Free math worksheets

- Order each sample before locating its center. - Use the same center for both groups. - Translate the numerical difference into the context.

1. Ordered P is \(18,19,20,21,22,23,24,25,26\), so its median is \(22\). 2. Ordered Q is \(27,28,29,30,31,32,33,34,35\), so its median is \(31\). 3. The median difference is \(31-22=9\) minutes. 4. The samples suggest commutes are typically longer in Neighborhood Q.

Median P: \(22\) minutes Median Q: \(31\) minutes The samples suggest Neighborhood Q has a typical commute about \(9\) minutes longer.

- Compare the measures of center first. - Then compare the measures of variability separately. - A complete inference should describe both typical value and consistency.

1. The sample mean difference is \(34.6-30.2=4.4\,\text{cm}\). 2. The MADs are similar, \(2.1\,\text{cm}\) and \(2.4\,\text{cm}\). 3. The samples suggest Variety A is generally taller, while the two varieties have similar within-group variability.

Variety A is estimated to be about \(4.4\,\text{cm}\) taller on average. The two samples have similar variability because their MADs differ by only \(0.3\,\text{cm}\).

- Compare the centers and the variability measures separately. - MAD describes a typical distance from the mean. - A smaller MAD indicates more tightly grouped values.

1. The equal means suggest that the groups have similar average reading times. 2. Group A’s MAD of \(3\) minutes means its times are typically much closer to the mean than Group B’s times are. 3. Group B’s MAD is \(9\) minutes, so its reading times show greater variability.

The groups have the same sample mean, \(28\) minutes, but Group A appears more consistent because its MAD is \(3\) minutes compared with \(9\) minutes for Group B.

- Use the center statistic to compare typical snowfall. - Use the middle-half spread to compare consistency. - Equal typical values do not require equal variability.

1. The medians are equal, so the samples have the same typical snowfall. 2. Town A's interquartile range is \(10-4=6\) inches smaller. 3. Town A's middle half is less spread out, so its snowfall appears more consistent.

The towns have the same sample median of \(7\) inches. Town A appears more consistent because its interquartile range is \(6\) inches smaller.

- Use the horizontal locations of the two dot clusters to compare their centers. - Identify score values where the displayed distributions occupy the same region. - Compute the same measure of center for both samples before making the population inference.

1. School A’s sample mean is \(\frac{626}{8}=78.25\). 2. School B’s sample mean is \(\frac{576}{8}=72\). 3. The mean difference is \(78.25-72=6.25\) points. The displayed score ranges overlap from about \(72\) to \(76\), but most School A scores lie to the right of most School B scores. 4. The samples suggest that School A generally has higher quiz scores, although the two populations may overlap.

School A mean: \(78.25\) School B mean: \(72\) The samples overlap visually from about \(72\) to \(76\), but School A’s distribution is shifted higher. The samples suggest School A generally scores about \(6.25\) points higher on average.

- Read the line inside each box to compare medians. - Find each box’s width by subtracting the first quartile from the third quartile. - Describe center and spread without claiming that the samples prove exact population values.

1. Warehouse Y’s median exceeds Warehouse X’s by \(9.5-9.0=0.5\,\text{lb}\). 2. Warehouse X’s IQR is \(9.3-8.7=0.6\,\text{lb}\). 3. Warehouse Y’s IQR is \(9.9-9.1=0.8\,\text{lb}\). 4. The samples suggest that Warehouse Y’s packages are typically heavier and have slightly more variability in the middle half.

Warehouse Y has a median \(0.5\,\text{lb}\) higher. Its IQR is \(0.8\,\text{lb}\), compared with \(0.6\,\text{lb}\) for Warehouse X, so Warehouse Y is also slightly more variable in the middle half.

- Compute the same center and spread for both samples. - Compare the center difference with the regions occupied by the two samples. - Distinguish an overall shift from complete separation.

1. Class A has mean \(\frac{277}{8}=34.625\) and range \(39-31=8\). 2. Class B has mean \(\frac{240}{8}=30\) and range \(33-27=6\). 3. Class A’s mean is \(34.625-30=4.625\) words per minute higher. 4. The sample values overlap from \(31\) to \(33\), so Class A appears faster overall, but the groups are not completely separated.

Class A: mean \(34.625\), range \(8\) Class B: mean \(30\), range \(6\) Class A is faster on average, while Class B has a slightly smaller range. The samples overlap from \(31\) to \(33\) words per minute.

- In this context, a smaller completion time means faster performance. - Compare where the centers of the two dot clusters lie. - Use the shared horizontal region to qualify the population inference.

1. Team Red’s mean time is \(\frac{118}{8}=14.75\) minutes. 2. Team Blue’s mean time is \(\frac{96}{8}=12\) minutes. 3. Team Blue’s mean is \(14.75-12=2.75\) minutes lower, but the displayed distributions overlap from \(12\) to \(15\) minutes. 4. The samples suggest that Team Blue is generally faster, although some completion times from the two teams are similar.

Team Blue appears faster: its sample mean is \(12\) minutes compared with \(14.75\) minutes for Team Red. The dot plots overlap from \(12\) to \(15\) minutes, so the evidence suggests an overall difference rather than complete separation.

- Locate the middle value in each ordered list. - Use the endpoints to compare overall spread. - Compare the median gap with the widths and overlap of the two samples.

1. Series M has median \(27\), and Series N has median \(30\). 2. Series M’s range is \(33-22=11\). 3. Series N’s range is \(42-18=24\). 4. Series N’s median is only \(3\) minutes higher, while its values spread across a much wider interval that contains the full range of Series M. 5. The samples suggest Series N may be slightly longer typically, but the difference in medians is modest relative to the variation.

Series M: median \(27\), range \(11\) Series N: median \(30\), range \(24\) Series N has a sample median \(3\) minutes longer, but that difference is small compared with the spread, especially in Series N.

