Statistics Quiz: Comparing Data Sets By Center Spread
Practice Comparing Data Sets By Center Spread in Statistics 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 By Center Spread, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics.
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
Two classes took the same 20-point quiz. The teacher summarized each class’s scores and described the shapes.
Data Set A (Class A): approximately symmetric with no outliers; xˉ=15.2, s=1.1, median =15, IQR =1.5.
Data Set B (Class B): approximately symmetric with no outliers; xˉ=14.6, s=2.0, median =15, IQR =2.5.
From the summary statistics, which statement best compares the typical score and variability of Class A and Class B using appropriate measures?
Use median and IQR; both classes have the same typical score, and Class A is more variable because its maximum score is higher.
Use mean and s; Class A typically scored higher and is more consistent (smaller s) than Class B.
Use mean and s; Class B typically scored higher and is more consistent (smaller s) than Class A.
Explanation: When comparing data sets like these quiz scores, we focus on the center to determine the typical value and the spread to assess consistency. Since both distributions are approximately symmetric with no outliers, mean and standard deviation are appropriate measures. The mean for Class A is 15.2, higher than Class B's 14.6, indicating Class A has a higher typical score. The standard deviation for Class A is 1.1, smaller than Class B's 2.0, showing Class A has less variability. The statement in choice A correctly matches by using mean and s to state Class A typically scored higher and is more consistent. A common misconception is relying on medians here, which are equal at 15, missing the difference captured by means due to symmetry. To compare any data sets, first determine if they are symmetric or skewed/outliers present, choose mean/SD or median/IQR accordingly, and then interpret the center and spread in the context of the data.
Question 2
Two different machines fill 1-liter bottles. The fill amounts (in mL) for a sample are shown below.
Data Set A (Machine A): 1001, 999, 1000, 1002, 998, 1001, 1000, 999, 1002, 998
Data Set B (Machine B): 1000, 1000, 1001, 999, 1000, 1001, 999, 1000, 1000, 1000
Based on these data, which statement best compares the typical fill amount and variability using appropriate measures?
Use mean and s; both machines have about the same typical fill, but Machine B is more consistent (smaller s).
Use median and IQR; Machine B has a higher typical fill because it has more 1000s, and Machine A is more consistent because its range is smaller.
Use mean and ; both machines have about the same typical fill, but Machine A is more consistent (smaller ).
Question 3
A company tracked the number of customer support tickets handled per day by two employees over 14 days.
Data Set A (Employee A): 18, 19, 20, 20, 21, 21, 22, 22, 22, 23, 23, 24, 24, 25
Data Set B (Employee B): 15, 16, 18, 19, 20, 20, 21, 21, 22, 23, 24, 26, 27, 29
Based on the data, which statement best compares the typical number of tickets and variability using appropriate measures?
Use median and IQR; Employee B typically handles more tickets and is more consistent (smaller IQR) than Employee A.
Use median and IQR; the employees have the same typical number of tickets, and Employee A is more variable because its range is larger.
Use mean and s; Employee B typically handles more tickets and is more consistent because B has a higher maximum.
Use mean and ; Employee A typically handles slightly more tickets and is more consistent (smaller ) than Employee B.
Question 4
Two teachers recorded the number of minutes it took students to finish a puzzle. Data Set A is roughly symmetric with no outliers; Data Set B is right-skewed with one unusually long time.
Summary statistics:
Data Set A: mean xˉ=18.0, median =18.0, standard deviation , IQR
Question 5
Two groups of students measured plant heights (in cm) after 4 weeks. Both data sets are approximately symmetric with no outliers.
Summary statistics:
Data Set A: mean xˉ=22.6, median =22.6, standard deviation , IQR
Question 6
Two vending machines were monitored for the number of items sold per day. Both data sets are approximately symmetric with no outliers.
Summary statistics:
Data Set A: mean xˉ=52.0, median =52.0, standard deviation , IQR
Question 7
A bookstore tracked the number of books customers bought in a single visit. Data Set A is right-skewed (a few customers bought many books). Data Set B is roughly symmetric with no outliers.
Summary statistics:
Data Set A: mean xˉ=4.8, median =3.0, standard deviation , IQR
Question 8
Two delivery routes were timed (in minutes) over several days. Data Set A is right-skewed with an outlier day due to a traffic accident. Data Set B is roughly symmetric.
