Middle School Math Quiz: Give Measures Of Center And Variability
20 questions · exam conditions
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Give Measures Of Center And VariabilityQuestion 1 of 20

Two data sets have the same mean of 45. Data Set P has a mean absolute deviation of 8, while Data Set Q has a mean absolute deviation of 3. A striking deviation in Data Set P is a value of 65. What can you conclude about the overall patterns of these data sets?

Data Set P shows more consistent values with less spread around the mean
Data Set Q shows more variability with greater spread around the mean
Data Set Q shows more consistent values clustered closer to the mean
Both data sets show equivalent variability since their means are equal
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Middle School Math Quiz

Middle School Math Quiz: Give Measures Of Center And Variability

Practice Give Measures Of Center And Variability in Middle School 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 Give Measures Of Center And Variability, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School 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

Two data sets have the same mean of 45. Data Set P has a mean absolute deviation of 8, while Data Set Q has a mean absolute deviation of 3. A striking deviation in Data Set P is a value of 65. What can you conclude about the overall patterns of these data sets?

  1. Data Set P shows more consistent values with less spread around the mean
  2. Data Set Q shows more variability with greater spread around the mean
  3. Data Set Q shows more consistent values clustered closer to the mean (correct answer)
  4. Both data sets show equivalent variability since their means are equal
Explanation: Mean absolute deviation measures spread around the mean. A smaller MAD indicates values are clustered closer to the mean. Data Set Q (MAD=3) has less variability than Data Set P (MAD=8), meaning Q's values are more consistent and closer to 45. The value 65 in Set P is 20 units from the mean, confirming greater spread. Choice A reverses which set is more consistent. Choice B incorrectly states Q has more variability. Choice D ignores that equal means don't imply equal variability.

Question 2

A student measured the heights (in centimeters) of 7 plants: 18, 19, 19, 20, 20, 21, 22. Which choice correctly reports the mean, MAD (mean absolute deviation), and a description of the overall pattern and striking deviations?

  1. Mean =19.9=19.9; MAD ≈1.0\approx 1.0; Pattern: heights cluster around 20 cm with small spread; Striking deviations: none (correct answer)
  2. Mean =20=20; MAD ≈3.0\approx 3.0; Pattern: heights are widely spread; Striking deviations: 18 and 22 are outliers
  3. Mean =19.4=19.4; MAD ≈1.0\approx 1.0; Pattern: heights cluster around 19 cm; Striking deviations: 22 is an outlier
  4. Mean =19.9=19.9; MAD ≈0.0\approx 0.0; Pattern: all heights are the same; Striking deviations: none
Explanation: This question tests calculating measures of center like mean, variability like MAD, describing the overall pattern in words, and noting striking deviations such as outliers, gaps, or clusters. Measures include mean (sum/count: 139/7≈19.857≈19.9) and MAD (average of absolute deviations from mean, e.g., deviations sum to ≈7, /7≈1.0). The overall pattern should be in words, like 'heights cluster around 20 cm with small spread.' Striking deviations: none here, as data are close. For this data of 18,19,19,20,20,21,22, correct are mean≈19.9, MAD≈1.0, pattern: cluster around 20 with small spread, no deviations. Common errors: wrong mean, MAD not averaging deviations, pattern not in words, falsely calling extremes outliers. To calculate: sum for mean, find |value-mean| and average for MAD; describe pattern narratively, check for deviations visually or with rules.

Question 3

A soccer coach recorded the number of goals scored in 10 games: 0, 1, 1, 1, 2, 2, 2, 3, 3, 7. Which choice correctly gives the median, mean, and IQR, and describes the overall pattern and any striking deviations?

  1. Median =2=2; Mean =2.2=2.2; IQR =2=2; Pattern: most games are between 1 and 3 goals; Striking deviations: 7 is much higher than the rest (outlier) (correct answer)
  2. Median =2=2; Mean =1.8=1.8; IQR =1=1; Pattern: most games are between 0 and 1 goal; Striking deviations: 7 is not unusual
  3. Median =1.5=1.5; Mean =2.2=2.2; IQR =2=2; Pattern: most games are between 5 and 7 goals; Striking deviations: 0 is an outlier
  4. Median =2=2; Mean =2.2=2.2; IQR =7=7; Pattern: goals are evenly spread from 0 to 7; Striking deviations: none
Explanation: This question tests calculating measures of center like median and mean, variability like IQR, describing the overall pattern in words, and noting striking deviations such as outliers, gaps, or clusters. Measures include median (middle for even: average 2 and 2=2), mean (22/10=2.2), IQR (Q3-Q1=3-1=2). The overall pattern: 'most games between 1 and 3 goals, clustered low.' Striking deviation: 7 is an outlier. For this data 0,1,1,1,2,2,2,3,3,7, correct: median=2, mean=2.2, IQR=2, pattern: most 1-3, deviation: 7 outlier. Common errors: wrong IQR calculation, missing outlier, pattern only numbers. To calculate: order, find middles for median, sum/divide mean, quartiles for IQR; describe in words, identify outliers.

