- 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.

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.

Group A has the higher center; Group B has less variability and is more consistent.

- Count how many observations are in each dotplot. - Find the middle dot after ordering the values in each group. - Compare those two middle values on the shared scale.

1. Each dotplot has five values. The median is the third value in order. 2. Class a)'s third value is \(3\); class b)'s is \(7\). 3. Class b) has the higher median.

Class b): median a) is \(3\) books and median b) is \(7\) books.

- Which two boundaries enclose the central half of the data? - Both plots share a horizontal scale; compare the lengths of their central boxes.

1. The walk-in box extends from \(5\) to \(15\) minutes, giving an IQR of \(10\) minutes. 2. The drive-through box extends from \(8\) to \(12\) minutes, giving an IQR of \(4\) minutes. 3. Walk-in waits have the wider middle half.

Walk-in pickup, with an IQR of \(10\,\text{min}\), compared with \(4\,\text{min}\) for drive-through pickup.

- Find the farthest-left and farthest-right nonempty bars in each graph. - Compare those locations, not the number or heights of the bars. - Does a bin show the exact positions of the measurements inside it?

1. In a), all six observations are in the interval from \(20\) up to \(30\). 2. In b), the first occupied bin is \([0,10)\) and the last is \([40,50)\). 3. Thus b) covers a wider horizontal stretch. Neither exact range is known because the individual values within the bins are hidden.

Distribution b) covers a wider horizontal stretch. In a), both the first and last occupied bins are \([20,30)\). In b), they are \([0,10)\) and \([40,50)\). Exact ranges cannot be read from grouped data.

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

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.

B has the higher median, \(28\) versus \(24\), and the larger IQR, \(16\) versus \(12\).

- Use the shared horizontal scale to compare the endpoints. - Compare how tightly the dots cluster around the common median. - Describe shape and spread separately.

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.

Both are approximately symmetric with median \(5\); a) has greater spread, while b) is more concentrated.

- 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.

1. A has IQR \(20-10=10\) and range \(40-0=40\). 2. B has IQR \(25-10=15\) and range \(30-5=25\). 3. B has the larger IQR, while A has the larger range.

Larger IQR: B, \(15\). Larger range: A, \(40\).

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

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.

A is likely left-skewed; B is likely right-skewed.

- 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.

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 \(40\).

A’s \(Q_3=40\) exceeds B’s \(Q_1=35\), so the middle halves overlap on \([35,40]\).

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

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.

No. Equal means do not imply equal shape or variability.

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

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.

Machine B has a median \(1\,\text{mL}\) higher; both machines have IQR \(6\,\text{mL}\).

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

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.

a) is unimodal with one cluster; b) is bimodal with two clusters separated by a large gap.

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

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.

The means and standard deviations will show the greatest change.

- 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.

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.

B is shifted \(10\) units higher than A, with the same IQR and range.

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

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.

B has the same shape and spread as A, while every measure of center and location is \(7\) units larger.

- 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.

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 \(2\cdot10=20\).

B has median \(24\), interquartile range \(6\), and range \(20\); each is twice the corresponding summary for A.

- 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.

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.

Student B has the higher raw score, but Student A has the higher relative standing within the corresponding distribution.

- 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.

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.

The difference between the means, both standard deviations, and both shapes remain unchanged; each mean simply decreases by \(5\).

- 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.

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.

B has a mean \(2\) units higher, and the standard deviations are equal. Exact overlap and shape cannot be determined from these summaries alone.

- 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.

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 \(45-40=5\).

Every B observation is greater than every A observation, with a gap of at least \(5\).

- Read each median from the ordered dots before comparing spread. - Compare both the overall horizontal span and the concentration around the center. - Look for asymmetry or isolated values that affect consistency.

1. Class A is approximately symmetric and concentrated around \(10\) minutes. Its median is \(10\), and its times range from \(8\) to \(12\) minutes. 2. Class B also has median \(10\), but it is right-skewed, with values extending from \(7\) to \(17\) minutes and a separated high value at \(17\). 3. Because the centers by median are the same, the main difference is spread and shape. 4. Class A has a much narrower range and stronger concentration near the center, so Class A is more consistent.

Both classes have median \(10\) minutes. Class A is approximately symmetric and tightly clustered from \(8\) to \(12\) minutes. Class B is right-skewed, spreads from \(7\) to \(17\) minutes, and has a separated high value at \(17\). Class A is more consistent.