Summary statistics:
Data Set A: mean xˉ=38.5, median =35.0, standard deviation , IQR
Question 9
A music app compared the number of songs streamed by users in one day. Data Set A is right-skewed with a few heavy users. Data Set B is also right-skewed but less extreme.
Summary statistics:
Data Set A: mean xˉ=36.0, median =22.0, standard deviation , IQR
Question 10
A coach compared sprint times (in seconds) for two training groups. Both distributions are approximately symmetric with no outliers.
Summary statistics:
Data Set A: mean xˉ=12.4, median =12.4, standard deviation , IQR
Question 11
A science teacher compared lab report scores (out of 50) from two classes. Data Set A is roughly symmetric with no outliers. Data Set B is left-skewed (a few very low scores).
Summary statistics:
Data Set A: mean xˉ=41.2, median =41.0, standard deviation , IQR
Question 12
Two brands of AA batteries were tested, and the lifetimes (in hours) were recorded.
Summary statistics:
Data Set A (Brand A): mean = 9.8, median = 9.7, s = 0.6, IQR = 0.8 (roughly symmetric)
Data Set B (Brand B): mean = 10.4, median = 9.9, s = 1.9, IQR = 0.9 (right-skewed with a few high values)
Which statement best compares the typical lifetime and variability using appropriate measures?
Using mean and s, Brand A typically lasts longer and is more variable than Brand B.
Using median and IQR, Brand B typically lasts slightly longer and has similar consistency to Brand A.
Question 13
A school compared the number of pages students read over a weekend.
Data Set A: 12, 14, 15, 16, 16, 17, 18, 18, 19, 20, 21
Data Set B: 5, 8, 10, 12, 14, 15, 16, 18, 22, 30, 45
Data Set B is right-skewed with a very large value (45). Which statement best compares the typical number of pages read and the variability using appropriate measures (median and IQR)?
Using mean and standard deviation, Data Set A has a higher typical value and is more consistent than Data Set B.
Using median and IQR, Data Set A has a higher typical value and is more consistent than Data Set B.
Using median and IQR, the two data sets have the same typical value, and Data Set A is more variable because its maximum is larger.
Using median and IQR, Data Set B has a higher typical value and is more consistent than Data Set A.
Explanation: When comparing distributions, we must identify their shape to choose appropriate measures. Data Set B is clearly right-skewed with a very large value (45 pages), while Data Set A appears more symmetric. With skewed data present, we use median and IQR for both sets. For center: Data Set A's median is 17 pages (6th value in ordered list), while Data Set B's median is 15 pages, so Data Set A has a higher typical value. For spread: Data Set A has a small IQR (roughly 4 pages), while Data Set B's IQR is larger due to its skewed distribution, making Data Set A more consistent. The outlier (45) in Data Set B doesn't affect these resistant measures. A common error is using maximum values to judge variability instead of IQR. The strategy is: identify skewness → use median/IQR → compare resistant measures.
Question 14
A fitness app recorded the time (in minutes) it took two groups to complete the same 2-mile walk.
Data Set A (Group A times): 22, 23, 24, 24, 25, 25, 26, 26, 27, 28, 60
Data Set B (Group B times): 24, 25, 25, 26, 26, 27, 27, 28, 28, 29, 30
Data Set A has a clear high outlier (60 minutes), while Data Set B does not. Which statement best compares the typical value and variability using appropriate measures (median and IQR)?
Using median and IQR, Group A has a lower typical time and similar variability to Group B.
Using median and IQR, Group A has a higher typical time and a larger variability than Group B.
Using median and IQR, Group B has a lower typical time and a larger variability than Group A.
Using mean and standard deviation, Group A has a higher typical time and much greater variability than Group B.
Explanation: When comparing distributions, we must first identify their shape to select appropriate measures. Data Set A has a clear high outlier (60 minutes), making it skewed, so we should use median and IQR for both sets to ensure fair comparison. For center: Group A's median is 25 minutes (middle of ordered data excluding outlier influence), while Group B's median is 27 minutes, so Group A has a lower typical time. For spread: Group A's IQR (Q3-Q1) is approximately 3 minutes, while Group B's IQR is also approximately 3 minutes, showing similar variability. The outlier in Group A doesn't affect these resistant measures. A common error is using mean with outliers, which would incorrectly inflate Group A's typical value. The strategy is: identify outliers/skewness → use median/IQR → compare resistant measures in context.