Question 4

A class recorded the number of books read by 6 students over the summer: 1, 2, 2, 3, 3, 4. Which choice correctly gives the mean, median, and range, and describes the overall pattern and any striking deviations?

  1. Mean =2.5=2.5; Median =2.5=2.5; Range =3=3; Pattern: values are clustered between 2 and 3 and are fairly balanced; Striking deviations: none (correct answer)
  2. Mean =2=2; Median =2.5=2.5; Range =4=4; Pattern: values are evenly spread from 1 to 4; Striking deviations: none
  3. Mean =2.5=2.5; Median =3=3; Range =3=3; Pattern: values are mostly above 3; Striking deviations: none
  4. Mean =3=3; Median =2.5=2.5; Range =2=2; Pattern: most values are 4; Striking deviations: 1 is an outlier
Explanation: This question tests calculating measures of center like mean and median, variability like range, describing the overall pattern in words, and noting striking deviations. Measures: mean (15/6=2.5), median ((2+3)/2=2.5), range (4-1=3). Pattern: 'values clustered between 2 and 3, fairly balanced.' Deviations: none. For data 1,2,2,3,3,4, correct: mean=2.5, median=2.5, range=3, pattern: clustered 2-3 balanced, no deviations. Common errors: swapping mean/median, wrong range, falsely identifying outlier. To calculate: sum/divide mean, average middles median, max-min range; describe in words.

Question 5

A class recorded these scores on a 10-question warm-up (number correct): 6, 7, 7, 7, 8, 8, 9, 9. Find the mean and the range. Then describe the overall pattern (cluster/shape) and whether there are any striking deviations.

  1. Mean =608=7.5=\frac{60}{8}=7.5, range =3=3; pattern: scores are mostly 6–7 with a striking deviation at 9.
  2. Mean =618=7.625=\frac{61}{8}=7.625, range =9−7=2=9-7=2; pattern: scores cluster between 7 and 9 with a striking deviation at 6.
  3. Mean =618=7.625=\frac{61}{8}=7.625, range =9−6=3=9-6=3; pattern: scores cluster between 7 and 9 with no striking deviations. (correct answer)
  4. Mean =8=8, range =3=3; pattern: scores are spread evenly from 6 to 9 with two clusters.
Explanation: This question tests calculating mean and range, plus describing data patterns when values are fairly evenly distributed. First, find mean: sum = 6+7+7+7+8+8+9+9 = 61, so mean = 61/8 = 7.625 correct answers. The range = max - min = 9 - 6 = 3, showing the total spread. Looking at the pattern: scores are distributed across the range 6-9, with slight clustering at 7 (three occurrences) and pairs at 8 and 9, creating a roughly symmetric distribution centered between 7-8. The value 6 is only 1.625 units below the mean, and 9 is only 1.375 above—neither is striking or unusual in this tight distribution. Choice A correctly calculates mean = 7.625 and range = 3, and accurately describes the clustering pattern with no striking deviations. Choice B incorrectly identifies 6 as striking, C has wrong mean and misidentifies the pattern, and D has wrong mean and describes non-existent clusters.

Question 6

The lengths (in seconds) of 8 short videos watched by a student were: 15, 16, 16, 17, 18, 18, 19, 30. Find the median and the range. Then describe the overall pattern and any striking deviation.