- Compare the totals and the number of observations before finding the means. - The range runs from the minimum to the maximum in each group. - Equal means do not force the distributions to have the same spread.

1. Each set totals \(100\) minutes for five deliveries, so both means are \(20\) minutes. 2. The Downtown range is \(22-18=4\) minutes; the Riverside range is \(30-10=20\) minutes. 3. Riverside delivery times vary more despite having the same mean.

Both means are \(20\) minutes. Downtown range: \(4\) minutes. Riverside range: \(20\) minutes. Riverside delivery times vary more.

- 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.

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.

Evening; both medians are \(12\), but its IQR is \(4\) versus \(9\), and its range is \(15\) versus \(21\).

- Describe each histogram's shape before comparing centers. - Treat each bin midpoint as the representative value for observations in that bin. - Weight each midpoint by its displayed frequency.

1. Histogram a) is approximately symmetric, while histogram b) is skewed right. 2. For a), the midpoint estimate is \(\frac{2(5)+6(15)+12(25)+6(35)+2(45)}{28}=25\). 3. For b), the midpoint estimate is \(\frac{10(5)+8(15)+5(25)+3(35)+2(45)}{28}=17.5\). 4. Histogram a) therefore has the higher estimated mean by \(25-17.5=7.5\) units.

a) is approximately symmetric; b) is right-skewed. The midpoint mean estimates are \(25\) for a) and \(17.5\) for b), so a) has the higher estimated mean by \(7.5\) units.

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

1. Route X has a lower median, \(30\) minutes versus \(35\) minutes. 2. However, the distributions overlap substantially, and Route X has a high outlier at \(50\) minutes. 3. A lower median does not imply every Route X trip is faster than every Route Y trip.

The claim is false. Route X has a lower median, but the distributions overlap and Route X includes a high outlier at \(50\) minutes.

- Read the ordered observations from each dotplot to locate the middle positions. - Compare the full horizontal extent of the two distributions. - Evaluate the center claim and variability claim separately.

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 \(6\), while group b) ranges from \(1\) to \(8\). 3. Group b) is visibly more spread out, so the variability part of the claim is also supported.

Yes. Group a) has a slightly higher median and a much smaller spread.

- Read the upper quartile of A and lower quartile of B from the plots. - Translate those quartiles into statements about broad portions of each distribution. - Check whether the corresponding value intervals overlap.

1. From the boxplots, A has \(Q_3=30\) and B has \(Q_1=25\). 2. Approximately \(75\%\) of A is at or below \(30\), while approximately \(75\%\) of B is at or above \(25\). 3. Because \(25<30\), these broad portions can overlap from \(25\) to \(30\). 4. The boxplots therefore do not establish the claimed near-complete ordering.

The claim is not established. A's \(Q_3=30\) exceeds B's \(Q_1=25\), so the distributions can overlap from \(25\) to \(30\).

- 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.

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).

a) is symmetric with a lower center; b) has a higher center and is somewhat left-skewed.

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

1. Both medians are \(20\). 2. A has IQR \(25-15=10\), while B has IQR \(30-10=20\). 3. A has two high outliers at \(40\) and \(45\); B has none displayed. 4. B has a wider middle half, while A has more extreme unusual values.

Same median \(20\); B has larger IQR \(20\) versus \(10\); A has high outliers at \(40\) and \(45\).

- 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.

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.

Distribution A, because its stronger right skew is likely to pull the mean farther above the median.

- Equal ranges do not imply equal standard deviations. - Compare how far most observations lie from the shared mean. - Standard deviation increases when more observations are concentrated farther from the center.

1. Both distributions have the same center and the same range, so those facts alone do not decide the comparison. 2. In A, four of the six observations lie within \(1\) unit of the mean, while only the two endpoints are \(10\) units away. 3. In B, every observation is \(10\) units from the mean. 4. Because B has much larger typical squared distances from the common mean, B has the larger standard deviation.

Distribution B has the larger standard deviation because all of its observations lie far from the common mean, whereas most of A is concentrated near the mean.

- Read an ogive height as the cumulative proportion at or below that x-value. - Locate where each curve first crosses the \(0.50\) level. - Compare the two curves at the same threshold.

1. Reading the ogive, B has the larger cumulative relative frequency at each listed threshold below \(40\). 2. A is at \(0.35\) by \(20\) and \(0.75\) by \(30\), so A's median lies in \((20,30]\). 3. B is at \(0.20\) by \(10\) and \(0.60\) by \(20\), 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.