- Compare the means and MADs separately. - Decide whether the mean gap is larger or smaller than a typical distance from the mean. - Use the relative size of the gap and spread to qualify the inference.

1. The Year 2 mean is \(7600-7100=500\) steps higher. 2. The MADs differ by only \(20\) steps, so the two samples have nearly the same variability. 3. The mean difference of \(500\) steps is smaller than either MAD, so the distributions likely have substantial overlap. 4. The samples suggest a modest increase in average daily steps, but the difference is not large relative to the within-year variation.

Year 2 is \(500\) steps higher on average, with nearly the same variability. Because \(500\) is smaller than both MADs, the samples likely overlap substantially, so the evidence for an increase should be stated cautiously.

- Since the means are equal, focus on how far the dots lie from \(12\). - Find the average of the absolute distances from the mean for each sample. - Connect the numerical MAD comparison to the clustering visible in the plot.

1. Zone A’s absolute deviations from \(12\) total \(20\), so its MAD is \(\frac{20}{7}\approx2.9\) minutes. 2. Zone B’s absolute deviations from \(12\) total \(12\), so its MAD is \(\frac{12}{7}\approx1.7\) minutes. 3. The dot plot for Zone B is more tightly clustered around \(12\), while Zone A includes a high value at \(20\). 4. Zone B appears more consistent because it has the smaller MAD and tighter visual spread.

Zone B appears more consistent. Its MAD is about \(1.7\) minutes, compared with about \(2.9\) minutes for Zone A, and its dots are more tightly clustered around the common mean of \(12\) minutes.

- Read the line inside each box to compare medians. - Compare the widths of the boxes to compare IQRs. - Use the amount of box overlap to qualify the center comparison.

1. Wetland P has median \(8\), first quartile \(7\), and third quartile \(10\), so its IQR is \(10-7=3\). 2. Wetland Q has median \(9\), first quartile \(6\), and third quartile \(13\), so its IQR is \(13-6=7\). 3. The median difference is only \(1\) call, while the boxes overlap substantially. 4. Wetland Q has a slightly higher sample median but much greater variability in the middle half.

Wetland P: median \(8\), IQR \(3\) Wetland Q: median \(9\), IQR \(7\) Wetland Q has a slightly higher typical count in the samples, but the center difference is small and the middle halves overlap substantially. Wetland Q is less consistent.

- Compare the means and MADs separately. - Judge the mean gap against a typical distance from each mean. - Use that comparison to avoid overstating the difference.

1. Route S has a mean travel time \(31-28=3\) minutes longer than Route R. 2. The MADs differ by only \(0.5\) minute, so the two routes have similar variability. 3. The mean gap of \(3\) minutes is about the size of one MAD, so the travel-time distributions likely overlap substantially. 4. The samples suggest Route S tends to take longer, but the difference is moderate rather than a complete separation.

Route S has a sample mean \(3\) minutes longer, and the routes have similar variability. Because the gap is about one MAD, the samples likely overlap substantially, so Route S should be described as tending to take longer rather than always taking longer.

- Decide what a mean measures and what a MAD measures. - Compare the mean difference with the sizes of the MADs. - Use “more variable” for a larger MAD, not “usually larger.”

1. MAD measures variability, not the typical fill amount. 2. The means differ by only \(501-499=2\,\text{mL}\). 3. The mean gap is smaller than either MAD, so there is no evidence of a large difference in typical fill amount. 4. Machine B’s larger MAD shows that its fill amounts are much less consistent.

The student confused spread with center. Machine B averages only \(2\,\text{mL}\) more, a difference smaller than both MADs, but its fills are much more variable because its MAD is \(12\,\text{mL}\) instead of \(4\,\text{mL}\).

- Compare the means and MADs separately. - Relate the \(90\)-foot gap to a typical deviation of about \(75\)–\(80\) feet. - State what the samples suggest without claiming that all trails differ.

1. South Park’s sample mean is \(510-420=90\,\text{ft}\) greater. 2. The MADs differ by only \(5\,\text{ft}\), so the samples have similar variability. 3. The mean gap of \(90\,\text{ft}\) is a little larger than one typical MAD of about \(78\,\text{ft}\). 4. The samples suggest South Park trails generally have greater elevation gain, although some overlap is still likely.

South Park trails have a sample mean about \(90\,\text{ft}\) higher, with similar variability. The gap is a little more than one MAD, so the evidence supports a general shift toward greater elevation gain, not complete separation.

- Use the common mean as the reference point for every dot. - Average the absolute distances from that reference point for each sample. - Connect the smaller MAD with the tighter cluster in the display.

1. League A’s absolute deviations from \(20\) sum to \(24\), so its MAD is \(\frac{24}{7}\approx3.4\) minutes. 2. League B’s absolute deviations from \(20\) sum to \(12\), so its MAD is \(\frac{12}{7}\approx1.7\) minutes. 3. The League B dots are more tightly clustered around \(20\), matching its smaller MAD. 4. The leagues have the same sample mean, but League B appears more consistent.

League A: MAD \(\approx3.4\) minutes League B: MAD \(\approx1.7\) minutes The leagues have the same sample mean, but League B appears less variable because its dots lie closer to \(20\) minutes.

- Read the median line in each box. - Compare the widths of the boxes to compare IQRs. - Use the shared part of the boxes to qualify the center difference.

1. Company K’s median is \(8-6=2\) miles greater than Company J’s median. 2. Company J’s IQR is \(8-4=4\) miles. 3. Company K’s IQR is \(12-3=9\) miles. 4. The boxes overlap from \(4\) to \(8\) miles, so the median difference is modest relative to the middle-half spreads. 5. Company K has a slightly higher typical distance in the samples and substantially greater variability.

