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Measure overlap and variability

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5416997
Random samples of sprint speeds from two training groups have similar variability. Group A: mean \(50\) meters per minute, MAD \(4\) meters per minute Group B: mean \(62\) meters per minute, MAD \(4\) meters per minute Express the difference in means as a multiple of the common MAD. What does this suggest about overlap?

Hints

- Find the distance between the two centers. - Compare that distance with the shared measure of variability. - A separation of several typical deviations suggests less overlap.

Solution

1. The difference in means is \(62-50=12\) meters per minute. 2. Relative to the common MAD, \(12\div4=3\). 3. The centers are separated by \(3\) MADs, so the distributions likely have limited overlap and a noticeable separation.

Answer

The means differ by \(3\) MADs. This suggests limited overlap and a noticeable difference between the groups.
5417007
Two random samples of reading scores have similar variability. School C: mean \(75\), MAD \(6\) School D: mean \(81\), MAD \(6\) Measure the difference in means in MAD units and describe the likely overlap.

Hints

- Compare the center difference with one group's typical deviation. - Express the comparison as a ratio. - A center difference about the same size as the variability usually allows considerable overlap.

Solution

1. The difference in means is \(81-75=6\) points. 2. Since the common MAD is \(6\), the difference is \(6\div6=1\) MAD. 3. A one-MAD separation suggests substantial overlap between the distributions.

Answer

The means differ by \(1\) MAD. The distributions likely overlap substantially.
5417017
Random samples of daily customer counts from two food trucks have similar variability. Truck A: mean \(24\), MAD \(3\) Truck B: mean \(30\), MAD \(3\) Express the difference in means as a multiple of the common MAD and describe the likely overlap.

Hints

- Find the distance between the two centers. - Compare that distance with the typical deviation in either group. - Interpret the ratio as a measure of separation.

Solution

1. The difference in means is \(30-24=6\) customers. 2. Relative to the common MAD, \(6\div3=2\). 3. A two-MAD separation suggests a noticeable difference with some overlap still possible.

Answer

The means differ by \(2\) MADs. The distributions likely have a noticeable separation but may still overlap.
5417027
Two random samples of science scores have the same MAD of \(4\) points. Their means are \(68\) and \(74\). How many MADs apart are the means? What does the result suggest about overlap?

Hints

- Subtract the smaller center from the larger center. - Measure that distance using the common spread as one unit. - A ratio between \(1\) and \(2\) usually indicates partial overlap.

Solution

1. The mean difference is \(74-68=6\) points. 2. The separation is \(6\div4=1.5\) MADs. 3. A separation of \(1.5\) MADs suggests moderate overlap rather than complete separation.

Answer

The means are \(1.5\) MADs apart, so the distributions likely overlap moderately.
5417117
The parallel box plots show random samples of commute speeds in two cities. Compare the medians, IQRs, overlap of the boxes, and overlap of the full displayed ranges.
Figure for problem 541711

Hints

- Read each median from the line inside its box. - Subtract the first quartile from the third quartile for each IQR. - Compare the boxes and the whiskers separately; they represent different portions of the samples.

Solution

1. City A has median \(40\) miles per hour and IQR \(44-36=8\) miles per hour. 2. City B has median \(52\) miles per hour and IQR \(56-48=8\) miles per hour. 3. The medians differ by \(52-40=12\) miles per hour, while the two samples have equal middle-half variability. 4. The boxes do not overlap because City A's third quartile is \(44\) and City B's first quartile is \(48\). 5. The whiskers overlap from \(44\) to \(48\), so the full sample ranges are not completely separated.

Answer

City A: median \(40\,\text{mph}\), IQR \(8\,\text{mph}\) City B: median \(52\,\text{mph}\), IQR \(8\,\text{mph}\) The boxes do not overlap, but the whiskers overlap from \(44\) to \(48\,\text{mph}\). City B has the higher center, and the middle halves have equal variability.
5417127
The parallel box plots show random samples of daily orders at two stores. Compare the medians, IQRs, and visible overlap. Which store appears more variable?
Figure for problem 541712

Hints

- Read the center line and both box edges for each store. - Use box width to compare the spread of the middle halves. - Check overlap of the boxes separately from overlap of the whiskers.