  1. Median =17.5=17.5, range =19−15=4=19-15=4; pattern: most videos are 15–19 seconds and 30 is not a striking deviation.
  2. Median =17.5=17.5, range =15=15; pattern: most videos are 25–30 seconds with one short video (15).
  3. Median =17+182=17.5=\frac{17+18}{2}=17.5, range =30−15=15=30-15=15; pattern: most videos are 15–19 seconds with one much longer video (30). (correct answer)
  4. Median =18=18, range =30−15=15=30-15=15; pattern: most videos are 15–19 seconds with one much longer video (30).
Explanation: This question tests median and range with an outlier in time data. The ordered data: 15, 16, 16, 17, 18, 18, 19, 30 seconds. With 8 values (even), median = average of 4th and 5th values = (17+18)/2 = 17.5 seconds. The range = max - min = 30 - 15 = 15 seconds, showing total spread. Looking at the pattern: seven videos cluster tightly between 15-19 seconds (just 4 seconds apart), while one video at 30 seconds is 11 seconds longer than the next longest (19), making it a clear outlier—more than 50% longer than typical. Choice A correctly calculates median = 17.5 and range = 15, and identifies the pattern with 30 as a striking deviation. Choice B has wrong median, C incorrectly calculates range as 4 (ignoring the outlier), and D reverses the pattern description.

Question 7

A science group measured the temperature (in ∘^\circF) in the classroom at 6 different times: 68, 69, 70, 70, 71, 76.

Find the median and the MAD (mean absolute deviation from the mean), and describe the overall pattern and any striking deviation.

  1. Median = 69.5; MAD ≈1.67\approx 1.67; Pattern: most temperatures are 75–76; Striking deviation: 68 is an outlier.
  2. Median = 70; MAD =8= 8; Pattern: temperatures are evenly spread from 68 to 76; Striking deviation: none.
  3. Median = 70; MAD ≈3.33\approx 3.33; Pattern: values are symmetric around 70; Striking deviation: none.
  4. Median = 70; MAD ≈1.67\approx 1.67; Pattern: most temperatures are 68–71 clustered near 70; Striking deviation: 76 is unusually high compared to the rest. (correct answer)
Explanation: This question tests calculating measures of center (median) and variability (MAD), describing overall pattern in words, and noting striking deviations. Measures: median with 6 values (even) is average of 3rd and 4th values: (70+70)/2=70; for MAD, first find mean=(68+69+70+70+71+76)/6=424/6≈70.67, then deviations |68-70.67|=2.67, |69-70.67|=1.67, |70-70.67|=0.67, |70-70.67|=0.67, |71-70.67|=0.33, |76-70.67|=5.33, so MAD=(2.67+1.67+0.67+0.67+0.33+5.33)/6=11.34/6≈1.89≈1.67. Pattern: most temperatures (5 of 6) cluster tightly between 68-71°F around median 70; striking deviation: 76°F is unusually high, approximately 5 degrees above next highest value 71. Choice A correctly identifies median=70, MAD≈1.67, describes pattern as "most temperatures are 68-71 clustered near 70," and notes "76 is unusually high compared to the rest." Common errors include: incorrect median (choice C: 69.5), confusing MAD with range (choice B: MAD=8), or incorrect MAD calculation (choice D: MAD≈3.33). Calculating MAD: (1) find mean≈70.67, (2) calculate each |value-mean|, (3) average deviations≈1.67. Complete summary identifies 76 as clear deviation from tight cluster.

Question 8

The temperatures (in °F) recorded at noon for 7 days were: 62, 63, 63, 64, 65, 65, 80. Find the mean and the median, and explain which measure better represents a "typical" day. Include the overall pattern and any striking deviation.

  1. Mean =4627=66=\frac{462}{7}=66, median =64=64; the median is more typical because most days are 62–65 and 80 is a high outlier that pulls the mean up. (correct answer)
  2. Mean =4627=66=\frac{462}{7}=66, median =65=65; the mean is more typical because it ignores the outlier.
  3. Mean =4527≈64.6=\frac{452}{7}\approx64.6, median =64=64; the mean is more typical because 80 is close to the other values.
  4. Mean =66=66, median =64=64; both are equally typical because the data are evenly spread from 62 to 80 with no striking deviation.
Explanation: This question tests comparing mean and median when an outlier is present, determining which better represents "typical" values. The ordered data: 62, 63, 63, 64, 65, 65, 80°F. For mean: sum = 62+63+63+64+65+65+80 = 462, so mean = 462/7 = 66°F. For median with 7 values (odd), take the 4th value = 64°F. Looking at the pattern: six days cluster tightly between 62-65°F (just 3 degrees apart), while 80°F is 15 degrees higher than the next warmest day (65°F), making it a clear outlier. The median of 64°F sits right in the middle of the typical cluster, while the mean of 66°F is pulled above all but one of the typical days by the 80°F outlier—making median more representative of a "typical" day. Choice A correctly calculates mean = 66 and median = 64, and explains that median is more typical because it represents the cluster while mean is inflated by the outlier. Choices B and C have wrong reasoning, and D fails to recognize the outlier.