Company K’s median is \(2\) miles greater. Company J has IQR \(4\) miles, while Company K has IQR \(9\) miles. The boxes overlap from \(4\) to \(8\) miles, so the typical-distance difference is modest, while Company K is clearly less consistent.

- Compare an “every student” claim with the actual horizontal overlap. - Compute the same center for both displayed samples. - Replace the absolute claim with language about an overall tendency.

1. Program A’s sample mean is \(\frac{154}{7}=22\) beats per minute. 2. Program B’s sample mean is \(\frac{196}{7}=28\) beats per minute. 3. The displayed distributions overlap from \(18\) to \(30\) beats per minute, so the samples do not show that every Program B value exceeds every Program A value. 4. Program B’s mean is \(6\) beats per minute greater, so the samples suggest a greater typical one-minute drop, with substantial individual overlap.

No. The dot plots overlap from \(18\) to \(30\) beats per minute. Program B has the higher sample mean, \(28\) compared with \(22\), so a reasonable inference is that Program B tends to produce a greater drop, not that every Program B student exceeds every Program A student.

- Find the middle value in each ordered sample. - Look for a value separated from the rest. - Consider which center is less affected by that value.

1. Team M's median is \(4\), while Team N's median is \(6\). 2. Team M has an unusually high value of \(16\), which raises its mean to \(6\). 3. Team N is more balanced around \(6\), so its mean and median both represent a typical shift well. 4. For Team M, the median \(4\) better represents a typical shift than the mean \(6\).

Team M has median \(4\); Team N has median \(6\). Team M's value of \(16\) pulls up its mean, so the median is more representative for Team M. Team N's mean and median both equal \(6\).

- Decide whether a larger or smaller center is preferable in this context. - Compare the mean gap with a typical MAD for the two samples. - Use both center and variability in the final inference.

1. Method Y’s mean reaction time is \(0.42-0.35=0.07\) second lower. 2. Method Y also has the smaller MAD, \(0.03\) second compared with \(0.04\) second. 3. The mean gap of \(0.07\) second is about twice a common MAD of roughly \(0.035\) second. 4. The samples provide fairly strong evidence that Method Y produces faster reaction times and slightly more consistency, although some overlap may remain.

Method Y has a sample mean \(0.07\) second lower and a slightly smaller MAD. Because the mean gap is about two typical MADs, the samples provide fairly strong evidence that Method Y is faster overall.

- Compare the means and MADs separately. - Judge whether the \(7\)-minute mean gap is large compared with typical deviations of \(6\) and \(11\) minutes. - Use the smaller MAD to identify the more consistent exhibit.

1. Exhibit A’s mean stay is \(42-35=7\) minutes longer. 2. Exhibit A’s MAD is \(11-6=5\) minutes smaller, so its stay times are more consistent. 3. The mean gap of \(7\) minutes is not large compared with MADs of \(6\) and \(11\) minutes. 4. The samples suggest a modest tendency for longer stays at Exhibit A, with substantial overlap likely, and clearly greater consistency at Exhibit A.

Exhibit A has a sample mean \(7\) minutes longer and a MAD \(5\) minutes smaller. The center difference is modest relative to the variability, so Exhibit A tends to have longer stays, while its stronger conclusion is that stay times are more consistent.

- Locate the balance point and middle dot for each distribution. - Compare how far the outer dots lie from the common center. - Look for a difference in spread after confirming that the centers match.

1. Each sample has mean \(9\) grams and median \(9\) grams. 2. Brand P’s values run from \(7\) to \(11\), while Brand Q’s values run from \(5\) to \(13\). 3. The Brand P dots are concentrated near \(9\), while Brand Q spreads farther on both sides of the same center. 4. The samples have the same center, but Brand Q has greater variability.

Both brands have mean \(9\) grams and median \(9\) grams. Brand P is more consistent because its dots are more tightly clustered, while Brand Q spreads from \(5\) to \(13\) grams.

- In a time context, a lower median means a faster typical result. - Compare the median gap with the sizes of the IQRs. - Use the smaller IQR to identify the more consistent shop.

1. Shop D has the lower median by \(3-2=1\) day. 2. The \(1\)-day median gap is smaller than either IQR, so it is modest relative to the middle-half variability. 3. Shop C has the smaller IQR, \(2\) days compared with \(5\) days, so its middle half is more tightly grouped. 4. Shop D may be slightly faster typically, while Shop C appears clearly more consistent.

Shop D’s sample median is \(1\) day lower, but that gap is small compared with the IQRs. Shop C appears more consistent because its IQR is \(2\) days instead of \(5\) days.

- Compare the upper whisker of Species A with the lower whisker of Species B. - Use the median lines for a typical-length comparison. - Distinguish overlap of the full ranges from overlap of the middle halves.

1. Species A’s upper whisker is \(16\) inches, while Species B’s lower whisker is \(13\) inches. 2. The whisker intervals overlap from \(13\) to \(16\) inches, so the claim about every fish is not supported. 3. Species B’s median is \(17-12=5\) inches greater. 4. The boxes do not overlap, so the middle half of Species B lies above the middle half of Species A even though the full sample ranges overlap.

The claim is not supported because the whisker intervals overlap from \(13\) to \(16\) inches. Species B is typically longer: its median is \(5\) inches greater, and its entire box lies above Species A’s box.

- Match speed with the mean and consistency with the MAD. - Compare the \(3\)-minute mean gap with typical deviations of \(7\) and \(3\) minutes. - Let the relative size of the gap affect how strongly you state the speed advantage.

1. Service L has the lower mean by \(27-24=3\) minutes, so it appears faster on average. 2. The \(3\)-minute mean gap is smaller than Service L’s MAD and equal to Service M’s MAD, so the samples likely overlap substantially. 3. Service M has the smaller MAD by \(7-3=4\) minutes, so its delivery times appear more consistent. 4. A customer prioritizing average speed may prefer Service L, but the advantage is modest; a customer prioritizing predictability may prefer Service M.