Solution

1. Store A has median \(24\) orders and IQR \(26-22=4\) orders. 2. Store B has median \(34\) orders and IQR \(38-30=8\) orders. 3. Store B's sample median is \(34-24=10\) orders higher. 4. The boxes do not overlap, although the whiskers overlap from \(26\) to \(30\) orders. 5. Store B appears more variable in the middle half because its IQR is twice Store A's IQR.

Answer

Store A: median \(24\), IQR \(4\) Store B: median \(34\), IQR \(8\) Store B has the higher center and greater middle-half variability. The boxes do not overlap, but the full displayed ranges overlap from \(26\) to \(30\) orders.
5417157
The parallel dot plots show random samples of daily machine output. Both samples have MAD \(4\), and their means are \(30\) and \(38\). Which statement is best supported? a) The populations cannot overlap because the means are \(2\) MADs apart. b) The displayed samples overlap, but Machine B's distribution is shifted higher. c) Machine A has greater variability because its mean is lower. d) The two distributions have the same center.
Figure for problem 541715

Hints

- Measure the distance between the means using the common MAD. - Look for horizontal values that contain dots from both samples. - A standardized center difference describes separation; it does not create boundaries around either population.

Solution

1. The mean difference is \(38-30=8\), and \(8\div4=2\) MADs. 2. The plots share displayed values at \(32\), \(34\), and \(36\), so the samples overlap. 3. Machine B's dots are generally farther to the right, showing a higher center with the same variability. 4. Therefore, statement b is best supported. A two-MAD center difference does not prove that populations cannot overlap.

Answer

b) The displayed samples overlap, but Machine B's distribution is shifted higher.
5417167
Random samples of monthly data use have these summaries. Plan A: mean \(120\) gigabytes, MAD \(3\) gigabytes Plan B: mean \(132\) gigabytes, MAD \(5\) gigabytes Use the average of the two MADs as a common variability unit. How many such units apart are the means, and what does that suggest?

Hints

- Average the two MADs to create one common comparison unit. - Divide the distance between the means by that common MAD. - Interpret the result as evidence about likely separation, not as a boundary around either population.

Solution

1. The average MAD is \(\frac{3+5}{2}=4\) gigabytes. 2. The mean difference is \(132-120=12\) gigabytes. 3. The separation is \(12\div4=3\) common-MAD units. 4. A separation of three common MADs suggests a strong difference in centers and limited likely overlap, though it does not prove that the populations never overlap.

Answer

The means are \(3\) common-MAD units apart. This suggests a strong difference in typical data use and limited likely overlap.
5417187
Population A has a sample mean of \(50\) and MAD \(6\). Using \(6\) as the common variability unit, Population B is to have a mean at least \(2\) MADs higher than Population A. What is the least possible mean for Population B?

Hints

- Convert the required number of common variability units into an actual distance. - Add that distance to Population A's mean because Population B must be higher. - “Least possible” means use exactly the required separation.

Solution

1. Two common MADs equal \(2\cdot6=12\). 2. The least mean at least that far above \(50\) is \(50+12=62\).

Answer

The least possible mean is \(62\).
5417197
The parallel box plots show random samples from two populations. Compare the medians, IQRs, overlap of the middle halves, and overlap of the full displayed ranges.
Figure for problem 541719

Hints

- Read each median and subtract the first quartile from the third quartile. - The box itself represents the middle half of a sample. - Compare the whiskers separately when discussing the full displayed ranges.

Solution

1. Population A has median \(18\) and IQR \(22-14=8\). 2. Population B has median \(24\) and IQR \(28-20=8\). 3. The middle halves overlap from \(20\) to \(22\). 4. The full displayed ranges overlap from \(16\) to \(26\). 5. Population B has the higher center, while the equal IQRs indicate equal middle-half variability in these samples.

Answer

Population A: median \(18\), IQR \(8\) Population B: median \(24\), IQR \(8\) The boxes overlap from \(20\) to \(22\), and the full displayed ranges overlap from \(16\) to \(26\). Population B has the higher center, and the samples have equal middle-half variability.
5417237
Two random samples have means \(80\) and \(70\), with a common MAD of \(8\). A student says the \(10\)-unit difference proves that the populations have little overlap. Evaluate the claim using MAD units.

Hints

- A raw difference cannot be interpreted without the variability scale. - Divide the center difference by the common MAD. - Decide whether the resulting number supports strong or modest separation.