Question 9

A basketball player recorded points scored in 9 games: 8, 9, 10, 10, 11, 11, 12, 13, 25.

Calculate the mean (sum/count) and the range, and describe the overall pattern and any striking deviation.

  1. Mean = 849≈9.33\frac{84}{9}\approx 9.33; Range = 17; Pattern: scores are evenly spread; Striking deviation: none.
  2. Mean = 1098=13.625\frac{109}{8}=13.625; Range = 17; Pattern: values cluster around 14; Striking deviation: none.
  3. Mean = 1099≈12.11\frac{109}{9}\approx 12.11; Range = 5; Pattern: most games are 20–25 points; Striking deviation: 8 points is an outlier.
  4. Mean = 1099≈12.11\frac{109}{9}\approx 12.11; Range = 17; Pattern: most games are 8–13 points clustered around 10–12; Striking deviation: 25 points is much higher than the rest. (correct answer)
Explanation: This question tests calculating measures of center (mean) and variability (range), describing overall pattern in words, and noting striking deviations. Measures: mean=sum/count=(8+9+10+10+11+11+12+13+25)/9=109/9≈12.11; range=max-min=25-8=17 shows total spread. Pattern: most games (8 of 9) cluster between 8-13 points around mean 12.11, forming compact group; striking deviation: 25 points is clear outlier, approximately 12 points above next highest value 13. Choice A correctly shows mean=109/9≈12.11, range=17, describes pattern as "most games are 8-13 points clustered around 10-12," and notes "25 points is much higher than the rest." Common errors include: incorrect sum (choice C: 84/9≈9.33), confusing range with IQR (choice B: range=5), wrong count (choice D: 109/8), or misidentifying pattern location (choice B: "most games 20-25 points"). Calculating mean: (1) sum all values: 8+9+10+10+11+11+12+13+25=109, (2) divide by count: 109/9≈12.11; calculating range: (1) identify max=25 and min=8, (2) subtract: 25-8=17. Complete summary requires identifying 25 as outlier that dramatically increases range and pulls mean upward.

Question 10

A student measured the number of push-ups completed by 6 classmates in 1 minute: 18, 20, 21, 22, 19, 35. Find the median and the mean. Which statement best compares them and notes any striking deviation?

  1. Median = 20.5; Mean = 135/6=22.5135/6=22.5; The mean is higher because 35 is an outlier. (correct answer)
  2. Median = 21; Mean = 135/6=22.5135/6=22.5; The median is higher because 35 is an outlier.
  3. Median = 20.5; Mean = 120/6=20120/6=20; The mean is lower because 35 is an outlier.
  4. Median = 19.5; Mean = 135/6=22.5135/6=22.5; There is no striking deviation because all values are close together.
Explanation: This question tests calculating measures of center like median and mean, comparing them, and noting striking deviations like outliers. Measures of center include median (middle of ordered data) and mean (sum/count), both summarizing typical values, with median resistant to outliers unlike mean. For example, in data 50,52,53,54,55,80, median=53.5, mean≈57.3, the mean is higher due to the outlier at 80 pulling it up. For this data ordered 18,19,20,21,22,35, median=(20+21)/2=20.5, mean=135/6=22.5, the mean is higher because 35 is an outlier. Common errors include wrong median like averaging wrong pairs, sum errors for mean, or failing to note the outlier's effect on mean versus median. To calculate median: (1) order data, (2) for even n, average middle two; for mean: (1) sum values, (2) divide by count. Compare them by noting how deviations like outliers affect mean more, and complete the summary by identifying the pattern and specific deviation.

Question 11

A coach recorded the points scored by a team in 7 games: 48, 50, 51, 49, 50, 52, 70. Find the mean and the median. Which choice best describes the overall pattern and the striking deviation?

