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Compare quantitative distributions

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53976412
Group A has mean \(62\) and standard deviation \(5\). Group B has mean \(58\) and standard deviation \(2\). Compare typical level and consistency.

Hints

- Interpret center and spread as separate features. - A higher mean indicates a higher typical level under that summary; a smaller standard deviation indicates tighter clustering.

Solution

1. Group A has the higher mean, so its typical level is higher by \(4\). 2. Group B has the smaller standard deviation, so its values are more consistent around its mean.

Answer

Group A has the higher center; Group B has less variability and is more consistent.
53975912
Compare the centers and IQRs of distributions A and B.
Figure for problem 539759

Hints

- Read matching summary landmarks from both plots before comparing. - Compare medians for center and subtract \(Q_1\) from \(Q_3\) for each IQR.

Solution

1. The medians are \(24\) for A and \(28\) for B, so B has the higher center. 2. A has IQR \(30-18=12\). 3. B has IQR \(36-20=16\), so B has greater middle-half variability.

Answer

B has the higher median, \(28\) versus \(24\), and the larger IQR, \(16\) versus \(12\).
53976312
The dotplots have the same median. Compare their spreads and shapes.
Figure for problem 539763

Hints

- Use the shared horizontal scale to compare the endpoints. - Use the stack pattern around the median to compare concentration and shape.

Solution

1. Both medians are \(5\). 2. a) spans \(1\) to \(9\), while b) spans \(3\) to \(7\), so a) is more spread out. 3. Both are approximately symmetric and unimodal, but b) is more concentrated around the center.

Answer

Both are approximately symmetric with median \(5\); a) has greater spread, while b) is more concentrated.
53976512
Which distribution has the larger IQR, and which has the larger range?
Figure for problem 539765

Hints

- Compute each spread measure separately; they can rank distributions differently. - Use the box width for IQR and the full endpoint distance for range; the measures need not agree.

Solution

1. A has IQR \(18-10=8\) and range \(40-0=40\). 2. B has IQR \(22-8=14\) and range \(30-5=25\). 3. B has the larger IQR, while A has the larger range.

Answer

Larger IQR: B, \(14\). Larger range: A, \(40\).
53976712
Compare the likely skewness of A and B.
Figure for problem 539767

Hints

- Interpret the tail and median position in each boxplot separately. - Then compare the two inferred skew directions.

Solution

1. A has a much longer left whisker and a median closer to \(Q_3\), suggesting left skew. 2. B has a much longer right whisker, suggesting right skew.

Answer

A is likely left-skewed; B is likely right-skewed.
53976812
Use the boxplots to compare \(Q_3\) of A with \(Q_1\) of B. What does this indicate about the middle halves?
Figure for problem 539768

Hints

- Compare the endpoints of the two boxes on the common scale. - Write each middle-half interval from \(Q_1\) to \(Q_3\), then find any common values.

Solution

1. For A, \(Q_3=40\). 2. For B, \(Q_1=35\). 3. Since \(35<40\), the middle \(50\%\) intervals \([20,40]\) and \([35,55]\) overlap from \(35\) to \(40\).

Answer

A’s \(Q_3=40\) exceeds B’s \(Q_1=35\), so the middle halves overlap on \([35,40]\).
53976912
Two distributions have the same mean. Must they have the same shape and standard deviation? Explain.

Hints

- Consider how much information one summary statistic leaves unspecified. - Imagine rearranging observations around a fixed balance point while changing their distances from it.

Solution

1. The mean describes only one aspect of center. 2. Distributions can place observations differently around the same balance point. 3. Their shapes and standard deviations can therefore differ.

Answer

No. Equal means do not imply equal shape or variability.
53977012
The boxplots show fill amounts in milliliters. Compare the medians and IQRs in context.
Figure for problem 539770

Hints

- Compare like landmarks and express differences with the variable’s units. - State both median differences and IQR differences in milliliters.

Solution

1. Machine A has median \(101\,\text{mL}\) and IQR \(104-98=6\,\text{mL}\). 2. Machine B has median \(102\,\text{mL}\) and IQR \(105-99=6\,\text{mL}\). 3. Machine B’s typical fill is \(1\,\text{mL}\) higher, while the middle-half variability is the same.

Answer

Machine B has a median \(1\,\text{mL}\) higher; both machines have IQR \(6\,\text{mL}\).
53977112
Compare the modality and clustering in histograms a) and b).
Figure for problem 539771

Hints

- Locate the prominent peaks in each histogram. - Check whether a low-frequency or empty interval separates the peaks enough to form distinct clusters.

Solution

1. a) has one main peak near \(12.5\), so it is unimodal with one cluster. 2. b) has two separated peaks near \(7.5\) and \(37.5\), with a large central gap. 3. b) is bimodal with two distinct clusters.

Answer

a) is unimodal with one cluster; b) is bimodal with two clusters separated by a large gap.
53977212
Two identical data sets are compared. In one set, the largest value is replaced by a much larger value. Which comparisons will show the greatest change: medians and IQRs, or means and standard deviations?

Hints

- Separate positional summaries from summaries that use every observation’s magnitude. - Decide which group of summaries is more sensitive to replacing one extreme value.