Solution

1. The difference in means is \(80-70=10\). 2. In MAD units, the separation is \(10\div8=1.25\). 3. A separation of \(1.25\) MADs is only a little larger than one typical deviation. 4. The samples suggest substantial overlap is plausible, so the student's claim is too strong.

Answer

The means are \(1.25\) MADs apart. That suggests substantial overlap may remain, so the claim of little overlap is not supported.
5417267
The parallel box plots show random samples of audition scores from two youth orchestra programs. Compare the medians, IQRs, overlap of the boxes, and overlap of the full displayed ranges.
Figure for problem 541726

Hints

- Read the median line and both edges of each box. - Compare the boxes before comparing the full whisker-to-whisker ranges. - Equal box widths indicate equal IQRs even when the boxes are centered differently.

Solution

1. Program A has median \(72\) and IQR \(76-68=8\). 2. Program B has median \(84\) and IQR \(88-80=8\). 3. The medians differ by \(84-72=12\) points, while the middle halves have equal variability. 4. The boxes do not overlap because Program A's third quartile is \(76\) and Program B's first quartile is \(80\). 5. The whiskers overlap from \(74\) to \(82\), so the full displayed ranges are not completely separated.

Answer

Program A: median \(72\), IQR \(8\) Program B: median \(84\), IQR \(8\) The boxes do not overlap, but the whiskers overlap from \(74\) to \(82\). Program B has the higher center, and the samples have equal middle-half variability.
5417277
Two robotics teams test how far, in centimeters, their robots stop from a target line. Team R: mean error \(14\,\text{cm}\), MAD \(1.5\,\text{cm}\) Team S: mean error \(14\,\text{cm}\), MAD \(4.5\,\text{cm}\) Compare the centers and variability. Which team is more consistent?

Hints

- Compare the measures of center separately from the measures of spread. - Decide what a smaller spread means in this context. - Do not use the mean alone to judge consistency.

Solution

1. The mean errors are equal, so the center difference is \(14-14=0\,\text{cm}\). 2. Team R has the smaller MAD because \(1.5<4.5\). 3. The equal centers show no separation between the typical errors, while Team R's smaller MAD shows less variability.

Answer

The teams have the same mean error. Team R is more consistent because its MAD is smaller.
5417347
The two panels show pairs of box plots. Within each pair, the samples have equal IQRs. Which pair has more overlap between its middle halves? Explain using the boxes and medians.
Figure for problem 541734

Hints

- Compare the box widths within and across the two panels. - Locate the numerical intersection of the two boxes in each panel. - When variability is held constant, a larger center shift generally reduces visible overlap.

Solution

1. In Pair 1, the medians are \(30\) and \(34\), and both IQRs are \(8\). The boxes \([26,34]\) and \([30,38]\) overlap from \(30\) to \(34\). 2. In Pair 2, the medians are \(30\) and \(42\), and both IQRs are \(8\). The boxes \([26,34]\) and \([38,46]\) do not overlap. 3. Pair 1 has more middle-half overlap because its centers are closer while the box widths are the same.

Answer

Pair 1 has more overlap between its middle halves: its boxes intersect from \(30\) to \(34\). Pair 2's boxes do not overlap because their medians are farther apart while the IQRs remain equal.
5417377
The parallel box plots show random samples of rehearsal times for two casts. Compare the medians, IQRs, and visible overlap. Which cast appears more variable?
Figure for problem 541737

Hints

- Read the line inside each box for the median. - Compare the widths of the boxes to compare IQRs. - Identify overlap of the boxes and whiskers directly from the display.

Solution

1. Cast X has median \(45\) minutes and IQR \(51-39=12\) minutes. 2. Cast Y has median \(51\) minutes and IQR \(53-49=4\) minutes. 3. Cast X is more variable in the middle half because its IQR is three times as large. 4. The boxes overlap from \(49\) to \(51\) minutes, and the full displayed ranges overlap from \(45\) to \(57\) minutes. 5. Cast Y has the higher median, but substantial overlap is visible.

Answer

Cast X is more variable: its IQR is \(12\) minutes compared with \(4\) minutes for Cast Y. Cast Y's median is \(6\) minutes higher. The boxes overlap from \(49\) to \(51\) minutes, and the full ranges overlap from \(45\) to \(57\) minutes.
5417497
The parallel box plots show two random samples. Compare their centers, IQRs, and full displayed ranges. What can be concluded about overlap in the middle halves?
Figure for problem 541749

Hints

- Compare the median lines and box edges first. - Use the whisker endpoints for the full displayed range. - Separate conclusions about the middle half from conclusions about the extremes.