Question 15
A store tracked the number of minutes customers spent waiting in line on two different days.
Data Set A (Day 1): 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8
Data Set B (Day 2): 1, 2, 2, 3, 3, 4, 4, 6, 7, 9, 12, 18
Data Set A is roughly symmetric. Data Set B is right-skewed with a few large values. Which measures are most appropriate for comparing the typical value and variability of these distributions?
Use the mean and standard deviation for both data sets, because both have the same sample size.
Use the mean and IQR for both data sets, because the mean measures center and IQR measures spread.
Use the median and IQR for both data sets, because at least one distribution is skewed.
Use the median and standard deviation for both data sets, because the median is resistant and standard deviation is always best for spread.
Explanation: To compare distributions effectively, we must choose measures that match the shape of our data. Data Set A is roughly symmetric, which would normally suggest mean and standard deviation. However, Data Set B is right-skewed with large values (12, 18), making it asymmetric. When at least one distribution is skewed, we should use median and IQR for both sets to ensure valid comparison. Using mean for skewed data gives misleading results because extreme values pull the mean away from the typical value. Similarly, standard deviation isn't appropriate for skewed data. A common misconception is mixing measures (like mean with IQR), but we need consistent measure pairs. The strategy is: examine both shapes → if any distribution is skewed, use median/IQR for all → ensure fair comparison.
Question 16
Two delivery services recorded the number of packages delivered per driver in a day.
Summary statistics:
Data Set A (Service A): mean = 52.0, median = 52.0, s = 4.0, IQR = 5 (roughly symmetric)
Data Set B (Service B): mean = 52.5, median = 50.0, s = 10.5, IQR = 6 (right-skewed with high values)
Which statement best compares the typical number of packages and consistency using appropriate measures?
Using mean and s, Service B typically delivers more packages and is more consistent than Service A.
Question 17
A science club measured the mass (in grams) of two types of small rocks collected from different locations.
Data Set A (Location A): 48, 49, 50, 50, 51, 51, 52, 52, 53, 54
Data Set B (Location B): 45, 46, 47, 49, 50, 51, 52, 54, 56, 58
Neither set has an obvious outlier, and both are roughly symmetric. Which statement best compares the typical mass and variability using appropriate measures (mean and standard deviation)?
Location B has a lower mean and a smaller standard deviation than Location A, so it is typically lighter and more consistent.
Location A has a higher mean and a smaller standard deviation than Location B, so it is typically heavier and more consistent.
Location A has a higher mean and a larger standard deviation than Location B, so it is typically heavier but less consistent.
Location B has a higher mean and a larger standard deviation than Location A, so it is typically heavier but less consistent.
Explanation: When comparing distributions, we first assess their shape to select appropriate measures. Both rock mass data sets are roughly symmetric with no obvious outliers, making mean and standard deviation the best choices. For center: Location A's mean is 51 grams (sum of 510 ÷ 10), while Location B's mean is 50.8 grams, so Location A typically has heavier rocks. For spread: Location A's values cluster tightly (48-54, range of 6), while Location B spans more widely (45-58, range of 13), giving Location A a smaller standard deviation and more consistency. Therefore, Location A has both higher typical mass and greater consistency. A common error is confusing larger numbers with larger variability. The strategy is: verify symmetric shape → use mean/SD → compare both center and spread in context.
Question 18
Two cafeterias tracked the number of students arriving during a 10-minute interval each day.
Data Set A: 30, 31, 31, 32, 33, 33, 34, 34, 35, 36, 36, 37
Data Set B: 20, 22, 25, 28, 30, 32, 34, 36, 38, 40, 42, 45
Both distributions are roughly symmetric with no clear outliers. Which statement best compares the typical number of students and the variability using appropriate measures (mean and standard deviation)?
Cafeteria B has a higher mean and a smaller standard deviation than Cafeteria A, so it is typically busier and more consistent.
Cafeteria A has a lower mean and a larger standard deviation than Cafeteria B, so it is typically less busy and less consistent.
Cafeteria A has a higher mean and a smaller standard deviation than Cafeteria B, so it is typically busier and more consistent.