Solution

1. The replacement creates or enlarges a high extreme value. 2. Means and standard deviations are nonresistant and will change substantially. 3. Medians and IQRs are resistant and may change little or not at all.

Answer

The means and standard deviations will show the greatest change.
53977412
Describe the relationship between the two distributions’ centers and spreads.
Figure for problem 539774

Hints

- Check whether corresponding landmarks differ by a constant. - A constant difference at every landmark signals a horizontal shift, while equal landmark gaps signal unchanged spread.

Solution

1. B’s five-number summary is exactly \(10\) greater at every landmark. 2. The medians differ by \(10\), but both IQRs are \(6\) and both ranges are \(12\). 3. B is a rightward shift of A with the same spread.

Answer

B is shifted \(10\) units higher than A, with the same IQR and range.
54881512
Distribution B is created by adding \(7\) to every observation in distribution A. Compare the two distributions' shape, measures of center, and measures of spread.

Hints

- Track what happens to pairwise differences under a common shift. - Separate location summaries from spread summaries. - A translation preserves relative positions.

Solution

1. Adding \(7\) shifts every observation the same distance to the right. 2. The mean, median, quartiles, minimum, and maximum all increase by \(7\). 3. Differences between observations do not change, so the range, interquartile range, standard deviation, and shape remain unchanged.

Answer

B has the same shape and spread as A, while every measure of center and location is \(7\) units larger.
54882012
Distribution B is created by multiplying every value in distribution A by \(2\). Distribution A has median \(12\), interquartile range \(3\), and range \(10\). Compare these summaries for A and B.

Hints

- Apply the scale factor to both values and distances. - Measures of spread are based on differences between values. - A positive scaling preserves ordering and shape.

Solution

1. Multiplying by \(2\) doubles every location value, so B's median is \(2\cdot12=24\). 2. All numerical distances double, so B's interquartile range is \(2\cdot3=6\). 3. B's range is \(2\cdot10=20\).

Answer

B has median \(24\), interquartile range \(6\), and range \(20\); each is twice the corresponding summary for A.
54882212
Student A scored \(70\) on Test A and was at the \(80\)th percentile. Student B scored \(75\) on Test B and was at the \(60\)th percentile. Compare their raw scores and their relative standing within the two score distributions.

Hints

- Compare the numerical scores separately from the percentile ranks. - A percentile describes position relative to other scores on the same assessment. - Do not treat scores from different distributions as interchangeable measures of standing.

Solution

1. Student B has the higher raw score because \(75>70\). 2. Student A has the higher relative standing because the \(80\)th percentile exceeds the \(60\)th percentile. 3. Raw scores from different tests and percentile ranks answer different comparison questions.

Answer

Student B has the higher raw score, but Student A has the higher relative standing within the corresponding distribution.
54883112
Every observation in both distributions A and B includes the same measurement offset of \(+5\). The offset is removed from both distributions. How do the difference between their means, their standard deviations, and their shapes change?

Hints

- Track the effect of the correction on each distribution separately. - Compare a difference of two quantities when both receive the same shift. - Location changes do not alter within-distribution distances.

Solution

1. Subtracting \(5\) from every value lowers each distribution's mean by \(5\). 2. Because both means change by the same amount, the difference between the means remains unchanged. 3. A common shift does not change either standard deviation or either shape.

Answer

The difference between the means, both standard deviations, and both shapes remain unchanged; each mean simply decreases by \(5\).
54883512
Distribution A has mean \(0\) and standard deviation \(1\). Distribution B has mean \(2\) and standard deviation \(1\). What comparisons are supported by these summaries, and what cannot be determined about overlap and shape?

Hints

- State the direct center and spread comparisons first. - Ask which features require more than two numerical summaries. - Avoid assuming a normal model unless one is given.

Solution

1. B's mean is \(2\) units greater than A's, so B has the higher center by mean. 2. The equal standard deviations indicate equal variability by that measure. 3. Means and standard deviations alone do not determine distribution shape, outliers, or the exact amount of overlap. 4. Different pairs of distributions can share these summaries while having very different forms.

Answer

B has a mean \(2\) units higher, and the standard deviations are equal. Exact overlap and shape cannot be determined from these summaries alone.
54883612
Distribution A has maximum \(40\). Distribution B has minimum \(45\). What can be concluded about every pair consisting of one observation from A and one from B?

Hints

- Use the most extreme value from each distribution. - Compare A's largest possible observation with B's smallest possible observation. - Nonoverlapping ranges support a statement about every cross-group pair.

Solution

1. Every observation in A is at most \(40\). 2. Every observation in B is at least \(45\). 3. Therefore every observation from B exceeds every observation from A by at least \(45-40=5\).

Answer

Every B observation is greater than every A observation, with a gap of at least \(5\).
53976012
Which distribution has the same center but less variability? Support the answer with two measures of spread.
Figure for problem 539760

Hints

- Compare the medians to verify whether the centers match. - Compare the box widths to evaluate the IQRs. - Compare the full endpoint spans as a second measure of variability.