Service L has a sample mean \(3\) minutes lower, but the gap is small relative to the MADs. Service M appears more consistent because its MAD is \(3\) minutes instead of \(7\) minutes.

- Be careful when subtracting a negative temperature. - Use the medians for typical temperature and the IQRs for middle-half spread. - Compare the size of the median gap with the sizes of the IQRs.

1. Station F’s median is \(3-(-4)=7\,^{\circ}\text{F}\) higher. 2. The IQRs differ by only \(6-5=1\,^{\circ}\text{F}\), so the middle-half spreads are similar. 3. The \(7\)-degree median gap is slightly larger than either IQR, so the samples support a meaningful typical-temperature difference, though the full distributions may still overlap.

Station F has a sample median \(7\,^{\circ}\text{F}\) higher. The stations have similar middle-half variability, with IQRs of \(6\,^{\circ}\text{F}\) and \(5\,^{\circ}\text{F}\). The median gap is somewhat larger than either IQR, supporting a meaningful difference in typical lows.

- Compare the size of the center difference with the typical spread. - A small center difference can be difficult to distinguish when both groups vary widely. - Use cautious language for an informal sample-based conclusion.

1. The sample means differ by only \(66-64=2\) points. 2. Both MADs are much larger than the \(2\)-point difference in means. 3. The distributions are likely to have substantial overlap, so the samples suggest only a small difference rather than a clear separation.

No. Program S's mean is only \(2\) points higher, while the MADs are \(8\) and \(9\) points. The samples suggest a small difference with substantial variability, not a clear separation.

- Compare each dot cluster’s center with the target line. - Compare how far the outer dots extend on each side. - The stronger printer should address both closeness to target and consistency.

1. Printer A’s mean is \(\frac{100.2}{5}=20.04\,\text{mm}\). 2. Printer B’s mean is \(\frac{100.0}{5}=20.0\,\text{mm}\). 3. Printer A’s values span \(20.2-19.8=0.4\,\text{mm}\), while Printer B’s values span \(20.1-19.9=0.2\,\text{mm}\). 4. Printer B’s mean matches the target line, and its dots are more tightly clustered.

Printer A: mean \(20.04\,\text{mm}\), displayed span \(0.4\,\text{mm}\) Printer B: mean \(20.0\,\text{mm}\), displayed span \(0.2\,\text{mm}\) Printer B appears closer to the target and more consistent.

- Compare the center difference with the size of the spreads. - Similar, fairly large spreads can weaken a claim of clear separation. - Use cautious language when the evidence shows only a small difference.

1. Lake J's sample median is only \(8.9-8.4=0.5\,\text{m}\) greater. 2. The interquartile ranges are nearly equal and are more than three times the median difference. 3. The samples suggest Lake J may be slightly deeper typically, but they do not show a clear separation.

It is not reasonable to say “clearly.” Lake J's median is only \(0.5\,\text{m}\) greater, while both samples have middle-half spreads near \(1.6\) to \(1.7\,\text{m}\). Lake J appears only slightly deeper typically.

- For germination time, a smaller mean indicates faster germination. - Compare the \(0.7\)-day mean gap with typical deviations of \(0.8\) and \(2.1\) days. - Use the smaller MAD to identify the more predictable mix.

1. Mix B has the smaller mean by \(6.2-5.5=0.7\) day. 2. The \(0.7\)-day gap is smaller than Mix A’s MAD and much smaller than Mix B’s MAD, so the samples likely overlap substantially. 3. Mix A has the smaller MAD by \(2.1-0.8=1.3\) days, so it appears more predictable. 4. Mix B may germinate slightly faster on average, but Mix A has the clearer advantage in consistency.

Mix B’s sample mean is \(0.7\) day lower, but that difference is small relative to the MADs. Mix A appears more predictable because its MAD is \(0.8\) day instead of \(2.1\) days.

- Read the center from the line inside each box. - Use the box edges for the middle-half spread. - Use the whisker endpoints for the full displayed spread.

1. Both models have a median battery range of \(26\) miles. 2. Model A has IQR \(30-22=8\) miles, while Model B has IQR \(36-20=16\) miles. 3. Model A has range \(32-18=14\) miles, while Model B has range \(42-16=26\) miles. 4. The median battery ranges are equal, but Model A is more consistent because both of its spread measures are smaller.

Both median battery ranges are \(26\) miles. Model A has IQR \(8\) miles and range \(14\) miles; Model B has IQR \(16\) miles and range \(26\) miles. Model A appears more consistent.

- Compare each type of center with the same type in the other course. - Judge the center gaps against the typical deviations shown by the MADs. - Use the smaller MAD to discuss consistency separately.

1. Course B’s mean is \(85-82=3\) points higher. 2. Course B’s median is \(85-84=1\) point higher. 3. Both center comparisons point toward Course B, but the gaps are small relative to MADs of \(8\) and \(5\). 4. Course B’s smaller MAD suggests greater consistency, while the evidence for a higher typical score is modest.

Both the mean and median are slightly higher for Course B, but the differences of \(3\) and \(1\) points are small relative to the MADs. Course B appears more consistent because its MAD is \(5\) instead of \(8\).

- Find each sample’s balance point and average absolute distance from it. - Compare the \(1\)-kWh mean gap with MADs of \(1.2\) and \(0.8\). - State consistency and typical output as separate conclusions.

1. Group A’s mean is \(100\div5=20\), and its MAD is \(\frac{6}{5}=1.2\). 2. Group B’s mean is \(105\div5=21\), and its MAD is \(\frac{4}{5}=0.8\). 3. Group B’s mean is \(1\,\text{kWh}\) higher, which is about the size of a typical MAD for the two groups. 4. Group B appears slightly higher in output and more consistent, but the output distributions likely overlap substantially.