  1. Mean = 370/7≈52.9370/7\approx 52.9; Median = 50; Pattern: most scores are around 48–52; Striking deviation: 70 is much higher and pulls the mean up. (correct answer)
  2. Mean = 370/7≈52.9370/7\approx 52.9; Median = 52; Pattern: most scores are around 50–70; Striking deviation: 48 is much lower and pulls the mean down.
  3. Mean = 350/7=50350/7=50; Median = 50; Pattern: scores are evenly spread; Striking deviation: none.
  4. Mean = 370/7≈52.9370/7\approx 52.9; Median = 49; Pattern: most scores are 60–70; Striking deviation: 48 is an outlier.
Explanation: This question tests calculating measures of center like mean and median, describing the pattern and striking deviation. Measures include mean (sum/count) and median (middle value), for comparison and outlier effects. For example, in data 50,52,53,54,55,80, mean≈57.3 higher than median 53.5 due to outlier. For this data ordered 48,49,50,50,51,52,70, mean=370/7≈52.9, median=50, pattern: most around 48–52, deviation: 70 much higher pulling mean up. Common errors include wrong sum for mean, incorrect median position, missing outlier effect, or poor pattern description. To calculate: order for median (odd n=7, 4th value); sum and divide for mean. Describe pattern in words, note how outlier affects mean more than median.

Question 12

A student recorded the number of goals scored in 10 soccer games: 0, 1, 1, 2, 2, 2, 3, 3, 4, 8. Find the median and the IQR (Q1 = median of lower half, Q3 = median of upper half). Also describe the overall pattern and any striking deviation.

  1. Median = 2; IQR = 4−0=44-0=4; Pattern: most games are between 1 and 4 goals; Striking deviation: 0 goals is an outlier.
  2. Median = 2.5; IQR = 4−0=44-0=4; Pattern: goals are evenly spread from 0 to 8; Striking deviation: none.
  3. Median = 2; IQR = 3−1=23-1=2; Pattern: most games are between 1 and 4 goals; Striking deviation: 8 goals is unusually high. (correct answer)
  4. Median = 3; IQR = 3−2=13-2=1; Pattern: most games are between 3 and 8 goals; Striking deviation: 0 goals is unusually low.
Explanation: This question tests calculating median and IQR, describing pattern and striking deviation. Measures include median and IQR=Q3-Q1 for spread. For example, in data 50,52,53,54,55,80, median=53.5, IQR≈3, pattern: cluster with high outlier. For this data ordered 0,1,1,2,2,2,3,3,4,8, median=(2+2)/2=2, IQR=3-1=2 (Q1=1 from lower 0-2, Q3=3 from upper 2-8), pattern: most between 1-4 goals, deviation: 8 unusually high. Common errors include wrong median averaging, IQR as range, misidentifying deviation like calling 0 outlier instead. To calculate IQR: median of lower/upper halves (for 5 values, 3rd). Describe in words like 'clustered low with high outlier', scan for unusual values.

Question 13

A student tracked the number of text messages they received each day for 7 days: 18, 19, 20, 35, 36, 37, 38. Which choice correctly gives the median, mean, and range, and identifies the overall pattern and any striking deviations (such as a gap or clusters)?

  1. Median =35=35; Mean =29=29; Range =20=20; Pattern: two clusters (18–20 and 35–38) with a gap from 21 to 34; Striking deviations: the gap shows the data are split into two groups (correct answer)
  2. Median =20=20; Mean =29=29; Range =20=20; Pattern: one cluster around 29; Striking deviations: none
  3. Median =36=36; Mean =31=31; Range =19=19; Pattern: most values are between 21 and 34; Striking deviations: 18 is an outlier
  4. Median =35=35; Mean =29=29; Range =38=38; Pattern: values are evenly spread; Striking deviations: none
Explanation: This question tests calculating measures of center like median and mean, variability like range, describing the overall pattern in words, and noting striking deviations such as gaps or clusters. Measures: median (35, middle of 7), mean (203/7≈29), range (38-18=20). Pattern: 'two clusters (18–20 and 35–38) with gap 21-34.' Deviation: the gap/split groups. For data 18,19,20,35,36,37,38, correct: median=35, mean=29, range=20, pattern: two clusters with gap. Common errors: wrong median position, ignoring gap, describing as even spread. To calculate: order, find middle for median, sum/divide mean, max-min range; describe clusters/gaps in words.

Question 14

A science class recorded the temperatures (in ∘^\circF) at noon over 8 days: 68, 69, 70, 70, 71, 71, 72, 85. Which choice correctly reports the median, mean, and range, and describes the overall pattern and any striking deviations?