Solution

1. Both medians are \(12\). 2. Morning has IQR \(17-8=9\) and range \(25-4=21\). 3. Evening has IQR \(14-10=4\) and range \(20-5=15\). 4. Evening has less variability.

Answer

Evening; both medians are \(12\), but its IQR is \(4\) versus \(9\), and its range is \(15\) versus \(21\).
53976112
Compare the shapes of histograms a) and b), and state which has the higher likely mean.
Figure for problem 539761

Hints

- Describe the shape of each histogram first. - Locate where most of each distribution lies on the common horizontal scale. - Use both the shape and the location of the bulk when comparing the likely means.

Solution

1. Histogram a) is approximately symmetric and centered near \(25\). 2. Histogram b) is skewed right, with most observations in lower bins. 3. Despite b)’s right tail, its mass is concentrated below a)’s center, so a) has the higher likely mean.

Answer

a) is approximately symmetric; b) is right-skewed. Histogram a) likely has the higher mean.
53976212
A planner claims Route X is always faster because its median is lower. Evaluate the claim using the boxplots.
Figure for problem 539762

Hints

- Distinguish a statement about typical values from a statement about every observation. - Translate “always faster” into a claim about every pair of observations. - Use overlap and any unusual points to test that universal claim.

Solution

1. Route X has a lower median, \(25\) versus \(27\). 2. However, the distributions overlap substantially, and Route X has a high outlier at \(48\). 3. A lower median does not imply every Route X trip is faster than every Route Y trip.

Answer

The claim is false. Route X has a lower median, but the distributions overlap and Route X includes a high outlier.
53976612
A teacher claims group a) has a higher median and less variability than group b). Is the claim supported?
Figure for problem 539766

Hints

- Compare both the middle positions and the full extent of the observations. - Recover each group’s ordered values or key landmarks from the dotplots. - Evaluate the center claim and variability claim separately before giving one verdict.

Solution

1. Group a) has median \(3.5\), while group b) has median \(3\), so a) has a slightly higher center. 2. Group a) ranges from \(2\) to \(6\); group b) ranges from \(1\) to \(8\). 3. Group b) is visibly more spread out. 4. The claim is supported.

Answer

Yes. Group a) has a slightly higher median and a much smaller spread.
53977312
A student claims, “Because A has \(Q_3=30\) and B has \(Q_1=25\), nearly all values in A are lower than nearly all values in B.” Evaluate the claim using only the boxplots.
Figure for problem 539773

Hints

- Translate the two quartiles into statements about proportions of observations. - Write the middle-half interval for each distribution. - Check whether those intervals overlap and what that overlap prevents you from concluding.

Solution

1. Approximately \(75\%\) of A is at or below \(30\). 2. Approximately \(75\%\) of B is at or above \(25\). 3. Because \(25<30\), those broad portions can overlap from \(25\) to \(30\). 4. The boxplots do not establish the claimed near-complete ordering.

Answer

The claim is not established. Since A’s \(Q_3=30\) exceeds B’s \(Q_1=25\), the distributions can overlap from \(25\) to \(30\).
53977512
The groups have different sample sizes, but the charts use relative frequencies. Compare their centers and shapes without using counts.
Figure for problem 539775

Hints

- Use the relative-frequency scales to compare the panels directly despite different sample sizes. - Locate the main concentration to compare centers. - Inspect which side tapers farther to compare shapes.

Solution

1. a) is symmetric and centered in the middle interval. 2. b) places more relative frequency in higher intervals and has a longer lower-value side. 3. b) has a higher center and is somewhat left-skewed compared with a).

Answer

a) is symmetric with a lower center; b) has a higher center and is somewhat left-skewed.
53977612
Compare the two distributions with respect to center, middle spread, and unusual values.
Figure for problem 539776

Hints

- Compare the median lines for center. - Compare the box widths for middle-half spread. - Inspect points beyond the whiskers separately from the box comparison.

Solution

1. Both medians are \(18\). 2. A has IQR \(21-15=6\), while B has IQR \(22-14=8\). 3. A has two high outliers at \(40\) and \(42\); B has none displayed. 4. B has a slightly wider middle half, while A has more extreme unusual values.

Answer

Same median \(18\); B has larger IQR \(8\) versus \(6\); A has high outliers at \(40\) and \(42\).
53977712
Which distribution is likely to have the larger difference between mean and median? Explain.
Figure for problem 539777

Hints

- Identify the direction and strength of asymmetry in each plot. - Recall that the mean is pulled toward a long tail. - Use the stronger asymmetry to predict which mean–median gap is larger.

Solution

1. A has a pronounced long right whisker, indicating stronger right skew. 2. Right skew tends to pull the mean above the median. 3. B is more nearly symmetric, so its mean and median are likely closer. 4. A likely has the larger difference.

Answer

Distribution A, because its stronger right skew is likely to pull the mean farther above the median.
53977812
Data set A and data set B both range from \(0\) to \(20\). Most A values are near \(10\), while most B values are near \(0\) or \(20\). Which likely has the larger standard deviation? Explain.

Hints

- The range alone does not determine how observations are distributed inside the endpoints. - Standard deviation depends on typical distance from the center, not only on the common range. - Compare whether the mass lies near the center or near the endpoints.