Solution

1. Both samples have median \(30\). 2. Both samples have IQR \(40-20=20\). 3. Sample A has range \(100-0=100\), while Sample B has range \(50-10=40\). 4. Their boxes are identical, \([20,40]\), so the middle-half intervals overlap completely. 5. Equal medians and IQRs do not imply equal full ranges.

Answer

Both medians are \(30\), and both IQRs are \(20\). Sample A's range is \(100\), while Sample B's range is \(40\). Their middle-half intervals overlap completely because the boxes are identical.
5417577
The parallel box plots show two random samples. Compare the medians, IQRs, and containment of the boxes. What does the display show about center and middle-half variability?
Figure for problem 541757

Hints

- Read the median lines and box edges. - Compare the positions of the boxes as well as their widths. - Containment of one box in another indicates a narrower middle-half interval, not identical data.

Solution

1. Sample A has median \(24\) and IQR \(34-14=20\). 2. Sample B has median \(26\) and IQR \(30-20=10\). 3. Sample B's box \([20,30]\) is entirely inside Sample A's box \([14,34]\). 4. The medians differ by only \(2\), while Sample B's middle-half spread is half as large. 5. The samples have similar centers, but Sample B is much more consistent in the middle half.

Answer

Sample A: median \(24\), IQR \(20\) Sample B: median \(26\), IQR \(10\) Sample B's box is contained within Sample A's. Their centers are similar, but Sample B has much less middle-half variability.
5417597
Sample A has mean \(20\) and MAD \(5\). Sample B has MAD \(3\), and its mean is greater than \(20\). Which Sample B mean would make the two means exactly \(2\) average-MAD units apart? a) \(24\) b) \(26\) c) \(28\) d) \(30\)

Hints

- Average the two MADs to create the common unit. - Convert two common units into a raw center difference. - Use the direction “greater than \(20\)” to choose addition.

Solution

1. The average MAD is \(\frac{5+3}{2}=4\). 2. Two average-MAD units correspond to a center difference of \(2\cdot4=8\). 3. Because Sample B's mean is greater, its mean must be \(20+8=28\). 4. Therefore, choice c is correct.

Answer

c) \(28\)
5417607
The two panels show pairs of box plots. In both pairs, the sample medians differ by \(8\). Which pair has more overlap between its middle halves? Explain how the box widths affect the comparison.
Figure for problem 541760

Hints

- Verify that the median gaps are equal in the two panels. - Compare the numerical intersections of the boxes. - Hold the center difference constant and focus on how box width changes overlap.

Solution

1. In Pair A, each IQR is \(8\). The boxes \([36,44]\) and \([44,52]\) meet only at \(44\). 2. In Pair B, each IQR is \(24\). The boxes \([58,82]\) and \([66,90]\) overlap from \(66\) to \(82\). 3. The median gap is the same in both pairs, but Pair B has much wider middle-half intervals. 4. Pair B therefore shows more overlap between its middle halves.

Answer

Pair B has more middle-half overlap. Pair A's boxes meet only at \(44\), while Pair B's boxes overlap from \(66\) to \(82\). The same center gap produces more overlap when variability is greater.
5419617
The frequency polygons show two random samples with means \(20\) and \(32\) and a common MAD of \(4\). Express the mean difference in MAD units. How does the visual overlap compare with the standardized separation?
Figure for problem 541961

Hints

- Divide the distance between the means by the common MAD. - Compare where each frequency polygon reaches its largest values. - Look for class intervals in which both samples have observations.

Solution

1. The mean difference is \(32-20=12\). 2. The standardized separation is \(12\div4=3\) MADs. 3. The frequency polygons are centered in different regions, with Sample B shifted to the right. 4. The polygons still occupy some common class intervals around the high end of Sample A and the low end of Sample B. 5. A three-MAD separation suggests strong overall separation, but the display confirms that overlap need not be zero.