Cafeteria B has a higher mean and a larger standard deviation than Cafeteria A, so it is typically busier but less consistent.
Explanation: To compare distributions effectively, we must first check their shape to choose appropriate measures. Both cafeteria data sets are roughly symmetric with no clear outliers, making mean and standard deviation appropriate. For center: Cafeteria A's mean is approximately 33.3 students, while Cafeteria B's mean is approximately 31.8 students, so Cafeteria A is typically busier. For spread: Cafeteria A's values cluster tightly (30-37, range of 7), while Cafeteria B spans widely (20-45, range of 25), giving Cafeteria A a much smaller standard deviation and more consistency. Therefore, Cafeteria A has both a higher mean and smaller standard deviation, making it typically busier and more consistent. A misconception is assuming wider spread means higher center. The strategy is: check shape → use mean/SD for symmetric data → interpret both measures correctly.
Question 19
A game developer compared the number of attempts players needed to beat a level in two versions of the game.
Data Set A (Version A): 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 7
Data Set B (Version B): 2, 3, 3, 4, 4, 5, 5, 6, 7, 9, 15
Data Set B is right-skewed with a high outlier (15). Which statement best compares the typical number of attempts and the variability using appropriate measures (median and IQR)?
Using median and IQR, Version A typically takes about the same number of attempts as Version B, but Version B is more variable.
Using range, Version A is more variable because its minimum is larger than Version B’s minimum.
Using mean and standard deviation, Version B typically takes more attempts and is more consistent than Version A.
Using median and IQR, Version A typically takes more attempts and is more variable than Version B.
Explanation: When comparing distributions, we must identify their shape to select appropriate measures. Version A appears roughly symmetric, but Version B is right-skewed with a high outlier (15 attempts), so we use median and IQR for both versions. For center: Version A's median is 5 attempts (6th value), and Version B's median is also 5 attempts, showing similar typical performance. For spread: Version A's IQR is approximately 2 (from Q1=4 to Q3=6), while Version B's IQR is larger due to the outlier's influence on the upper quartile, making Version B more variable. The outlier in Version B doesn't affect the median but does increase spread. A common error is using range with outliers or comparing minimums alone. The strategy is: identify skewness/outliers → use median/IQR → compare resistant measures appropriately.
Question 20
A cafeteria tracked how many items students bought at lunch.
Data Set A (items per student): 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4
Data Set B (items per student): 1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 7
Which statement best compares the typical value and variability of Data Set A and Data Set B? Use measures appropriate to the distribution shapes.
Use the mean and standard deviation: Data Set B has a higher typical number of items, and Data Set A is more variable.
Use the median and IQR: The typical number of items is about the same for both, but Data Set B is more variable.
Use the median and IQR: Data Set A has a higher typical number of items, and Data Set B is more consistent because most values are small.
Use the range: Data Set B is more variable and also has a much higher typical value because it has a 7.
Explanation: We compare data sets by their centers and spreads to assess typical values and consistency. Data Set A is symmetric, but B is right-skewed with an outlier, so median and IQR are suitable for both. The medians are 2.5 for A and 2 for B, which are about the same for typical items bought. B's IQR of 2 exceeds A's 1, indicating B has more variability. Choice B accurately uses median and IQR to note similar centers but greater spread in B. A misconception is using mean for skewed data, as B's outlier inflates its mean, suggesting a falsely higher typical value. To proceed, check shape, choose median/IQR for skewness, and compare in the lunch purchasing context.
Use median and IQR; Class B typically scored higher and is more consistent (smaller IQR) than Class A.
s
s
Use median and IQR; Machine A has a higher typical fill and Machine B is more variable (larger IQR).
Explanation: When comparing data sets like these fill amounts, we focus on the center to determine the typical value and the spread to assess consistency. Both distributions appear symmetric with no outliers, so mean and standard deviation are appropriate measures. The mean for both Machine A and Machine B is 1000 mL, indicating about the same typical fill amount. Calculations show Machine A has a larger standard deviation due to values ranging from 998 to 1002, while Machine B is tighter around 1000. The statement in choice B correctly matches by using mean and s to state both have the same typical fill, but Machine B is more consistent. A common misconception is using range, where Machine A has a range of 4 mL and Machine B has 2 mL, but standard deviation better quantifies the overall spread. To compare any data sets, first determine if they are symmetric or skewed/outliers present, choose mean/SD or median/IQR accordingly, and then interpret the center and spread in the context of the data.