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 each listed threshold.

- 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.

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.

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.

- Read each set of observations from its dotplot before computing summaries. - Compare where observations are concentrated within the common range. - A few numerical summaries do not encode the full shape.

1. Reading the dotplots, A is \(0,0,5,10,10\) and B is \(0,4,5,6,10\). 2. Both distributions have mean \(5\), median \(5\), and range \(10\). 3. A is concentrated at the two ends with a middle value, while B is concentrated near \(5\) and tapers toward the endpoints. 4. Equal center and range do not determine a distribution's shape or concentration.

Both distributions have mean \(5\), median \(5\), and range \(10\), but A is concentrated at the ends while B is concentrated near the center.

- 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.

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.

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.

- Read each plotted dot as one wait, including the stack at the largest recorded time. - How does correcting one wait change the ordered list? - Check the measures based on middle positions separately from those involving all values or the extremes.

1. The recorded waits are \(10,12,14,16,18,18\). The actual waits are \(10,12,14,16,18,100\). 2. The recorded mean is \(\frac{88}{6}\approx14.67\) minutes; the actual mean is \(\frac{170}{6}\approx28.33\) minutes. 3. Both medians are \(\frac{14+16}{2}=15\). 4. The recorded range is \(18-10=8\) minutes; the actual range is \(100-10=90\) minutes. 5. For both, \(Q_1=12\) and \(Q_3=18\), so \(\text{IQR}=6\). 6. The median and IQR stay the same; the mean and range change.

Recorded: mean \(14.67\) minutes, median \(15\) minutes, range \(8\) minutes, IQR \(6\) minutes. Actual: mean \(28.33\) minutes, median \(15\) minutes, range \(90\) minutes, IQR \(6\) minutes. Unchanged: median and IQR.

- 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.

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.

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.

- 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.

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.

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.

- Apply the stated recording rule to the smallest actual observations. - Read both dotplots as numerical lists before computing the summaries. - Distinguish a stack created by recording from a real cluster of equal values.

1. A has values \(0,1,2,3,4,5\), with mean \(2.5\), median \(2.5\), and range \(5\). 2. B has recorded values \(2,2,2,3,4,5\), with mean \(3\), median \(2.5\), and range \(3\). 3. A has one observation at each value from \(0\) through \(5\), while B has three recorded at \(2\). 4. The rule raises the recorded mean, leaves the median unchanged here, reduces the range, and creates an artificial stack at \(2\).

A: mean \(2.5\,\text{ppm}\), median \(2.5\,\text{ppm}\), range \(5\,\text{ppm}\), evenly spaced. B: mean \(3\,\text{ppm}\), median \(2.5\,\text{ppm}\), range \(3\,\text{ppm}\), with a stack at \(2\,\text{ppm}\). Recording small values as \(2\) distorts the mean, range, and visible shape.

- Read the observations from each plot and compare their internal spacing. - Then measure each observation's distance from the dashed target line. - Internal variability and closeness to an external target are different comparisons.

1. Reading the plots, A is \(48,50,52,54\) and B is \(52,54,56,58\). B is obtained by adding \(4\) to every value in A, so the two distributions have the same range and standard deviation. 2. A's absolute distances from \(55\) are \(7,5,3,1\), with mean distance \(4\). 3. B's absolute distances from \(55\) are \(3,1,1,3\), with mean distance \(2\). 4. The distributions have equal internal spread, but B is more tightly centered on the external target.

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.

- 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.

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.

No. Standard deviation \(6\) and interquartile range \(9\) are not directly comparable measures of spread.

- 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.

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.

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.

- 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.

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.

Both standardized distributions have mean \(0\) and standard deviation \(1\), but their shapes and unusual features can still differ.

- 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.

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.

Both medians are \(15\), and both interquartile ranges are \(10\). B has much greater lower-tail spread, and its boxplot suggests stronger left skew.

- Which values fall inside both middle-half intervals? - Use each service's own interquartile range when finding a percentage. - What extra information about individual delivery times would be needed to count matching times?

1. The common interval is \([22, 30]\), with length \(30-22=8\) minutes. 2. Cedar's IQR length is \(30-10=20\) minutes, so the overlap is \(\frac{8}{20}\cdot100\%=40.00\%\) of that length. 3. Harbor's IQR length is \(40-22=18\) minutes, so the overlap is \(\frac{8}{18}\cdot100\%\approx44.44\%\) of that length. 4. These percentages compare lengths of intervals, not counts of deliveries. The intervals alone cannot tell how many deliveries took identical amounts of time.