Solution

1. Standard deviation measures typical distance from the mean. 2. A’s values are concentrated near its center. 3. B’s values are concentrated far from the center near the endpoints. 4. B likely has the larger standard deviation.

Answer

Data set B likely has the larger standard deviation.
54881112
The table gives cumulative relative frequencies for two distributions. Lower values are preferable. <table><thead><tr><th>Threshold</th><th>A: proportion at or below</th><th>B: proportion at or below</th></tr></thead><tbody><tr><td>\(10\)</td><td>\(0.10\)</td><td>\(0.20\)</td></tr><tr><td>\(20\)</td><td>\(0.35\)</td><td>\(0.60\)</td></tr><tr><td>\(30\)</td><td>\(0.75\)</td><td>\(0.90\)</td></tr><tr><td>\(40\)</td><td>\(1.00\)</td><td>\(1.00\)</td></tr></tbody></table> Compare the distributions across the listed thresholds. Identify the interval containing each median and state which distribution generally tends to have lower values.
Figure for problem 548811

Hints

- Read each cumulative entry as a proportion not exceeding the threshold. - Locate where each cumulative proportion first reaches one-half. - Compare the two columns at the same threshold rather than comparing row totals.

Solution

1. At every listed threshold below \(40\), B has the larger cumulative relative frequency. 2. A first reaches at least \(0.50\) between \(20\) and \(30\), so A's median lies in \((20,30]\). 3. B first reaches at least \(0.50\) between \(10\) and \(20\), so B's median lies in \((10,20]\). 4. The consistently larger cumulative proportions for B indicate that B generally has lower values across these thresholds.

Answer

A's median lies in \((20,30]\), and B's median lies in \((10,20]\). Distribution B generally tends lower because a larger proportion of B is at or below every listed threshold.
54881212
The back-to-back stem-and-leaf plot compares distributions A and B. A key is \(1\mid2=12\). <table><thead><tr><th>A leaves</th><th>Stem</th><th>B leaves</th></tr></thead><tbody><tr><td>\(8\ 5\ 4\ 2\)</td><td>\(1\)</td><td>\(8\ 9\)</td></tr><tr><td>\(8\ 5\ 3\ 1\)</td><td>\(2\)</td><td>\(2\ 4\ 7\ 9\)</td></tr><tr><td></td><td>\(3\)</td><td>\(1\ 3\)</td></tr></tbody></table> Compare the medians, interquartile ranges, and overall locations of the two distributions.

Hints

- Read the left leaves in the correct order for distribution A. - Reconstruct each ordered list before locating quartiles. - Compare location and spread as separate features.

Solution

1. A is \(12,14,15,18,21,23,25,28\), with median \(\frac{18+21}{2}=19.5\). 2. B is \(18,19,22,24,27,29,31,33\), with median \(\frac{24+27}{2}=25.5\). 3. For A, \(Q_1=\frac{14+15}{2}=14.5\) and \(Q_3=\frac{23+25}{2}=24\), so \(\text{IQR}=9.5\). 4. For B, \(Q_1=\frac{19+22}{2}=20.5\) and \(Q_3=\frac{29+31}{2}=30\), so \(\text{IQR}=9.5\). 5. B is shifted toward higher values while the middle-half spreads are equal.

Answer

A has median \(19.5\) and IQR \(9.5\). B has median \(25.5\) and IQR \(9.5\). B is generally located higher, with the same middle-half variability.
54881312
Data set A is \(0,0,5,10,10\). Data set B is \(0,4,5,6,10\). Compare their centers, ranges, and shapes. Explain why the matching numerical summaries do not make the distributions equivalent.
Figure for problem 548813

Hints

- Compute matching summaries before inspecting arrangement. - Compare where observations are concentrated within the common range. - A few numerical summaries cannot encode the entire distribution.

Solution

1. Both data sets have mean \(5\), median \(5\), and range \(10\). 2. Data set A is concentrated at the two ends with a middle value, giving a strongly two-ended or bimodal pattern. 3. Data set B is concentrated near \(5\), with values tapering toward the endpoints. 4. Equal center and range do not determine the distribution's shape or concentration.

Answer

Both sets have mean \(5\), median \(5\), and range \(10\), but A is concentrated at the ends while B is concentrated near the center.
54881412
Two distributions have the following percentile profiles. <table><thead><tr><th>Percentile</th><th>A</th><th>B</th></tr></thead><tbody><tr><td>\(10\)th</td><td>\(5\)</td><td>\(5\)</td></tr><tr><td>\(25\)th</td><td>\(12\)</td><td>\(8\)</td></tr><tr><td>\(50\)th</td><td>\(20\)</td><td>\(20\)</td></tr><tr><td>\(75\)th</td><td>\(28\)</td><td>\(32\)</td></tr><tr><td>\(90\)th</td><td>\(35\)</td><td>\(35\)</td></tr></tbody></table> Compare the center, the spread of the middle \(50\%\), and the spread in the two outer percentile bands from the \(10\)th to \(25\)th and from the \(75\)th to \(90\)th percentiles.