s
s
Explanation: When comparing data sets like these tickets handled, we focus on the center to determine the typical value and the spread to assess consistency. Both distributions appear approximately symmetric with no outliers, so mean and standard deviation are appropriate measures. The mean for Employee A is approximately 21.7 tickets, slightly higher than Employee B's 21.5. Employee A has a smaller standard deviation, as values range from 18 to 25 compared to B's 15 to 29. The statement in choice A correctly matches by using mean and s to state Employee A handles slightly more and is more consistent. A common misconception is using range for spread, which is 7 for A and 14 for B, but standard deviation accounts for all data points better. To compare any data sets, first determine if they are symmetric or skewed/outliers present, choose mean/SD or median/IQR accordingly, and then interpret the center and spread in the context of the data.
s=
2.1
=3
Data Set B: mean xˉ=20.6, median =19.0, standard deviation s=5.8, IQR =4
From the summary statistics, which statement best compares the typical value and variability of Data Set A and Data Set B using appropriate measures?
Use the median and IQR: Data Set A typically takes longer (higher median) and is more variable (larger IQR) than Data Set B.
Use the mean and standard deviation: Data Set B typically takes longer (higher mean) and is more variable (larger s) than Data Set A.
Use the mean and standard deviation: Data Set B has a higher typical time and is more consistent because its mean is larger.
Use the median and IQR: Data Set B typically takes longer (higher median) and is more variable (larger IQR) than Data Set A.
Explanation: When comparing data sets, we focus on center (typical value) and spread (variability) to understand differences in distributions. For symmetric data without outliers like Data Set A, mean and standard deviation are appropriate, but for skewed data with outliers like Data Set B, median and IQR are better to avoid distortion. Here, Data Set B's median of 19 is higher than Data Set A's 18, indicating B typically takes longer. Data Set B's IQR of 4 is larger than A's 3, showing more variability in B. The correct statement uses median and IQR, matching the higher center and greater spread in B due to its skew. A misconception is using mean for skewed data, as B's mean is inflated by the outlier, unlike the robust median. To compare, assess shape first, choose measures accordingly, then evaluate center and spread in the puzzle completion context.
s
=
1.8
=2.4
Data Set B: mean xˉ=24.0, median =24.0, standard deviation s=1.8, IQR =2.4
Which statement best compares the typical value and variability of Data Set A and Data Set B using appropriate measures?
Use the mean and standard deviation: Data Set B typically has taller plants (higher mean) and the two groups are equally consistent (same s).
Use the median and IQR: Data Set B typically has taller plants (higher median) and is less consistent (larger IQR) than Data Set A.
Use the median and IQR: Data Set A typically has taller plants (higher median) and is less consistent (larger IQR) than Data Set B.
Use the mean and standard deviation: Data Set A typically has taller plants (higher mean) and is more consistent (smaller s) than Data Set B.
Explanation: We compare data by center and spread to gauge typical outcomes and variability. Both symmetric without outliers, so mean and standard deviation work well. Data Set B's mean of 24 exceeds A's 22.6, indicating taller typical plants in B. Both have the same SD of 1.8, showing equal consistency. The correct statement uses mean and SD, reflecting B's higher center and identical spread. Misconception: assuming different IQRs imply variability differences when SD is equal in symmetric data. Tactic: verify symmetry, select mean/SD, and contextualize for plant growth comparisons.
s
=
4.0
=6
Data Set B: mean xˉ=49.5, median =49.5, standard deviation s=2.0, IQR =3
Which statement best compares the typical value and variability of Data Set A and Data Set B using appropriate measures?
Use the mean and standard deviation: Data Set A typically sells more items (higher mean) but is less consistent (larger s) than Data Set B.
Use the median and IQR: Data Set A typically sells more items (higher median) and is more consistent (smaller IQR) than Data Set B.
Use the median and IQR: Data Set B typically sells more items (higher median) and is less consistent (larger IQR) than Data Set A.
Use the mean and standard deviation: Data Set B typically sells more items (higher mean) and is less consistent (larger s) than Data Set A.