Answer

The means are \(3\) MADs apart. The frequency polygons show strong separation of the centers with limited visual overlap in nearby class intervals.
5417047
Random samples of monthly utility costs have these summaries. Neighborhood P: mean \(\$40\), MAD \(\$5\) Neighborhood Q: mean \(\$52\), MAD \(\$7\) Use the average of the two MADs as a common measure of variability. How many average-MAD units apart are the means, and what does that suggest?

Hints

- First combine the two similar spread measures into one representative value. - Compare the center difference with that representative spread. - Use cautious language about overlap.

Solution

1. The average MAD is \(\frac{5+7}{2}=6\) dollars. 2. The mean difference is \(52-40=12\) dollars. 3. The separation is \(12\div6=2\) average-MAD units. 4. The samples suggest a noticeable difference in typical cost, with some overlap still possible.

Answer

The means are \(2\) average-MAD units apart. This suggests a noticeable difference, though the distributions may still overlap.
5417057
Two random samples of typing scores have means \(90\) and \(102\), with a common MAD of \(4\). A student says the means are \(\frac{1}{3}\) of a MAD apart because \(4\div12=\frac{1}{3}\). Explain the error and find the correct separation.

Hints

- Decide which quantity is being measured and which quantity is the unit. - The answer should tell how many spread-units fit into the center difference. - Check whether the result is reasonable from \(12\) compared with \(4\).

Solution

1. The difference in means is \(102-90=12\) points. 2. Separation in MAD units requires dividing the center difference by the MAD, not the reverse. 3. The correct ratio is \(12\div4=3\) MADs. 4. A three-MAD separation suggests little overlap.

Answer

The student reversed the ratio. The means are \(3\) MADs apart, suggesting little overlap.
5417067
The parallel dot plots show random samples of daily bicycle counts on two paths. Compute each mean and MAD. Express the difference in means in common-MAD units, and compare that result with the visible overlap.
Figure for problem 541706

Hints

- Find the balance point and average absolute deviation for each dot plot. - Divide the difference between the means by the common MAD. - Compare the ratio with the horizontal region occupied by both distributions.

Solution

1. Path A has mean \(35\) bicycles and MAD \(\frac{40}{8}=5\) bicycles. 2. Path B has mean \(42\) bicycles and MAD \(\frac{40}{8}=5\) bicycles. 3. The mean difference is \(42-35=7\) bicycles, so the separation is \(7\div5=1.4\) MADs. 4. The two dot plots occupy the same horizontal region from about \(34\) to \(43\), so substantial visual overlap remains. 5. The ratio and the display both suggest a moderate shift rather than clear separation.

Answer

Path A: mean \(35\), MAD \(5\) Path B: mean \(42\), MAD \(5\) The means are \(1.4\) MADs apart. The dot plots overlap visibly from about \(34\) to \(43\), so the samples show a moderate difference with substantial overlap.
5417077
The parallel dot plots show random samples of repair times at two shops. Compute each mean and MAD. Then describe the sample overlap and make a cautious inference about the populations.
Figure for problem 541707

Hints

- Compute the mean and average absolute deviation for each displayed sample. - Check whether any part of the two plotted sample ranges occupies the same horizontal region. - Distinguish what the samples show from what can be concluded with certainty about entire populations.

Solution

1. Shop A has mean \(20\) minutes and MAD \(\frac{12}{6}=2\) minutes. 2. Shop B has mean \(27\) minutes and MAD \(\frac{12}{6}=2\) minutes. 3. The means are \((27-20)\div2=3.5\) MADs apart. 4. The displayed samples do not overlap: Shop A's largest time is \(23\) minutes, and Shop B's smallest time is \(24\) minutes. 5. The samples suggest strong population separation, but they do not prove that the full populations never overlap.

Answer

Shop A: mean \(20\) minutes, MAD \(2\) minutes Shop B: mean \(27\) minutes, MAD \(2\) minutes The means are \(3.5\) MADs apart, and the displayed samples do not overlap. This suggests strong separation, but it does not prove zero overlap in the populations.
5417087
Two random samples have a common MAD of \(4\). Population B has the higher mean, and the means are \(2.5\) MADs apart. Population A's mean is \(64\). Find Population B's mean and describe the likely overlap.

Hints

- Convert the number of spread-units into an actual distance. - Use the direction of the comparison to decide whether to add or subtract. - Interpret a separation greater than two variability units.