Hints

- Use matching percentile positions to define comparable intervals. - Separate the middle-half width from the two outer-band widths. - Equal endpoints at the \(10\)th and \(90\)th percentiles do not force equal internal spread.

Solution

1. Both medians are \(20\). 2. A has \(\text{IQR}=28-12=16\), while B has \(\text{IQR}=32-8=24\). B is more spread out in the middle \(50\%\). 3. For A, each outer percentile band has width \(12-5=7\) and \(35-28=7\). 4. For B, each outer percentile band has width \(8-5=3\) and \(35-32=3\). 5. A has more spread in the outer bands, while B has more spread in the middle half.

Answer

Both medians are \(20\). B has the larger IQR, \(24\) versus \(16\). A has wider outer percentile bands, \(7\) units each versus \(3\) units each for B.
54881812
The ordered values of distribution A are \(10,20,30,40,50\), and the ordered values of distribution B are \(12,19,35,44,60\). Pair observations with the same rank and calculate \(B-A\) at each rank. Is B a constant translation of A? Find the median rank-by-rank difference and describe how the separation changes across the distributions.

Hints

- Match smallest with smallest, second-smallest with second-smallest, and so on. - A pure translation would produce the same difference at every rank. - Summarize the set of rank differences after calculating all of them.

Solution

1. The rank-by-rank differences are \(12-10=2\), \(19-20=-1\), \(35-30=5\), \(44-40=4\), and \(60-50=10\). 2. The differences are not constant, so B is not a translation of A. 3. In order, the differences are \(-1,2,4,5,10\), so their median is \(4\). 4. The upper end of B is farther above A than the lower end, while B's second ordered value is slightly below A's.

Answer

The rank differences are \(2,-1,5,4,10\), with median \(4\). B is not a constant translation of A because the differences vary, especially at the upper end.
54881912
Distribution A has \(5\)th percentile \(40\), median \(50\), and \(95\)th percentile \(60\). Distribution B has \(5\)th percentile \(35\), median \(50\), and \(95\)th percentile \(65\). Compare the typical values and the widths of the central \(90\%\) intervals. State one feature that cannot be determined from these three percentiles alone.

Hints

- Compare matching percentile locations across the two distributions. - Subtract the lower percentile from the upper percentile to measure the stated central interval. - Distinguish information supplied by selected percentiles from information requiring the full distribution.

Solution

1. Both medians are \(50\), so their typical values by median are equal. 2. A's central \(90\%\) interval has width \(60-40=20\). 3. B's central \(90\%\) interval has width \(65-35=30\). 4. B has greater spread across its central \(90\%\). 5. The three percentiles do not determine features such as exact shape, standard deviation, sample size, or the most extreme values.

Answer

Both medians are \(50\). The central \(90\%\) widths are \(20\) for A and \(30\) for B, so B is more spread out over that percentile interval. Exact shape or standard deviation cannot be determined.
54882112
Distribution A records the values \(0,1,2,3,4,5\). An instrument cannot report values below \(2\), so distribution B records the same observations as \(2,2,2,3,4,5\). Compare the means, medians, ranges, and visible shapes. Explain how the detection limit changes the comparison.
Figure for problem 548821

Hints

- Treat B as a changed data set rather than as missing observations. - Compare resistant and nonresistant summaries separately. - Look for repeated values created by the reporting rule.

Solution

1. A has mean \(\frac{15}{6}=2.5\), median \(\frac{2+3}{2}=2.5\), and range \(5\). 2. B has mean \(\frac{18}{6}=3\), median \(\frac{2+3}{2}=2.5\), and range \(5-2=3\). 3. A is evenly spaced across \(0\) through \(5\), while B has a pileup of three observations at the detection limit \(2\). 4. The detection limit raises the mean, leaves the median unchanged here, reduces the recorded range, and creates an artificial cluster at \(2\).

Answer

A: mean \(2.5\), median \(2.5\), range \(5\). B: mean \(3\), median \(2.5\), range \(3\). The detection limit raises low observations to \(2\), producing a spike and changing the recorded center and spread.
54882312
Distribution A is \(48,50,52,54\), and distribution B is \(52,54,56,58\). A target value is \(55\). Compare the two distributions' ordinary variability and their average absolute distance from the target. Which distribution is more tightly centered on the target?
Figure for problem 548823

Hints

- First compare the spacing among values within each distribution. - Then measure every observation's distance from the common target. - Internal consistency and closeness to an external target are different comparisons.

Solution

1. B is obtained by adding \(4\) to every value in A, so both distributions have the same range and standard deviation. 2. A's absolute distances from \(55\) are \(7,5,3,1\), with mean distance \(\frac{16}{4}=4\). 3. B's absolute distances from \(55\) are \(3,1,1,3\), with mean distance \(\frac{8}{4}=2\). 4. The distributions have equal internal spread, but B is more tightly centered on the external target.

Answer

A and B have equal ordinary variability. Their average absolute distances from \(55\) are \(4\) and \(2\), respectively, so B is more tightly centered on the target.
54882412
Distribution A is approximately symmetric and is summarized by mean \(40\) and standard deviation \(6\). Distribution B is right-skewed and is summarized by median \(40\) and interquartile range \(9\). Can you determine which distribution is more variable from these summaries alone? Explain.