Group A: mean \(20\,\text{kWh}\), MAD \(1.2\,\text{kWh}\) Group B: mean \(21\,\text{kWh}\), MAD \(0.8\,\text{kWh}\) Group B has a slightly higher sample mean and is more consistent. The \(1\)-kWh center gap is only about one MAD, so the output difference is modest.

- Check whether the horizontal regions occupied by the two dot plots overlap. - Compute the same center for both samples. - Compare the widths of the displayed distributions before describing consistency.

1. Band A’s sample mean is \(\frac{644}{7}=92\) minutes. 2. Band B’s sample mean is \(\frac{693}{7}=99\) minutes. 3. The displayed ranges overlap from \(84\) to \(102\) minutes, so the samples do not support a statement about every concert. 4. Band B’s mean is \(7\) minutes greater, and its dots extend across a wider interval, from \(79\) to \(119\) minutes. 5. Band B appears longer on average but also more variable.

The “every concert” statement is not supported because the displayed ranges overlap. Band B has a sample mean \(7\) minutes higher, but its concert lengths also show greater visual spread.

- Put both summaries in minutes before comparing them. - Compare the mean gap with typical deviations of \(3\) and \(4\) minutes. - Use the smaller MAD to discuss consistency separately.

1. Park A’s mean is \(0.4\cdot60=24\) minutes. 2. Park A’s MAD is \(0.05\cdot60=3\) minutes. 3. Park A’s mean is \(27-24=3\) minutes shorter, and its MAD is \(4-3=1\) minute smaller. 4. The \(3\)-minute mean gap is about one MAD, so the samples likely overlap substantially. 5. Park A appears slightly faster and slightly more consistent.

Park A has mean \(24\) minutes and MAD \(3\) minutes. Park B has mean \(27\) minutes and MAD \(4\) minutes. Park A appears slightly faster and more consistent, but the \(3\)-minute gap is only about one MAD.

- Read the median lines to compare typical braking distance. - Compare the widths of the boxes to compare IQRs. - Distinguish overlap of the boxes from overlap of the whisker intervals.

1. Tire Type C has a median \(24-20=4\) feet shorter than Tire Type D. 2. Tire Type C’s IQR is \(22-18=4\) feet. 3. Tire Type D’s IQR is \(26-22=4\) feet. 4. The boxes meet at \(22\) feet, so the middle halves have almost no overlap, while the whisker intervals overlap from \(18\) to \(27\) feet. 5. Tire Type C appears to stop sooner typically, with the same middle-half variability.

Tire Type C has a median braking distance \(4\) feet shorter. Both tire types have IQR \(4\) feet. Their boxes only touch at \(22\) feet, supporting a clear typical difference, although the full displayed ranges overlap.

- Use the common mean as the center for both samples. - Find each value's distance from the center before averaging. - The smaller average distance indicates greater consistency.

1. Theater A's absolute deviations from \(4\) total \(8\), so its MAD is \(8\div5=1.6\) minutes. 2. Theater B's absolute deviations from \(4\) total \(2\), so its MAD is \(2\div5=0.4\) minute. 3. Theater B's wait times are more consistent because its MAD is smaller.

Theater A: MAD \(1.6\) minutes Theater B: MAD \(0.4\) minute Theater B appears more consistent, even though the sample means are equal.

- Compare the centers using percentage points. - Judge the \(3\)-point gap against typical deviations of \(2.5\) and \(5\) points. - Use the smaller MAD for the stronger consistency conclusion.

1. School A’s mean attendance is \(91\%-88\%=3\) percentage points higher. 2. The \(3\)-point mean gap is only slightly larger than School A’s MAD and smaller than School B’s MAD, so the distributions likely overlap substantially. 3. School A’s MAD is \(5-2.5=2.5\) percentage points smaller. 4. School A appears slightly higher in attendance and clearly more consistent.

School A has a sample mean \(3\) percentage points higher, but that gap is modest relative to the MADs. School A appears more consistent because its MAD is \(2.5\) percentage points instead of \(5\).

- In a charging-time context, a smaller median indicates faster typical charging. - Compare the \(4\)-minute median gap with IQRs of \(8\) and \(18\) minutes. - Use the smaller IQR to identify the more predictable model.

1. Model D has the lower median by \(72-68=4\) minutes. 2. The \(4\)-minute median gap is smaller than either IQR, so the typical-time difference is modest relative to the middle-half variability. 3. Model C has the smaller IQR by \(18-8=10\) minutes. 4. Model D may be slightly faster typically, while Model C appears much more predictable.

Model D’s sample median is \(4\) minutes lower, but that difference is small compared with the IQRs. Model C appears more predictable because its IQR is \(8\) minutes instead of \(18\) minutes.

- Compare the mean gap with the common MAD. - Interpret \(3\) years as a fraction of a typical \(6\)-year deviation. - Separate the ability to compare statistics from the stability associated with sample size.

1. Reserve N’s sample mean is \(37-34=3\) years greater. 2. The MADs are equal at \(6\) years. 3. The mean gap is only \(\frac{3}{6}=0.5\) MAD, so the samples likely overlap substantially and do not support a strong age difference. 4. The unequal sample sizes do not prevent comparing the means and MADs, but the larger sample from Reserve N would generally be expected to give a more stable estimate.

Reserve N’s sample mean is \(3\) years higher, with the same MAD of \(6\) years. Because the gap is only \(0.5\) MAD, the samples suggest at most a small typical-age difference. The unequal sample sizes do not block the comparison, though the larger sample estimate is generally more stable.

- Compare the mean gap with a typical MAD for the two samples. - Use the smaller MAD to identify the more consistent greenhouse. - State the center and consistency conclusions separately.