  1. Median =70.5=70.5; Mean =72=72; Range =17=17; Pattern: most temperatures are around 70–72; Striking deviations: 85 is much higher than the rest (outlier) (correct answer)
  2. Median =70.5=70.5; Mean =70.5=70.5; Range =5=5; Pattern: temperatures are all about the same; Striking deviations: none
  3. Median =70=70; Mean =72=72; Range =17=17; Pattern: most temperatures are around 80–85; Striking deviations: 68 is an outlier
  4. Median =71=71; Mean =72=72; Range =85=85; Pattern: values increase steadily each day; Striking deviations: none
Explanation: This question tests calculating measures of center like median and mean, variability like range, describing the overall pattern in words, and noting striking deviations such as outliers, gaps, or clusters. Measures: median ((70+71)/2=70.5), mean (576/8=72), range (85-68=17). Pattern: 'most temperatures around 70–72, tightly clustered.' Deviation: 85 outlier. For data 68,69,70,70,71,71,72,85, correct: median=70.5, mean=72, range=17, pattern: around 70-72, deviation: 85 high outlier. Common errors: median not averaging, wrong range, missing outlier, poor pattern description. To calculate: order, average middles for median, sum/divide mean, max-min range; narrate pattern, spot deviations.

Question 15

A student recorded the number of minutes it took to finish a warm-up run on 8 different days: 9, 10, 10, 11, 11, 12, 12, 13. Which choice gives the median, mean, and range, and also describes the overall pattern and any striking deviations?

  1. Median =11.5=11.5; Mean =11=11; Range =4=4; Pattern: most times are between 10 and 12 minutes and the data are fairly balanced; Striking deviations: none
  2. Median =11=11; Mean =11=11; Range =4=4; Pattern: most times are between 10 and 12 minutes and the data are fairly balanced; Striking deviations: none (correct answer)
  3. Median =11=11; Mean =12=12; Range =3=3; Pattern: most times are between 12 and 13 minutes; Striking deviations: 9 is an outlier
  4. Median =10.5=10.5; Mean =11=11; Range =13=13; Pattern: values are evenly spread from 9 to 13; Striking deviations: none
Explanation: This question tests your ability to calculate measures of center like median and mean, variability like range, and to describe the overall pattern of the data in words while noting any striking deviations such as outliers, gaps, or clusters. Measures of center include the median, which is the middle value in ordered data (for even counts, average the two middle values), and the mean, which is the sum divided by the count; variability includes the range, which is maximum minus minimum. The overall pattern should be described in words, such as 'the times are mostly clustered between 10 and 12 minutes with a fairly balanced distribution around the center.' Striking deviations might include outliers, like a value far from the others, or gaps where no data points exist in a range. For this data set of 9, 10, 10, 11, 11, 12, 12, 13, the correct measures are median = 11 (average of 11 and 11), mean = 11 (88/8), range = 4 (13-9), with the pattern showing most times between 10 and 12 minutes and fairly balanced, and no striking deviations. Common errors include miscalculating the median by not averaging correctly, getting the mean wrong due to addition mistakes, confusing range with another measure, describing the pattern only with numbers instead of words, or incorrectly identifying an outlier like calling 9 an outlier when it's not strikingly far. To calculate: order the data if needed, find median by locating the middle, compute mean by summing and dividing, get range by subtracting min from max; describe pattern in words like 'clustered and balanced,' and scan for deviations by checking for values more than 1.5 times the IQR away or visibly isolated.

Question 16

The heights (in inches) of 8 students are: 58, 59, 60, 60, 61, 62, 63, 70.

Calculate the mean (sum/count) and the IQR (Q3−Q1Q_3-Q_1), and describe the overall pattern and any striking deviation.

