=60, new median=60. Both new means and medians are equal. A is correct.

Question 9

The frequency table shows the number of minutes spent on homework by students in two classes. Use the table to answer the question.

Which of the following best describes the relationship between the means and medians of the two classes?

  1. Class 1 has a higher mean and a higher median than Class 2.
  2. Class 1 has a higher mean but a lower median than Class 2.
  3. Class 2 has a higher mean and a higher median than Class 1. (correct answer)
  4. Class 2 has a higher mean but a lower median than Class 1.
Explanation: Class 1: 30 min (5 students), 45 min (8), 60 min (4), 75 min (3). Total students=20. Sum=30·5+45·8+60·4+75·3=150+360+240+225=975. Mean=975/20=48.75. Median: 10th and 11th values. Cumulative: 5 at 30, 13 at 45. Both 10th and 11th are 45. Median=45. Class 2: 30 min (2), 45 min (5), 60 min (9), 75 min (4). Total=20. Sum=60+225+540+300=1125. Mean=56.25. Median: 10th+11th. Cumulative: 2,7,16. Both are 60. Median=60. Class 2 has higher mean AND higher median.

Question 10

The table lists the heights (in cm) of plants grown with two different fertilizers. Use the table to answer the question.

What value of xx would make the mean height of Fertilizer B equal to the median height of Fertilizer A?

  1. 1818
  2. 2222
  3. 2626
  4. 3030 (correct answer)
Explanation: Fertilizer A heights in order: 15, 18, 20, 22, 25 (5 values). The median is the middle value: 20 cm. For Fertilizer B, the mean is (14 + 16 + 19 + 21 + x) ÷ 5 = (70 + x) ÷ 5. Setting this equal to 20: (70 + x) ÷ 5 = 20, so 70 + x = 100, therefore x = 30.

Question 11

The table shows the weights (in pounds) of fish caught at two lakes. Use the table to answer the question.

A researcher claims that the 'typical' fish at Lake East is heavier than at Lake West. Which of the following best supports the researcher's claim?

  1. The mean weight of Lake East is greater than the mean weight of Lake West. (correct answer)
  2. The median weight of Lake East is greater than the median weight of Lake West.
  3. The range of weights at Lake East is greater than the range of weights at Lake West.
  4. The maximum weight at Lake East is greater than the maximum weight at Lake West.
Explanation: Lake East weights: 2, 3, 3, 4, 4, 5, 5, 20. Mean = 46/8 = 5.75. Median = (4+4)/2 = 4. Lake West weights: 3, 3, 4, 4, 5, 5, 6, 6. Mean = 36/8 = 4.5. Median = (4+5)/2 = 4.5. The mean of Lake East (5.75) is greater than Lake West (4.5), supporting the claim. However, the median of Lake East (4) is actually less than Lake West (4.5). Since the question asks what best supports the researcher's claim, only the mean provides numerical support, even though it's influenced by the 20-pound outlier.

Question 12

The two graphs show the number of cars sold per month over 6 months at Dealership A and Dealership B. Use the graphs to answer the question.

How much greater is the mean monthly sales of the dealership with the higher median than the mean monthly sales of the other dealership?

  1. 11
  2. 22
  3. 33 (correct answer)
  4. 44
Explanation: Dealership A: 10, 15, 20, 22, 25, 30. Median = (20+22)/2 = 21. Mean = 122/6 ≈ 20.33. Dealership B: 12, 18, 22, 25, 28, 35. Median = (22+25)/2 = 23.5. Mean = 140/6 ≈ 23.33. Dealership B has the higher median. Difference in means: 23.33 − 20.33 = 3. Answer C.

Question 13

The stem-and-leaf plots below show the ages of people attending two workshops. Refer to the plots.

Which of the following correctly orders the medians of the two workshops and the combined data set from least to greatest?

  1. Workshop A < Combined < Workshop B (correct answer)
  2. Workshop B < Combined < Workshop A
  3. Combined < Workshop A < Workshop B
  4. Workshop A < Workshop B < Combined
Explanation: Workshop A ages: 22, 25, 28, 31, 34, 37, 40 (n=7), median=31. Workshop B ages: 35, 38, 42, 45, 48, 51, 55 (n=7), median=45. Combined (14 values sorted): 22,25,28,31,34,35,37,38,40,42,45,48,51,55. Median = average of 7th and 8th values = (37+38)/2 = 37.5. Order: 31 < 37.5 < 45, so Workshop A < Combined < Workshop B.

Question 14

The histogram below shows the distribution of exam scores for two sections of a class. Based on the histogram, which statement correctly compares the medians?

  1. The median of Section 1 is in the 70–79 interval, and the median of Section 2 is in the 80–89 interval. (correct answer)
  2. The median of Section 1 is in the 80–89 interval, and the median of Section 2 is in the 70–79 interval.
  3. Both medians are in the 70–79 interval.
  4. Both medians are in the 80–89 interval.
Explanation: Section 1 frequencies: 60–69: 3; 70–79: 8; 80–89: 5; 90–99: 4. Total=20, so median is between the 10th and 11th values. Cumulative: 3, 11, 16, 20. Both 10th and 11th fall in 70–79. Section 2: 60–69: 2; 70–79: 4; 80–89: 9; 90–99: 5. Total=20. Cumulative: 2, 6, 15, 20. The 10th and 11th values fall in 80–89. Answer A.