1. Greenhouse A’s mean humidity is \(62.5\%-59.0\%=3.5\) percentage points higher. 2. A common MAD is about \(\frac{1.8+1.2}{2}=1.5\) percentage points, so the mean gap is about \(\frac{3.5}{1.5}\approx2.3\) MADs. 3. Greenhouse B’s MAD is \(1.8-1.2=0.6\) percentage point smaller. 4. The samples provide fairly strong evidence that Greenhouse A is more humid typically, while Greenhouse B is more consistent.

Greenhouse A averages \(3.5\) percentage points higher humidity, a gap of about \(2.3\) common MADs. Greenhouse B appears more consistent because its MAD is \(1.2\) percentage points instead of \(1.8\).

- Compare the means and MADs separately. - Express the mean gap in units of the common MAD. - Ask which features cannot be recovered from only two summary statistics.

1. Warehouse B’s sample mean is \(56-48=8\) pounds greater. 2. The equal MADs show that the samples have the same typical distance from their means. 3. The mean gap is \(\frac{8}{5}=1.6\) common MADs, supporting a noticeable shift in center with possible overlap. 4. A mean and MAD do not determine the exact shape, clustering, or individual values of either distribution.

Supported: Warehouse B packages are about \(8\) pounds heavier on average, a gap of \(1.6\) common MADs, and the samples have equal typical variability. Not determined: the exact distributions or whether every Warehouse B package is heavier than every Warehouse A package.

- Identify which number counts observations and which number describes a typical observation. - Compare the \(2\)-item mean gap with the common MAD of \(3\) items. - Separate sample-size stability from the size of the measured baskets.

1. Sample size tells how many customers were measured, not the typical basket size. 2. Store P’s sample mean is \(12-10=2\) items greater. 3. The MADs are equal at \(3\) items, and the mean gap is only \(\frac{2}{3}\) MAD. 4. The samples likely overlap substantially, so they suggest at most a small tendency toward larger baskets at Store P. 5. The larger sample from Store Q would generally give a more stable estimate, but it does not make the baskets larger.

The student confused sample size with a measure of center. Store P’s sample mean is \(2\) items higher, but the gap is only about \(0.67\) MAD, so the typical-basket difference appears small. Store Q’s larger sample may give a more stable estimate.

- Use the means for average lifetime and the MADs for predictability. - Compare the \(6\)-day mean gap with both typical deviations. - Distinguish an average comparison from a statement about every filter.

1. Filter B’s mean lifetime is \(98-92=6\) days greater. 2. The \(6\)-day gap is larger than Filter A’s MAD but much smaller than Filter B’s MAD, so substantial overlap is plausible. 3. Filter A’s MAD is \(15-4=11\) days smaller, so its lifetimes are much more predictable. 4. A higher mean does not order every individual value, especially when one group has large variability.

Filter B has a sample mean \(6\) days longer, but the gap is small compared with its \(15\)-day MAD. Filter A is much more predictable because its MAD is only \(4\) days. The summaries do not imply that every Filter B lasts longer.

- Match “middle half” with the statistic that uses quartiles. - Match “overall spread” with the statistic that uses the endpoints. - Different measures of variability can favor different groups.

1. Group A has the smaller interquartile range, so its middle half is more consistent. 2. Group B has the smaller range, so its overall sample spread is smaller. 3. The two variability measures emphasize different parts of the distributions.

Group A is more consistent in the middle half because its interquartile range is \(8\) minutes. Group B has the smaller overall spread because its range is \(20\) minutes.

- Use the mean to compare sample centers. - Use the MAD to compare typical spread. - Reject choices that turn a sample-center difference into a claim about every individual tree.

1. Orchard X has the larger mean, \(18\) compared with \(15\). 2. Orchard X has the smaller MAD, \(2\) compared with \(6\). 3. The mean gap is \(3\) apples, which is not larger than both MADs, so overlap may remain. 4. These statistics support option b), not a claim about every tree.

b) Orchard X has a higher sample mean and less variability, although the populations may still overlap.

- Include the negative values when finding each total. - Compare both measures of center and the endpoint spreads. - Identify the interval occupied by values from both samples.

1. Greenhouse A has mean \(0\), median \(0\), and range \(3-(-3)=6\) centimeters. 2. Greenhouse B has mean \(\frac{10}{5}=2\), median \(2\), and range \(5-(-1)=6\) centimeters. 3. Greenhouse B is \(2\) centimeters higher by both centers, while the total sample spreads are equal. 4. The sample values overlap from \(-1\) to \(3\), so Greenhouse B shows higher typical growth without complete separation.

Greenhouse A: mean \(0\,\text{cm}\), median \(0\,\text{cm}\), range \(6\,\text{cm}\) Greenhouse B: mean \(2\,\text{cm}\), median \(2\,\text{cm}\), range \(6\,\text{cm}\) Greenhouse B has higher sample centers, but the samples overlap from \(-1\) to \(3\) centimeters and have the same range.

- Find the difference between the two means. - Average the two MADs to create one common variability unit. - Compare the mean difference with that common MAD before describing how strong the difference is.

1. Program B's mean is \(5.6-4.8=0.8\) hour greater. 2. A common MAD for comparison is \(\frac{1.1+1.0}{2}=1.05\) hours. 3. The mean difference is \(\frac{0.8}{1.05}\approx0.76\) common MAD, which is less than one typical deviation. 4. The MADs differ by only \(0.1\) hour, so the groups have nearly the same variability. 5. Program B shows a modest tendency toward more volunteer time, but substantial overlap is plausible.

Program B's mean is \(0.8\) hour higher, which is about \(0.76\) common MAD. This is a modest difference rather than a clear separation. The programs have nearly equal variability because their MADs differ by only \(0.1\) hour.

- Find a center and an endpoint difference for each sample. - Compare the differences between the centers with the spreads of the samples. - Use cautious language when the center difference is small and the samples occupy much of the same number region.