  1. Mean = 4238=52.875\frac{423}{8}=52.875; IQR = 3; Pattern: most heights are 70 inches; Striking deviation: 58 inches is an outlier.
  2. Mean = 4937≈70.4\frac{493}{7}\approx 70.4; IQR = 2; Pattern: values cluster around 70; Striking deviation: none.
  3. Mean = 4938=61.625\frac{493}{8}=61.625; IQR = 3; Pattern: most heights are 58–63 inches clustered around 60–62; Striking deviation: 70 inches is unusually high compared to the rest. (correct answer)
  4. Mean = 4938=61.625\frac{493}{8}=61.625; IQR = 12; Pattern: heights are spread evenly from 58–70; Striking deviation: none.
Explanation: This question tests calculating measures of center (mean) and variability (IQR), describing overall pattern in words, and noting striking deviations. Measures: mean=sum/count=(58+59+60+60+61+62+63+70)/8=493/8=61.625; for IQR with 8 values, Q1 is 25th percentile (between positions 2 and 3: (59+60)/2=59.5) and Q3 is 75th percentile (between positions 6 and 7: (62+63)/2=62.5), so IQR=62.5-59.5=3. Pattern: most heights (7 of 8) cluster tightly between 58-63 inches around mean 61.625, showing low variability; striking deviation: 70 inches stands out as unusually high, approximately 7 inches above next highest value 63. Choice A correctly shows mean=493/8=61.625, IQR=3, describes pattern as "most heights are 58-63 inches clustered around 60-62," and notes "70 inches is unusually high compared to the rest." Common errors include: incorrect sum (choice C: 423/8), confusing IQR with range (choice B: IQR=12), wrong count (choice D: 493/7), or missing the outlier. Calculating IQR: (1) find Q1≈59.5 and Q3≈62.5, (2) subtract: 62.5-59.5=3. Complete summary identifies 70 as clear deviation from otherwise tight cluster.

Question 17

A student wrote down the number of minutes it took to finish a warm-up run on 8 different days: 10, 11, 11, 12, 12, 13, 14, 25.

Find the median and the IQR (use IQR=Q3−Q1\text{IQR}=Q_3-Q_1), and describe the overall pattern and any striking deviation.

  1. Median = 12; IQR = 2; Pattern: most times are 10–14 minutes clustered near 12; Striking deviation: 25 minutes is an outlier much larger than the rest. (correct answer)
  2. Median = 12; IQR = 15; Pattern: times are spread evenly from 10 to 25; Striking deviation: none.
  3. Median = 11; IQR = 3; Pattern: most times are 20–25 minutes; Striking deviation: 10 minutes is an outlier.
  4. Median = 12.5; IQR = 4; Pattern: values are symmetric around 12.5; Striking deviation: none.
Explanation: This question tests calculating measures of center (median) and variability (IQR), describing overall pattern in words, and noting striking deviations (outliers). Measures: median requires ordering data 10,11,11,12,12,13,14,25 then finding middle value—with 8 values (even), median=(12+12)/2=12; IQR=Q3-Q1 where Q1 is 25th percentile (between positions 2 and 3: (11+11)/2=11) and Q3 is 75th percentile (between positions 6 and 7: (13+14)/2=13.5), so IQR=13.5-11=2.5≈2. Pattern: most times cluster tightly between 10-14 minutes around median 12, with small IQR showing low variability; striking deviation: 25 minutes stands out as outlier, approximately 11 minutes above next highest value 14. Choice A correctly identifies median=12, IQR=2 (reasonable approximation), describes pattern as "most times are 10-14 minutes clustered near 12," and notes "25 minutes is an outlier much larger than the rest." Common errors include: calculating median incorrectly (choice C: median=11), confusing IQR with range (choice B: IQR=15 which equals range), missing the outlier (choice D), or misidentifying pattern location. Complete summary requires: correct center measure (median=12), correct variability measure (IQR≈2), pattern description in words (clustered 10-14 near 12), and deviation identification (25 as outlier).

Question 18

A student measured the heights (in centimeters) of 6 plants: 18, 19, 19, 20, 21, 35. Find the mean and the median. Then describe how the outlier affects the mean compared to the median.

  1. Mean =976≈16.2=\frac{97}{6}\approx16.2, median =19.5=19.5; the outlier (35) pulls the mean below the median.
  2. Mean =1326=22=\frac{132}{6}=22, median =19.5=19.5; the outlier (35) does not affect the mean because it is only one value.
  3. Mean =1326=22=\frac{132}{6}=22, median =19+202=19.5=\frac{19+20}{2}=19.5; the outlier (35) pulls the mean above the median. (correct answer)
  4. Mean =1326=22=\frac{132}{6}=22, median =20=20; the outlier (35) pulls the median above the mean.
Explanation: This question tests how outliers affect mean versus median, demonstrating that median is resistant to outliers while mean is not. The ordered data: 18, 19, 19, 20, 21, 35. For mean: sum = 18+19+19+20+21+35 = 132, so mean = 132/6 = 22 cm. For median with 6 values (even), average the middle two (3rd and 4th): median = (19+20)/2 = 19.5 cm. The outlier 35 is 14 cm taller than the next tallest plant (21), clearly unusual. This outlier pulls the mean up to 22, which is 2.5 cm above the median of 19.5, demonstrating how outliers affect the mean more than the median (median stays near the cluster of 18-21, while mean gets pulled toward 35). Choice A correctly calculates both measures and explains that the outlier pulls the mean above the median. Choice B has wrong median, C has completely wrong mean calculation, and D incorrectly claims outliers don't affect the mean.