Hints

- Identify which spread statistic is used for each distribution. - Ask whether the two numbers measure the same feature in the same way. - A valid comparison requires a common metric or more information.

Solution

1. A's spread is reported with standard deviation, while B's spread is reported with interquartile range. 2. These measures describe different aspects of variability and have different numerical scales. 3. Without a common spread measure or additional distribution information, \(6\) and \(9\) cannot be compared directly to determine which distribution is more variable.

Answer

No. Standard deviation \(6\) and interquartile range \(9\) are not directly comparable measures of spread.
54882512
In distribution A, \(Q_3=70\). In distribution B, the median is \(70\). Compare the proportions at or above \(70\) that these summary positions suggest. Explain why ties at \(70\) prevent an exact comparison.

Hints

- Translate each summary position into an approximate percentile statement first. - Distinguish observations strictly above \(70\) from observations equal to \(70\). - Test whether repeated values at the threshold could change the at-or-above proportions.

Solution

1. As a quartile-position interpretation, \(Q_3=70\) suggests that about the upper \(25\%\) of A is at or above \(70\). 2. A median of \(70\) suggests that about \(50\%\) of B is at or above \(70\). 3. These percentages are exact only under additional assumptions about the data and the percentile convention. Repeated observations at \(70\) can make the proportion at or above \(70\) larger. 4. Therefore the summary positions suggest a larger share for B when there are no relevant ties, but the exact shares—and even which at-or-above share is larger—cannot be determined from the two summary values alone.

Answer

Without relevant ties, the quartile positions suggest about \(25\%\) of A and \(50\%\) of B at or above \(70\). With ties at \(70\), the exact proportions may be larger, so the two shares cannot be ordered with certainty from the given information alone.
54882612
Two distributions with positive standard deviations are each standardized by subtracting their own mean and dividing by their own standard deviation. What center and spread will the standardized distributions share? Which features can still differ?

Hints

- Recall what the two parts of standardization do to location and scale. - A linear transformation does not rearrange observations. - Equal center and spread do not determine shape.

Solution

1. Standardization makes each transformed distribution have mean \(0\). 2. It makes each transformed distribution have standard deviation \(1\). 3. The transformation is linear with a positive scale factor, so each distribution keeps its own shape, skewness, clusters, gaps, and outlier pattern in standardized units. 4. Therefore two standardized distributions can share center and spread while still having different shapes.

Answer

Both standardized distributions have mean \(0\) and standard deviation \(1\), but their shapes and unusual features can still differ.
54882812
The two boxplots have the same quartiles and median but different lower whiskers. Compare their center, middle spread, and likely shape.
Figure for problem 548828

Hints

- Compare the common box before inspecting the whiskers. - Measure tail lengths on both sides of each box. - Equal center and middle spread do not imply equal shape.

Solution

1. Both distributions have median \(15\) and interquartile range \(20-10=10\). 2. A has whiskers of lengths \(10-0=10\) and \(25-20=5\), so it is only mildly longer on the left. 3. B has a much longer lower whisker, \(10-(-20)=30\), and upper whisker length \(5\). 4. B's boxplot suggests much stronger left skew, even though the two distributions share the same median and interquartile range.

Answer

Both medians are \(15\), and both interquartile ranges are \(10\). B has much greater lower-tail spread, and its boxplot suggests stronger left skew.
54882912
Distribution A has interquartile interval \([10,30]\), and distribution B has interquartile interval \([22,40]\). Find the numerical overlap of the two middle-half intervals. Express the overlap length as a percentage of each distribution's interquartile range. Explain why these percentages are not percentages of observations that the two distributions share.

Hints

- Find the intersection of the two quartile intervals. - Compare the intersection length with each full box length separately. - Distinguish overlap on a number line from overlap of individual observations.

Solution

1. The intervals overlap from \(22\) to \(30\), so the overlap interval is \([22,30]\) with length \(8\). 2. A's interquartile range is \(30-10=20\), so the overlap is \(\frac{8}{20}=40\%\) of A's IQR length. 3. B's interquartile range is \(40-22=18\), so the overlap is \(\frac{8}{18}\approx44.44\%\) of B's IQR length. 4. These percentages compare numerical interval lengths. They do not identify which observations are the same or how many observations from either distribution lie in the overlap beyond the quartile constraints.

Answer

The overlap is \([22,30]\), with length \(8\). This is \(40\%\) of A's IQR length and approximately \(44.44\%\) of B's IQR length. These are geometric interval comparisons, not shared-observation percentages.
54883012
Distribution A has minimum \(0\) and maximum \(10\). Distribution B has minimum \(8\) and maximum \(20\). a) Identify the interval in which observations from the two distributions could overlap. b) Is every observation from B necessarily greater than every observation from A? c) State the guaranteed comparisons for any A-value below \(8\) and any B-value above \(10\).

Hints

- Treat each minimum-to-maximum span as an interval of possible values. - Find the intersection of the two intervals. - Use each distribution's extreme value to make guaranteed comparisons outside the overlap.