1. Site A has mean \(2.0\) meters, median \(2.0\) meters, and range \(2.5-1.5=1.0\) meter. 2. Site B has mean \(\frac{11.0}{5}=2.2\) meters, median \(2.1\) meters, and range \(3.0-1.2=1.8\) meters. 3. The mean difference is only \(0.2\) meter, and the median difference is only \(0.1\) meter. Both are small compared with the sample ranges. 4. The values from the two samples occupy much of the same region around \(2\) meters, so their typical depths appear similar. 5. Site B is more variable in these equal-size samples because its range is larger.

Site A: mean \(2.0\,\text{m}\), median \(2.0\,\text{m}\), range \(1.0\,\text{m}\) Site B: mean \(2.2\,\text{m}\), median \(2.1\,\text{m}\), range \(1.8\,\text{m}\) The typical depths are very similar, with Site B only slightly higher. Site B is more variable in these equal-size samples.

- Compare the center difference with the spread of each group. - A small difference can be hard to distinguish when one group varies widely. - Replace an absolute-sounding word with a cautious sample-based conclusion.

1. Group D's median is only \(54-52=2\) minutes greater. 2. Group D's interquartile range is \(12\) minutes, much larger than the \(2\)-minute center difference. 3. The samples likely overlap substantially, so “clearly” is too strong. 4. A careful conclusion is that Group D has a slightly higher sample median but much greater variability.

The claim is too strong. Group D's median is only \(2\) minutes higher, while its interquartile range is \(12\) minutes. Group D appears slightly higher typically but much less consistent.

- Convert both centers and both MADs to centimeters. - Average the two MADs to create one common variability unit. - Compare the mean difference with that unit, then use the smaller MAD to identify greater consistency.

1. Farm B's mean is \(22\,\text{mm}=2.2\,\text{cm}\). 2. Farm B's MAD is \(5\,\text{mm}=0.5\,\text{cm}\). 3. Farm B's mean growth is \(2.2-1.8=0.4\,\text{cm}\) greater. 4. The average MAD is \(\frac{0.3+0.5}{2}=0.4\,\text{cm}\), so the mean difference is one common MAD. 5. Farm B shows a modest tendency toward greater growth, but overlap is plausible. Farm A is more consistent because its MAD is smaller.

Farm B has mean \(2.2\,\text{cm}\) and MAD \(0.5\,\text{cm}\). Its mean is \(0.4\,\text{cm}\) higher, equal to one common MAD, so the difference is modest rather than a clear separation. Farm A is more consistent because its MAD is \(0.3\,\text{cm}\).

- Use a statistic from the middle of a sample to describe a typical score. - Compare the median difference with the endpoint spread before deciding whether the difference is large. - Remember that a range uses only two values and does not describe the full distribution.

1. A maximum describes one extreme value, not a typical score. 2. Team A's median is \(70-68=2\) points higher, so Team A has the higher sample median. 3. Team A's range is \(85-55=30\), and Team B's range is \(90-60=30\). 4. The \(2\)-point median difference is small compared with the \(30\)-point sample ranges, so the typical scores appear similar rather than clearly separated. 5. Equal ranges describe only the endpoint spread of these samples; they do not prove that the full distributions have equal variability.

The student used an extreme value to describe the center. Team A's median is \(2\) points higher, but that difference is small compared with the \(30\)-point range of each sample. The typical scores appear similar, and the equal sample ranges alone do not establish equal overall consistency.

- Find the difference between the means. - Average the two MADs, then measure the mean difference in that common unit. - Compare the MADs separately to decide which region is more consistent.

1. Region A's mean swing is \(14-11=3\,^{\circ}\text{F}\) greater. 2. The average MAD is \(\frac{3+1.5}{2}=2.25\,^{\circ}\text{F}\). 3. The mean difference is \(\frac{3}{2.25}\approx1.33\) common MADs, so Region A shows a moderate tendency toward larger swings, but overlap is still plausible. 4. Region B is more consistent because its MAD is \(1.5\,^{\circ}\text{F}\) smaller.

Region A's mean swing is \(3\,^{\circ}\text{F}\) higher, about \(1.33\) common MADs. This is a moderate difference, not a complete separation. Region B is more consistent because its MAD is \(1.5\,^{\circ}\text{F}\) smaller.

- Average the two middle values in each sample. - Compare each value in the first list with the value in the same position in the second. - A uniform shift changes location but not distances between values.

1. Company A's median is \(\frac{8+9}{2}=8.5\) hours, and its range is \(20-6=14\) hours. 2. Company B's median is \(\frac{11+12}{2}=11.5\) hours, and its range is \(23-9=14\) hours. 3. Every Company B value is \(3\) hours greater than the corresponding Company A value. 4. The shift raises the median by \(3\) hours but leaves the range unchanged.

Company A: median \(8.5\) hours, range \(14\) hours Company B: median \(11.5\) hours, range \(14\) hours Company B's sample is Company A's sample shifted up by \(3\) hours.

- Compute both centers for each group. - Look for a value far from the rest of its sample. - Choose the center that is less changed by that unusual value.

1. Shop A's mean is \(\frac{499}{7}\approx71.3\) dollars, and its median is \(\$49\). 2. Shop B's mean is \(\$63\), and its median is \(\$63\). 3. The unusually high \(\$210\) cost raises Shop A's mean. 4. The medians suggest a typical repair costs less at Shop A, while the means are distorted by Shop A's outlier.

Shop A: mean about \(\$71.3\), median \(\$49\) Shop B: mean \(\$63\), median \(\$63\) The medians give the clearer typical-cost comparison because Shop A has a \(\$210\) outlier.

- Use the first sample to determine the common mean. - Convert the second sample's mean into the total it must have. - After finding the missing value, compare the endpoints of both samples.