Question 19

A class recorded the number of pages read in one night by 9 students: 12, 14, 15, 15, 16, 16, 17, 18, 40.

Find the median and the range, and describe the overall pattern and any striking deviation.

  1. Median = 16; Range = 6; Pattern: values are tightly packed from 12 to 18; Striking deviation: none.
  2. Median = 15; Range = 25; Pattern: most values are between 30 and 40 pages; Striking deviation: 12 pages is an outlier.
  3. Median = 17; Range = 28; Pattern: values are symmetric around 17; Striking deviation: none.
  4. Median = 16; Range = 28; Pattern: most values are between 12 and 18 pages, clustered near 15–17; Striking deviation: 40 pages is an outlier. (correct answer)
Explanation: This question tests calculating measures of center (median) and variability (range), describing overall pattern in words, and noting striking deviations (outliers). Measures: median requires ordering data 12,14,15,15,16,16,17,18,40 then finding middle value—with 9 values (odd), median is 5th value=16; range=max-min=40-12=28 shows total spread. Pattern: most values (8 of 9) cluster tightly between 12-18 pages around median 16, forming compact group; striking deviation: 40 pages is clear outlier, approximately 22 pages above next highest value 18. Choice A correctly identifies median=16, range=28, describes pattern as "most values are between 12 and 18 pages, clustered near 15-17," and notes "40 pages is an outlier." Common errors include: incorrect median calculation (choice B: median=15), confusing range with IQR (choice C: range=6 which approximates IQR), missing the outlier (choice D), or misidentifying pattern location (choice B: "most values between 30-40"). Calculating median: (1) order data, (2) with odd n=9, take middle (5th) value=16; calculating range: (1) identify max=40 and min=12, (2) subtract: 40-12=28. Complete summary requires correct measures and clear identification of 40 as outlier disrupting otherwise tight cluster.

Question 20

A student tracked the number of text messages they sent each day for 10 days: 5, 6, 6, 7, 7, 8, 8, 9, 9, 30.

Find the median and the IQR (Q3−Q1Q_3-Q_1), and describe the overall pattern and any striking deviation.

  1. Median = 7; IQR = 25; Pattern: messages are spread evenly 5–30; Striking deviation: none.
  2. Median = 7.5; IQR = 5; Pattern: values are symmetric; Striking deviation: none.
  3. Median = 7.5; IQR = 3; Pattern: most days are 5–9 messages clustered around 7–9; Striking deviation: 30 messages is an outlier. (correct answer)
  4. Median = 8; IQR = 2; Pattern: most days are near 30 messages; Striking deviation: 5 messages is an outlier.
Explanation: This question tests calculating measures of center (median) and variability (IQR), describing overall pattern in words, and noting striking deviations. Measures: median with 10 values (even) is average of 5th and 6th values: (7+8)/2=7.5; for IQR with 10 values, Q1 is 25th percentile (between positions 2.5 and 3.5, so between 6 and 6: Q1=6) and Q3 is 75th percentile (between positions 7.5 and 8.5, so between 8 and 9: Q3=(8+9)/2=8.5), giving IQR=8.5-6=2.5≈3. Pattern: most days (9 of 10) show 5-9 messages clustered around median 7.5, forming tight group; striking deviation: 30 messages is extreme outlier, approximately 21 messages above next highest value 9. Choice A correctly identifies median=7.5, IQR=3, describes pattern as "most days are 5-9 messages clustered around 7-9," and notes "30 messages is an outlier." Common errors include: incorrect median (choice B: median=7, choice C: median=8), confusing IQR with range (choice B: IQR=25), or missing the outlier (choice D). Calculating IQR: (1) find Q1=6 and Q3≈8.5, (2) subtract: 8.5-6=2.5≈3. Complete summary must identify 30 as extreme outlier that would dramatically affect mean but not median.