Overlap: \([22, 30]\), length \(8\) minutes. This is \(40.00\%\) of Cedar's IQR length and \(44.44\%\) of Harbor's IQR length. The percentages compare interval lengths, not the share of deliveries with identical times.

- 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.

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 \(10\).

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.

- 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.

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.

B has mean \(71.6^\circ\text{F}\) and standard deviation \(7.2^\circ\text{F}\). B has the higher center and smaller spread.

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

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.

B has mean \(-10\), standard deviation \(6\), and is left-skewed.

- 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.

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.

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.

- Compare the two shapes before comparing centers. - Use the common horizontal scale to judge where most trips fall and how far each distribution extends. - Separate “typically faster” from “more consistent.”

1. Route A is approximately symmetric and unimodal, with its highest frequency in the \([20,25)\)-minute interval and most trips between about \(15\) and \(30\) minutes. 2. Route B is right-skewed: most trips are in the lower intervals, with a tail extending toward \(40\) minutes. 3. Route B has the lower typical travel time because much of its distribution lies below Route A's center. 4. Route A occupies a narrower effective span, about \(10\) to \(35\) minutes, while Route B extends from about \(5\) to \(40\) minutes and has the longer right tail. 5. Neither histogram shows an interior gap. Route A therefore appears more predictable, while Route B is typically faster but more variable.

Route A is approximately symmetric and unimodal, centered around the low \(20\)s, with a narrower spread. Route B is right-skewed, centered lower, and extends across a wider range with a long high-time tail. Route B appears to have the lower typical travel time, while Route A appears more predictable.

- Calculate the average of each list, then locate the two middle values. - Do the mean and median rank the shuttles the same way? - Which single value has an unusually large effect on one mean?

1. East mean is \(\frac{102}{6}=17\) minutes and median is \(\frac{16+18}{2}=17\) minutes. 2. West mean is \(\frac{121}{6}\approx20.17\) minutes, while median is \(\frac{15+16}{2}=15.5\) minutes. 3. West has the shorter median trip, but its unusually long \(44\)-minute trip raises its mean above East's.

East: mean \(17\) minutes, median \(17\) minutes. West: mean about \(20.17\) minutes, median \(15.5\) minutes. West has the shorter median; its \(44\)-minute trip makes its mean higher.

- Read A from the first dotplot and check how squaring changes each position. - Reorder the transformed values before finding quartiles. - Compare whether different A-values map to the same B-value.

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

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 it is not a simple positive rescaling.

- Read and order the values from each plot. - Compare the totals and squared totals before doing separate variance calculations. - For quartiles, find the medians of the lower and upper halves separately.

1. Read A as \(0,0,0,3,3\) and B as \(0,0,1,1,4\). 2. Both lists have sum \(6\) and sum of squares \(18\), so the mean is \(1.2\) and the sample variance is \(\frac{18-5(1.2)^2}{4}=2.7\). 3. Both sample standard deviations are \(\sqrt{2.7}\approx1.643\). 4. A has \(Q_1=0\), \(Q_3=3\), and IQR \(3\). B has \(Q_1=0\), \(Q_3=2.5\), and IQR \(2.5\). 5. Equal standard deviations do not guarantee equal IQRs.

Both means are \(1.2\); both sample standard deviations are approximately \(1.643\). IQR A is \(3\); IQR B is \(2.5\). Different measures of spread can differ even when another agrees.

- Read the three values in each distribution from the dotplots. - Treat each cross-group pairing as one equally likely outcome. - Compare the pairwise result with, rather than replacing it by, the median comparison.

1. From the dotplots, A is \(1,4,5\) with median \(4\), and B is \(2,3,6\) with median \(3\). 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. Thus B exceeds A in \(5\) of the \(9\) pairs, or \(\frac{5}{9}\), even though B's median is lower.

The proportion is \(\frac{5}{9}\). A has median \(4\), while B has median \(3\), so the median comparison and the all-pairs comparison point in different directions.

- Read both displayed data sets before calculating summaries. - Recalculate the rounded distribution instead of assuming every measure gets smaller. - Compare endpoints and quartile positions separately.

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

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 change different spread measures in different directions.