1. Class A's total is \(12+14+15+16+18=75\), so its mean is \(75\div5=15\). 2. Class B must also total \(5\cdot15=75\). 3. The known Class B values total \(10+13+15+17=55\), so \(x=75-55=20\). 4. Class A's range is \(18-12=6\), and Class B's range is \(20-10=10\). 5. The centers are equal, but Class A is less variable by range.

\(x=20\). Class A has range \(6\) minutes, and Class B has range \(10\) minutes. Class A is more consistent.

- Use the stacked dots to count repeated values. - Find the common center before measuring absolute distances from it. - Connect the smaller MAD with the tighter dot cluster.

1. Each sample has \(10\) students and a total of \(20\) books, so each mean is \(20\div10=2\) books. 2. School A’s absolute deviations from \(2\) total \(8\), so its MAD is \(8\div10=0.8\) book. 3. School B’s absolute deviations from \(2\) total \(12\), so its MAD is \(12\div10=1.2\) books. 4. The samples have the same center, but School A’s dots are more tightly clustered and its MAD is smaller.

School A: mean \(2\), MAD \(0.8\) School B: mean \(2\), MAD \(1.2\) The typical number checked out is the same, but School A’s sample is more consistent.

- Locate the main cluster and any isolated dot before choosing a typical value. - Calculate both centers for each displayed sample. - Decide which center is less affected by City A’s long trip.

1. City A has mean \(\frac{20}{7}\approx2.9\) miles and median \(2\) miles. 2. City B has mean \(\frac{18}{7}\approx2.6\) miles and median \(3\) miles. 3. City A’s isolated \(9\)-mile trip raises its mean, so the median better represents the main cluster in City A. 4. By medians, City B has the longer typical trip; by means, the outlier makes City A appear longer.

City A: mean \(\approx2.9\) miles, median \(2\) miles City B: mean \(\approx2.6\) miles, median \(3\) miles The median better represents City A because the isolated \(9\)-mile trip pulls up its mean. The medians suggest City B has the longer typical trip.

- Think about how adding the same amount to every value affects a center. - Compare the distance between the smallest and largest values before and after the shift. - You can use the transformation instead of rewriting the full second sample.

1. Grove A's mean is \(\frac{72}{5}=14.4\) feet, median is \(14\) feet, and range is \(18-11=7\) feet. 2. Adding \(5\) to every value adds \(5\) to the mean and median but does not change the range. 3. Grove B has mean \(19.4\) feet, median \(19\) feet, and range \(7\) feet.

Grove A: mean \(14.4\,\text{ft}\), median \(14\,\text{ft}\), range \(7\,\text{ft}\) Grove B: mean \(19.4\,\text{ft}\), median \(19\,\text{ft}\), range \(7\,\text{ft}\) Grove B is \(5\) feet taller by both centers, with the same variability by range.

- With an even number of values, average the two middle values for the median. - Find the median of each lower and upper half for the quartiles. - Compare the median gap with the sizes of the IQRs before stating an inference.

1. Town A’s median is \(\frac{5+6}{2}=5.5\) miles. Its quartiles are \(3.5\) and \(7.5\), so its IQR is \(4\) miles. 2. Town B’s median is \(\frac{4+8}{2}=6\) miles. Its quartiles are \(2.5\) and \(9.5\), so its IQR is \(7\) miles. 3. The median gap is only \(0.5\) mile, much smaller than either IQR. 4. The samples show nearly the same typical commute distance, while Town B has greater middle-half variability.

Town A: median \(5.5\) miles, IQR \(4\) miles Town B: median \(6\) miles, IQR \(7\) miles The \(0.5\)-mile median difference is very small relative to the spreads, so the typical commute distances appear similar. Town B is more variable.

- With eight values, average the two middle values for the median. - Find the center of each four-value half to locate the quartiles. - Equal medians can still come with different middle-half spreads.

1. Each sample has median \(\frac{4+4}{2}=4\) minutes. 2. League A has first quartile \(2\) and third quartile \(7\), so its interquartile range is \(5\) minutes. 3. League B has first quartile \(3\) and third quartile \(6\), so its interquartile range is \(3\) minutes. 4. The typical penalty time is the same, but League B is more consistent in the middle half.

Both leagues have median \(4\) minutes. League A has interquartile range \(5\) minutes, while League B has interquartile range \(3\) minutes. League B appears more consistent.

- Think about how multiplying every observation affects a center. - The distance between two scaled values changes by the same factor. - Compare the transformed summaries with the originals.

1. Multiplying every value by \(1.2\) multiplies the mean by \(1.2\), so Greenhouse B’s mean is \(45\cdot1.2=54\,\text{cm}\). 2. The distance between the largest and smallest values also scales by \(1.2\), so Greenhouse B’s range is \(10\cdot1.2=12\,\text{cm}\). 3. Greenhouse B’s sample is taller on average and has a range \(2\,\text{cm}\) greater.

Greenhouse B has mean \(54\,\text{cm}\) and range \(12\,\text{cm}\). Its sample is taller on average and has a slightly larger overall spread than Greenhouse A’s sample.

- Use the middle value of each ordered list for typical performance. - Find the middle-half spread separately from the full spread. - Compare the size of the median difference with the spreads before describing one team as faster.

1. Team A has median \(23\), \(Q_1=21\), and \(Q_3=35\), so its IQR is \(35-21=14\) and its range is \(36-20=16\). 2. Team B has median \(24\), \(Q_1=22\), and \(Q_3=26\), so its IQR is \(26-22=4\) and its range is \(30-18=12\). 3. Team A's sample median is only \(1\) minute lower, so the typical times appear similar rather than clearly different. 4. Team B is more consistent because both its IQR and range are smaller.

Team A: median \(23\) minutes, IQR \(14\) minutes, range \(16\) minutes Team B: median \(24\) minutes, IQR \(4\) minutes, range \(12\) minutes The typical times are similar, with Team A's sample median only \(1\) minute lower. Team B is clearly more consistent.