Solution

1. A occupies values from \(0\) through \(10\), and B occupies values from \(8\) through \(20\), so possible overlap is from \(8\) through \(10\). 2. It is not guaranteed that every B-value exceeds every A-value because both distributions may contain values in the overlap interval. 3. Any A-value below \(8\) is less than every B-value, since B's minimum is \(8\). 4. Any B-value above \(10\) is greater than every A-value, since A's maximum is \(10\).

Answer

a) Possible overlap: \([8,10]\) b) No. c) Every A-value below \(8\) is below all B-values, and every B-value above \(10\) is above all A-values.
54883212
A variable-width boxplot is explicitly designed so that each box's vertical width is proportional to the square root of its sample size. Box A is \(2\,\text{cm}\) wide and represents \(64\) observations. Box B is \(3\,\text{cm}\) wide. Find the sample size represented by box B.

Hints

- Use the graph's stated width rule rather than the ordinary boxplot convention. - Compare widths with square roots of sample sizes. - Undo the square root only after finding the proportional value.

Solution

1. Since width is proportional to \(\sqrt{n}\), the width ratio is \(\frac{3}{2}=\frac{\sqrt{n_B}}{\sqrt{64}}\). 2. Because \(\sqrt{64}=8\), \(\sqrt{n_B}=\frac{3}{2}\cdot8=12\). 3. Therefore, \(n_B=12^2=144\).

Answer

Box B represents \(144\) observations.
54883312
Distribution A is measured in degrees Fahrenheit and has mean \(68\) and standard deviation \(9\). Distribution B is measured in degrees Celsius and has mean \(22\) and standard deviation \(4\). Convert B's summaries to degrees Fahrenheit and compare the distributions' centers and spreads.

Hints

- Convert both summaries to one common unit before comparing. - The additive part of a linear conversion affects center but not spread. - Compare center and variability separately.

Solution

1. B's mean in degrees Fahrenheit is \(1.8\cdot22+32=71.6\). 2. B's standard deviation is multiplied only by \(1.8\), giving \(1.8\cdot4=7.2\). 3. B has the higher mean by \(71.6-68=3.6\) degrees Fahrenheit. 4. B has the smaller standard deviation, \(7.2\) versus \(9\), so it is less variable.

Answer

B has mean \(71.6^\circ\text{F}\) and standard deviation \(7.2^\circ\text{F}\). B has the higher center and smaller spread.
54883412
Distribution A has mean \(5\), standard deviation \(3\), and is right-skewed. Distribution B is formed by replacing every value \(x\) with \(-2x\). Compare B's mean, standard deviation, and shape with A's.

Hints

- Apply the signed scale factor to the center. - Use the magnitude of the factor for spread. - A negative factor reverses left and right.

Solution

1. B's mean is \(-2\cdot5=-10\). 2. Multiplying by \(-2\) multiplies distances by \(2\), so B's standard deviation is \(2\cdot3=6\). 3. The negative factor reflects the distribution, so A's right skew becomes left skew in B.

Answer

B has mean \(-10\), standard deviation \(6\), and is left-skewed.
54883912
Distribution A is \(10,20,30,40,50\), and distribution B is \(100,101,102,103,1000\). Within each distribution, replace each observation by its percentile rank using \(100\cdot\frac{\text{rank}}{5}\). Compare the two percentile-rank distributions and explain what information this transformation preserves and loses.

Hints

- Assign ranks separately within each distribution. - Apply the stated rank-to-percentile rule to each ordered position. - Compare what rank records with what raw numerical differences record.

Solution

1. Each distribution has five distinct ordered observations with ranks \(1,2,3,4,5\). 2. Both percentile-rank distributions are \(20,40,60,80,100\). 3. The transformation preserves the within-distribution order and relative rank of each observation. 4. It removes the original units and numerical distances. In particular, B's extreme gap from \(103\) to \(1000\) is no longer visible.

Answer

Both transformed distributions are \(20,40,60,80,100\). Percentile ranks preserve order and relative standing but discard original units, spacing, and magnitude of gaps.
54881612
Distribution A is \(-3,-2,-1,0,1,2,3\). Distribution B is formed by squaring every value in A. Compare the medians, interquartile ranges, ranges, and shapes of A and B. Explain why this transformation behaves differently from a positive linear rescaling.
Figure for problem 548816

Hints

- Transform each observation before calculating B's summaries. - Reorder the transformed values because the rule is not increasing on all real numbers. - Compare whether distinct original values become identical after transformation.

Solution

1. A is symmetric with median \(0\), \(Q_1=-2\), \(Q_3=2\), \(\text{IQR}=4\), and range \(6\). 2. B in order is \(0,1,1,4,4,9,9\). 3. B has median \(4\), \(Q_1=1\), \(Q_3=9\), \(\text{IQR}=8\), and range \(9\). 4. B is concentrated near the lower end with a right tail toward \(9\), unlike A's symmetry. 5. Squaring is not one-to-one on negative and positive values and is not increasing over all real numbers, so it can merge observations and change order and shape.