Question 15

The bar graph shows the number of goals scored per game by two soccer teams over 8 games. Based on the graph, which of the following is closest to the positive difference between the mean goals per game of Team Red and the mean goals per game of Team Blue?

  1. 0.250.25
  2. 0.500.50 (correct answer)
  3. 0.750.75
  4. 1.001.00
Explanation: Team Red goals across 8 games: 2,3,1,4,2,3,5,2. Sum=22, mean=2.75. Team Blue: 1,2,3,2,4,1,3,2. Sum=18, mean=2.25. Difference=0.50.

Question 16

The dot plots below show the number of books read during the summer by students in two reading clubs. Use the dot plots to answer the question.

Which statement is true?

  1. The mean of Club X is greater than the mean of Club Y, but the median of Club X is less than the median of Club Y. (correct answer)
  2. The mean of Club Y is greater than the mean of Club X, and the median of Club Y is greater than the median of Club X.
  3. The mean of Club X equals the mean of Club Y, but the median of Club X is greater than the median of Club Y.
  4. The mean of Club X is greater than the mean of Club Y, and the median of Club X is greater than the median of Club Y.
Explanation: Club X data: 2,2,3,3,3,4,4,5,10 (n=9). Sum=36, mean=4. Median=3 (5th value). Club Y data: 1,2,3,3,4,4,4,5,5 (n=9). Sum=31, mean≈3.44. Median=4. So Club X has a higher mean (due to the outlier 10) but a lower median. Choice A is correct. B and D misinterpret the influence of the outlier; C incorrectly equates the means.

Question 17

The double dot plot below shows the number of hours spent reading per week by students in two different grades. Based on the dot plots, which measure best shows that Grade 7 students tend to read more than Grade 6 students, given the presence of an outlier?

  1. The mean, because it accounts for every value.
  2. The median, because it is resistant to the outlier. (correct answer)
  3. The range, because it shows the spread of the data.
  4. The mode, because it shows the most frequent value.
Explanation: Grade 6 data: 1,2,2,3,3,3,4,4,5. Grade 7 data: 2,3,4,4,5,5,5,6,15 (15 is an outlier). Grade 6 mean ≈ 3.0, median = 3. Grade 7 mean = 49/9 ≈ 5.44, median = 5. The mean of Grade 7 is inflated by the outlier 15, but even without it, Grade 7 reads more. To fairly represent the center without distortion from the outlier, the median is best. Mean is distorted (A wrong). Range shows spread, not center (C). Mode may not reliably represent center (D).

Question 18

The box plots below display the monthly rainfall (in inches) for City X and City Y over the past year. Based on the box plots, which statement must be true?

  1. The mean rainfall of City X is greater than the mean rainfall of City Y.
  2. The median rainfall of City X is less than the median rainfall of City Y. (correct answer)
  3. The interquartile range of City X equals the interquartile range of City Y.
  4. At least 25% of the months in City X had more rainfall than the median month in City Y.
Explanation: From the box plots: City X has median=4, City Y has median=6, so B is true. A cannot be determined from box plots (means aren't shown and can differ from medians). C: City X IQR=6−2=4; City Y IQR=8−4=4 — wait these match; but the question asks what MUST be true. Actually C is also true based on these values. Let me recalibrate: City X Q1=2, Q3=5, IQR=3; City Y Q1=4, Q3=8, IQR=4. So C is false. D: City X's Q3=5 < City Y's median=6, so the top 25% of City X is still below Y's median—so fewer than 25% of City X months exceed Y's median. D is false. B is correct.

Question 19

Two sets of quiz scores are shown.

Set P: {6,7,7,8,9}\{6,7,7,8,9\}
Set Q: {4,7,7,10,13}\{4,7,7,10,13\}

Which statement correctly compares the means and medians of the two sets?

  1. P and Q have the same mean but different medians.
  2. P has the greater mean and the greater median.
  3. Q has the greater mean and the greater median.
  4. The two sets have the same median but different means. (correct answer)
Explanation: When comparing data sets, you need to calculate both the mean (average) and median (middle value) for each set to see how they relate. For Set P: {6,7,7,8,9}\{6,7,7,8,9\}, the mean is 6+7+7+8+95=375=7.4\frac{6+7+7+8+9}{5} = \frac{37}{5} = 7.4. Since there are 5 values, the median is the 3rd value when ordered: 7. For Set Q: {4,7,7,10,13}\{4,7,7,10,13\}, the mean is 4+7+7+10+135=415=8.2\frac{4+7+7+10+13}{5} = \frac{41}{5} = 8.2. The median is also the 3rd value: 7. Comparing results: Set P has mean 7.4 and median 7, while Set Q has mean 8.2 and median 7. The medians are the same (both 7), but the means are different (7.4 vs 8.2). Choice A incorrectly claims the means are the same when they're actually different (7.4 ≠ 8.2). Choice B wrongly states that P has both the greater mean and median, but P's mean is smaller and the medians are equal. Choice C incorrectly says Q has the greater median when both medians equal 7, though Q does have the greater mean. Choice D correctly identifies that the medians are equal but the means differ. Study tip: Always calculate both statistics completely before comparing. Don't assume that sets with the same median will have the same mean—outliers and distribution shape affect the mean much more than the median.