Explanation: We compare data sets by examining center and spread to assess typical values and variability. Both sets being symmetric without outliers makes mean and standard deviation the right choices. Data Set A's mean of 52 surpasses B's 49.5, indicating higher typical sales in A. Data Set A's SD of 4 exceeds B's 2, showing A is less consistent. The correct statement uses mean and SD, aligning with A's higher center and greater spread. A misconception is confusing IQR with SD in symmetric data, as both tell similar stories here but SD is precise for normality. Approach: confirm symmetry, pick mean/SD, and contextualize comparisons for vending machine sales.
s=
4.2
=3
Data Set B: mean xˉ=4.1, median =4.0, standard deviation s=1.3, IQR =2
Which statement best compares the typical value and variability of Data Set A and Data Set B using appropriate measures?
Use the mean and standard deviation: Data Set B typically has more books purchased (higher mean) and is more variable (larger s) than Data Set A.
Use the median and IQR: Data Set B typically has more books purchased (higher median) and is more consistent (smaller IQR) than Data Set A.
Use the mean and standard deviation: Data Set A typically has more books purchased (higher mean) and is more variable (larger s) than Data Set B.
Use the median and IQR: Data Set A typically has more books purchased (higher median) and is more consistent (smaller IQR) than Data Set B.
Explanation: The key concept is comparing distributions by their center and spread to interpret typical behavior and variability. Data Set A is skewed, so median and IQR are appropriate, while B is symmetric, but for fair comparison, use median/IQR consistently. Data Set B's median of 4 exceeds A's 3, suggesting more typical books purchased in B. Data Set B's IQR of 2 is smaller than A's 3, indicating greater consistency in B. The correct statement employs median and IQR, capturing B's higher center and lower spread aptly. Misconception: using mean for skewed A would overstate its center due to high outliers. Strategy: evaluate shape, choose robust measures like median/IQR for skew, then compare in the book purchasing context.
s
=
10.2
=5
Data Set B: mean xˉ=36.8, median =37.0, standard deviation s=3.1, IQR =4
Which statement best compares the typical value and variability of Data Set A and Data Set B using appropriate measures?
Use the median and IQR: Data Set B typically takes longer (higher median) and is more consistent (smaller IQR) than Data Set A.
Use the median and IQR: Data Set A typically takes longer (higher median) and is more variable (larger IQR) than Data Set B.
Use the median and IQR: Data Set B typically takes longer (higher median) and is more variable (larger IQR) than Data Set A.
Use the mean and standard deviation: Data Set A typically takes longer (higher mean) and is more variable (larger s) than Data Set B.
Explanation: The concept involves comparing centers and spreads to understand data distributions. With A skewed and having an outlier, median and IQR are appropriate, while B is symmetric, but use them consistently. Data Set B's median of 37 is higher than A's 35, meaning longer typical times in B. Data Set B's IQR of 4 is smaller than A's 5, suggesting more consistency in B. The correct statement uses median and IQR, fitting B's higher center and reduced spread. Common error: using mean for A, inflated by the outlier, misrepresents typical time. Strategy: assess shape, choose median/IQR for skew/outliers, and compare in delivery times context.
s
=
40.0
=18
Data Set B: mean xˉ=28.0, median =24.0, standard deviation s=18.0, IQR =12
Which statement best compares the typical value and variability of Data Set A and Data Set B using appropriate measures?
Use the median and IQR: Data Set A typically streams more songs (higher median) and is more variable (larger IQR) than Data Set B.
Use the median and IQR: Data Set B typically streams more songs (higher median) and is more consistent (smaller IQR) than Data Set A.
Use the mean and standard deviation: Data Set A typically streams more songs (higher mean) and is more consistent (smaller s) than Data Set B.
Use the mean and standard deviation: Data Set B typically streams more songs (higher mean) and is more variable (larger s) than Data Set A.
Explanation: Comparing distributions means analyzing center and spread for typical values and variability. Both sets skewed, so median and IQR are robust choices. Data Set B's median of 24 tops A's 22, showing more typical streams in B. Data Set B's IQR of 12 is less than A's 18, indicating higher consistency in B. The correct statement employs median and IQR, matching B's elevated center and lower spread. Error to avoid: using mean in skewed data, as A's is pulled high by extremes. Process: identify skew, pick median/IQR, and compare in streaming behavior context.
s
=
0.6
=0.8
Data Set B: mean xˉ=12.0, median =12.0, standard deviation s=1.1, IQR =1.5
Which statement best compares the typical value and variability of Data Set A and Data Set B using appropriate measures?