Solution

1. A separation of \(2.5\) MADs equals \(2.5\cdot4=10\) units. 2. Population B has the higher mean, so its mean is \(64+10=74\). 3. A \(2.5\)-MAD separation suggests limited overlap.

Answer

Population B's mean is \(74\). The distributions likely have limited overlap.
5417097
Two random samples have means \(18\) and \(24\). Their variabilities are similar, and the means are exactly \(2\) MADs apart. Find the common MAD and describe the degree of separation.

Hints

- Find the full distance between the centers. - Divide that distance into the stated number of equal variability units. - Use the number of units to describe overlap cautiously.

Solution

1. The difference in means is \(24-18=6\). 2. If \(6\) equals \(2\) MADs, then one MAD is \(6\div2=3\). 3. A two-MAD separation suggests noticeable separation with some overlap possible.

Answer

The common MAD is \(3\). The distributions are noticeably separated but may still overlap.
5417137
Two random samples of daily messages received are Group A: \(2,\ 4,\ 4,\ 6,\ 6,\ 8,\ 8,\ 10\) Group B: \(10,\ 12,\ 12,\ 14,\ 14,\ 16,\ 16,\ 18\) Compute each median and IQR. Do the middle halves overlap? Do the full sample ranges overlap?

Hints

- Find the median, then find the median of each half of each ordered list. - Use the quartile intervals to compare the middle halves. - Use the minimum and maximum only when comparing the full sample ranges.

Solution

1. Group A has median \(6\), first quartile \(4\), and third quartile \(8\), so its IQR is \(8-4=4\). 2. Group B has median \(14\), first quartile \(12\), and third quartile \(16\), so its IQR is \(16-12=4\). 3. The middle halves are \([4,8]\) and \([12,16]\), so they do not overlap. 4. The full sample ranges are \([2,10]\) and \([10,18]\), so they meet at \(10\). 5. The samples have equal middle-half variability and clearly different centers, but their full ranges are not separated by a gap.

Answer

Group A: median \(6\), IQR \(4\) Group B: median \(14\), IQR \(4\) The middle halves do not overlap. The full sample ranges meet at \(10\).
5417147
Two pairs of random samples are being compared. Pair A: means differ by \(10\), common MAD \(5\) Pair B: means differ by \(12\), common MAD \(8\) A student says Pair B has clearer separation because \(12>10\). Evaluate the reasoning.

Hints

- Raw center differences are not directly comparable when spreads differ. - Put each difference in units of its own variability. - The larger standardized ratio indicates clearer separation.

Solution

1. Pair A's standardized separation is \(10\div5=2\) MADs. 2. Pair B's standardized separation is \(12\div8=1.5\) MADs. 3. Although Pair B has the larger raw difference, Pair A has the larger difference relative to variability. 4. Pair A therefore has clearer likely separation.

Answer

The reasoning is incorrect. Pair A is \(2\) MADs apart, while Pair B is \(1.5\) MADs apart. Pair A has clearer separation relative to its variability.
5417207
The parallel dot plots show random samples of daily production from two lines. Compute the means and MADs. About how many MADs apart are the means, and how does that compare with the visible overlap?
Figure for problem 541720

Hints

- Find each dot plot's balance point before measuring absolute deviations. - Divide the distance between the means by the common MAD. - Compare the standardized result with the horizontal region occupied by both samples.

Solution

1. Line A has mean \(8\) and MAD \(\frac{12}{5}=2.4\). 2. Line B has mean \(13\) and MAD \(\frac{12}{5}=2.4\). 3. The mean difference is \(13-8=5\). 4. The separation is \(5\div2.4\approx2.1\) MADs. 5. The displayed sample ranges overlap from \(9\) to \(12\), so the plots show a noticeable shift with limited, not zero, overlap.

Answer

Line A: mean \(8\), MAD \(2.4\) Line B: mean \(13\), MAD \(2.4\) The means are about \(2.1\) MADs apart. The sample ranges overlap from \(9\) to \(12\), so the distributions are noticeably shifted but not completely separated.
5417217
Two pairs of random samples have the same raw difference between means. Pair X: means \(10\) and \(18\), common MAD \(2\) Pair Y: means \(30\) and \(38\), common MAD \(5\) Which pair likely has less overlap? Explain using MAD units.

Hints

- The raw center differences are equal, so compare the spreads. - Express each difference in units of its own MAD. - The larger ratio indicates less overlap.