Answer

A: median \(0\), IQR \(4\), range \(6\), symmetric. B: median \(4\), IQR \(8\), range \(9\), right-skewed. Squaring merges opposite values and changes order, so the comparison is not a simple rescaling.
54881712
Distribution A is \(10,12,14,16,18,100\). Distribution B is formed by replacing every value above \(18\) with \(18\), giving \(10,12,14,16,18,18\). Compare the means, medians, ranges, and interquartile ranges. Which summaries are unchanged by this upper-end capping?
Figure for problem 548817

Hints

- Compare the ordered positions that determine the median and quartiles. - Identify which summaries use the magnitude of the largest value directly. - Recalculate both resistant and nonresistant measures.

Solution

1. A has mean \(\frac{170}{6}=\frac{85}{3}\approx28.33\), while B has mean \(\frac{88}{6}=\frac{44}{3}\approx14.67\). 2. Both medians are \(\frac{14+16}{2}=15\). 3. A's range is \(100-10=90\), while B's range is \(18-10=8\). 4. For both distributions, \(Q_1=12\) and \(Q_3=18\), so both interquartile ranges are \(6\). 5. Capping changes the mean and range but leaves the median and IQR unchanged in this case.

Answer

A: mean \(\frac{85}{3}\approx28.33\), median \(15\), range \(90\), IQR \(6\). B: mean \(\frac{44}{3}\approx14.67\), median \(15\), range \(8\), IQR \(6\). The median and IQR are unchanged.
54882712
Distribution A is \(0,0,0,3,3\), and distribution B is \(0,0,1,1,4\). Show that the two distributions have the same mean and sample standard deviation but different interquartile ranges. Explain what this demonstrates about comparing spread.
Figure for problem 548827

Hints

- Compare totals and sums of squares before expanding deviation calculations. - Compute quartiles from ordered positions separately for each distribution. - Different spread measures respond to different parts of a distribution.

Solution

1. Both distributions have sum \(6\), so both means are \(\frac{6}{5}=1.2\). 2. Both have sum of squares \(18\). Thus each has sum of squared deviations \(18-5(1.2)^2=10.8\). 3. Both sample variances are \(\frac{10.8}{4}=2.7\), so both sample standard deviations are \(\sqrt{2.7}\approx1.643\). 4. For A, \(Q_1=0\) and \(Q_3=3\), so \(\text{IQR}=3\). 5. For B, \(Q_1=0\) and \(Q_3=\frac{1+4}{2}=2.5\), so \(\text{IQR}=2.5\). 6. Equal standard deviations do not force equal interquartile ranges because the measures summarize spread differently.

Answer

Both distributions have mean \(1.2\) and sample standard deviation \(\sqrt{2.7}\approx1.643\). A has IQR \(3\), while B has IQR \(2.5\).
54883712
Distribution A is \(1,4,5\), and distribution B is \(2,3,6\). One observation is selected at random from each distribution. Find the proportion of the \(9\) possible A–B pairs for which the B observation is greater than the A observation. Compare this result with the two medians.
Figure for problem 548837

Hints

- Treat each cross-group pairing as one equally likely outcome. - Count favorable comparisons separately for each possible B-value. - Compare the pairwise result with, rather than replacing it by, the median comparison.

Solution

1. A's median is \(4\), and B's median is \(3\). 2. If B contributes \(2\), it exceeds only A's \(1\), giving \(1\) favorable pair. 3. If B contributes \(3\), it again exceeds only A's \(1\), giving \(1\) favorable pair. 4. If B contributes \(6\), it exceeds all three A-values, giving \(3\) favorable pairs. 5. Therefore B exceeds A in \(1+1+3=5\) of the \(9\) pairs, or \(\frac{5}{9}\). 6. Although B's median is lower, a random B observation exceeds a random A observation more than half the time in these empirical distributions.

Answer

The proportion is \(\frac{5}{9}\). A has median \(4\), while B has median \(3\), showing that a median comparison and an all-pairs comparison can point in different directions.
54883812
Distribution A is \(14,16,24,26,34,36\). Distribution B is formed by rounding every A-value to the nearest multiple of \(10\), giving \(10,20,20,30,30,40\). Compare the means, medians, ranges, interquartile ranges, and visible effects of rounding.
Figure for problem 548838

Hints

- Recalculate summaries from the rounded values rather than assuming every measure shrinks. - Compare the endpoints and quartile positions separately. - Look for repeated values introduced by rounding.

Solution

1. Both distributions have sum \(150\), so both means are \(25\). 2. A's median is \(\frac{24+26}{2}=25\), and B's median is \(\frac{20+30}{2}=25\). 3. A's range is \(36-14=22\), while B's range is \(40-10=30\). 4. For A, \(Q_1=16\) and \(Q_3=34\), so \(\text{IQR}=18\). 5. For B, \(Q_1=20\) and \(Q_3=30\), so \(\text{IQR}=10\). 6. Rounding creates ties, leaves both centers unchanged here, increases the range, and decreases the IQR.

Answer

Both means and medians are \(25\). A has range \(22\) and IQR \(18\); B has range \(30\) and IQR \(10\). Rounding creates ties and can affect different spread measures in different directions.

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