Use the mean and standard deviation: Data Set A typically has faster times (lower mean) and is less consistent (larger s) than Data Set B.
Use the mean and standard deviation: Data Set B typically has faster times (lower mean) but is less consistent (larger s) than Data Set A.
Use the median and IQR: Data Set A typically has faster times (lower median) and is more consistent (smaller IQR) than Data Set B.
Use the median and IQR: Data Set B typically has slower times (higher median) and is more consistent (smaller IQR) than Data Set A.
Explanation: Comparing data sets involves analyzing center and spread to highlight typical values and consistency. Since both sets are symmetric without outliers, mean and standard deviation are suitable measures. Data Set B's mean of 12.0 is lower than A's 12.4, showing B has faster typical sprint times. Data Set B's standard deviation of 1.1 is larger than A's 0.6, indicating B is less consistent. The correct statement uses mean and SD, accurately reflecting B's lower center and greater spread. A common misconception is relying on range instead of SD, which might overlook the full variability in symmetric data. The strategy is to check for symmetry, select mean/SD, and compare center and spread in the sprint training context.
s
=
3.0
=4
Data Set B: mean xˉ=39.0, median =40.0, standard deviation s=5.5, IQR =6
Which statement best compares the typical value and variability of Data Set A and Data Set B using appropriate measures?
Use the median and IQR: Data Set B typically scores higher (higher median) and is more consistent (smaller IQR) than Data Set A.
Use the mean and standard deviation: Data Set A typically scores higher (higher mean) and is more consistent (smaller s) than Data Set B.
Use the mean and standard deviation: Data Set B typically scores higher (higher mean) and is more variable (larger s) than Data Set A.
Use the median and IQR: Data Set A typically scores higher (higher median) and is more consistent (smaller IQR) than Data Set B.
Explanation: Comparing distributions requires evaluating center and spread for insights into typical performance and consistency. Data Set B is skewed, so median and IQR are preferable, while A is symmetric, but use median/IQR for consistency. Data Set A's median of 41 is higher than B's 40, showing better typical scores in A. Data Set A's IQR of 4 is smaller than B's 6, indicating more consistency in A. The correct statement applies median and IQR, correctly identifying A's superior center and lower spread. Misconception: applying mean to skewed B underestimates its center due to low outliers. Method: determine shape, select median/IQR for skew, and compare in the lab scores context.
Using mean and s, Brand B typically lasts longer and is more consistent than Brand A.
Using median and IQR, Brand A typically lasts longer and is more variable than Brand B.
Explanation: When comparing distributions, we must match our measures to the data's shape. Brand A is roughly symmetric (mean ≈ median), so mean and standard deviation would work well. However, Brand B is right-skewed (mean 10.4 > median 9.9), indicating outliers pulling the mean up. With skewed data, we use median and IQR for both brands to ensure fair comparison. For center: Brand B's median (9.9) is slightly higher than Brand A's (9.7), so Brand B typically lasts longer. For spread: The IQRs are similar (0.8 vs 0.9), showing comparable consistency. The large standard deviation for Brand B (1.9) is inflated by outliers and doesn't represent typical variability. A common error is using mean with skewed data. The strategy is: check symmetry → use median/IQR for any skewness → interpret in context.
Using range only, Service B is more consistent because it has a higher maximum.
Using median and IQR, Service B typically delivers more packages and is slightly more consistent than Service A.
Using median and IQR, Service A typically delivers more packages and is slightly more consistent than Service B.
Explanation: To compare distributions accurately, we must match our measures to the data's shape. Service A is roughly symmetric (mean = median = 52), suggesting mean and standard deviation work well. However, Service B is right-skewed (mean 52.5 > median 50), with high outliers pulling the mean up. When one distribution is skewed, we use median and IQR for both to ensure fair comparison. For center: Service A's median (52) is higher than Service B's (50), so Service A typically delivers more packages. For spread: Service A's IQR (5) is slightly smaller than Service B's (6), making Service A slightly more consistent. The skewness in Service B inflates its mean and standard deviation. A misconception is using range alone, which ignores the distribution's shape. The strategy is: check symmetry → use median/IQR with any skewness → interpret appropriately.