Solution

1. Pair X has separation \((18-10)\div2=4\) MADs. 2. Pair Y has separation \((38-30)\div5=1.6\) MADs. 3. Pair X has the larger standardized separation, so it likely has less overlap.

Answer

Pair X likely has less overlap because its means are \(4\) MADs apart, compared with \(1.6\) MADs for Pair Y.
5417227
The parallel dot plots show random samples of weekly practice times for two groups. Compute each mean and MAD. Express the mean difference in MAD units, and compare that result with the visible overlap.
Figure for problem 541722

Hints

- Find the balance point and average absolute deviation for each dot plot. - Divide the difference between the means by the common MAD. - Look for values and horizontal regions occupied by both samples.

Solution

1. Group A has mean \(55\) minutes and MAD \(\frac{24}{6}=4\) minutes. 2. Group B has mean \(61\) minutes and MAD \(\frac{24}{6}=4\) minutes. 3. The mean difference is \(61-55=6\) minutes, so the separation is \(6\div4=1.5\) MADs. 4. The displayed samples share values at \(57\) and \(59\) minutes, and their full ranges overlap from \(55\) to \(61\) minutes. 5. The numerical and visual evidence both indicate a moderate shift with substantial overlap.

Answer

Group A: mean \(55\) minutes, MAD \(4\) minutes Group B: mean \(61\) minutes, MAD \(4\) minutes The means are \(1.5\) MADs apart. The samples share values at \(57\) and \(59\) minutes, so they overlap substantially.
5417247
Two random samples have these summaries. Population C: mean \(48\), MAD \(3\) Population D: mean \(54\), MAD \(5\) Use the average MAD as the common variability unit. Explain why this convention is more balanced than automatically using only the smaller or only the larger MAD, then measure the separation.

Hints

- Create one variability unit using both MADs. - Consider what would happen if only the tighter or only the wider group set the scale. - Divide the mean difference by the common unit and interpret the result cautiously.

Solution

1. The average MAD is \(\frac{3+5}{2}=4\). 2. This common unit uses information from both populations instead of favoring the tighter or wider distribution alone. 3. The mean difference is \(54-48=6\). 4. The separation is \(6\div4=1.5\) common MADs. 5. This suggests a moderate difference with substantial overlap still plausible.

Answer

The average MAD is \(4\), which treats both groups symmetrically. The means are \(1.5\) common MADs apart, suggesting a moderate difference with plausible overlap.
5417257
Random samples of weekly sales have these summaries. Store M: mean \(30\), MAD \(2\) Store N: mean \(39\), MAD \(4\) Use the average of the two MADs to measure separation. A manager calls the distributions “nearly identical.” Is that description reasonable?

Hints

- Form one representative variability unit from the two MADs. - Compare the center difference with that unit. - Test the manager's wording against the size of the ratio.

Solution

1. The average MAD is \(\frac{2+4}{2}=3\). 2. The mean difference is \(39-30=9\). 3. The separation is \(9\div3=3\) average-MAD units. 4. A three-unit separation is not consistent with “nearly identical”; it suggests little overlap.

Answer

No. The means are \(3\) average-MAD units apart, suggesting little overlap rather than nearly identical distributions.
5417287
Two random samples of daily quiz scores have these summaries. Class A: mean \(62\), MAD \(5\) Class B: mean \(69\), MAD \(4\) A student draws a segment from one MAD below to one MAD above each mean. The student then uses the percentage of segment length that overlaps to estimate the percentage of scores shared by the populations. Explain why that method is not valid. Then measure the mean difference using the average MAD.

Hints

- Recall what MAD measures: an average distance, not an endpoint. - Ask whether the summaries identify where every observation lies. - Use the average MAD only as a scale for the center difference, not as a distribution boundary.

Solution

1. A MAD is an average distance from the mean. It is not a boundary containing a fixed percentage of the data. 2. Therefore, overlap between mean-centered segments cannot be converted into a percentage of scores or a percentage of population overlap. 3. The average MAD is \(\frac{5+4}{2}=4.5\). 4. The mean difference is \(69-62=7\), so the standardized separation is \(7\div4.5\approx1.56\) common MADs. 5. This suggests a moderate difference with substantial overlap still plausible, but the summaries do not determine an exact overlap percentage.

Answer

The method is invalid because mean \(\pm\) MAD does not contain a fixed percentage of a distribution. Segment-length overlap therefore does not equal data overlap. The means are about \(1.56\) common MADs apart, suggesting a moderate difference with plausible overlap but no exact overlap percentage.
5417297
Random samples of library checkout times, in minutes, have these summaries. Group A: \(Q_1=14\), median \(18\), \(Q_3=22\) Group B: \(Q_1=21\), median \(25\), \(Q_3=29\) A student says, “The middle-half intervals overlap from \(21\) to \(22\), so only one minute of actual checkout data is shared.” Evaluate the statement and compare the summaries.

Hints

- Compute each IQR and compare the medians. - Distinguish an interval of possible values from a count of observations. - Decide what quartiles reveal and what individual-data information they leave unknown.

Solution

1. Each IQR is \(22-14=8\) minutes and \(29-21=8\) minutes. 2. The median difference is \(25-18=7\) minutes. 3. The numerical intervals for the middle halves intersect on \([21,22]\). 4. However, quartile summaries do not identify the individual observations or how many observations have similar values. 5. The samples have equal middle-half variability and different centers, but the amount or percentage of actual data overlap cannot be determined from these summaries alone.

Answer

The middle-half intervals intersect on \([21,22]\), but this does not mean that exactly one minute of observed data is shared or reveal how many observations overlap. Both IQRs are \(8\) minutes, and Group B's median is \(7\) minutes higher.
5417307
Two populations have sample means of \(40\) and \(52\). In Study 1, both samples have MAD \(3\). In Study 2, both samples have MAD \(6\). For each study, express the difference between the means in MAD units. Which study suggests more overlap between the populations?

Hints

- The raw difference between the centers is the same in both studies. - Compare that difference with the variability in each study. - A smaller standardized separation generally indicates more overlap.

Solution

1. The difference between the means is \(52-40=12\). 2. In Study 1, the separation is \(12\div3=4\) MADs. 3. In Study 2, the separation is \(12\div6=2\) MADs. 4. Study 2 suggests more overlap because the same center difference is smaller relative to its variability.

Answer

Study 1: \(4\) MADs apart. Study 2: \(2\) MADs apart. Study 2 suggests more overlap.
5417317
The box plots show random samples of delivery times for two courier services. Which service has the larger IQR? What numerical interval is shared by the two boxes, and what can that intersection tell you?
Figure for problem 541731

Hints

- Use the left and right edges of each box to find its IQR. - Find the intersection of the two numerical box intervals. - Distinguish the middle-half interval of a sample from the individual values or counts inside that interval.

Solution

1. Service A has IQR \(18-8=10\) minutes. 2. Service B has IQR \(22-14=8\) minutes, so Service A has the larger IQR. 3. The boxes cover \([8,18]\) and \([14,22]\). 4. Their intersection is \([14,18]\), an interval with length \(18-14=4\) minutes. 5. This interval is the numerical region common to both middle-half intervals. It does not show how many actual observations are shared or imply that \(50\%\) of either sample lies in the intersection.

Answer

Service A has the larger IQR: \(10\) minutes compared with \(8\) minutes. The boxes intersect on \([14,18]\), a \(4\)-minute interval. This is the region common to the two middle-half intervals, not a count or percentage of shared observations.
5417327
Two samples have means \(26\) and \(36\). Their MADs are \(2\) and \(5\). Measure the mean difference using the smaller MAD, the larger MAD, and the average MAD. Explain why the average MAD is the most balanced default when one common variability unit is needed.

Hints

- Measure the same center difference with each proposed unit. - Notice how a smaller measuring unit produces a larger numerical ratio. - Choose the convention that incorporates both groups equally when no special reason favors one group.

Solution

1. The mean difference is \(36-26=10\). 2. Using the smaller MAD, the separation is \(10\div2=5\) units. 3. Using the larger MAD, the separation is \(10\div5=2\) units. 4. The average MAD is \(\frac{2+5}{2}=3.5\), so the separation is \(10\div3.5\approx2.86\) common-MAD units. 5. The smaller MAD makes the separation look largest, while the larger MAD makes it look smallest. The average MAD uses information from both samples and treats them symmetrically.

Answer

Smaller-MAD scale: \(5\) units Larger-MAD scale: \(2\) units Average-MAD scale: \(\frac{10}{3.5}\approx2.86\) units The average MAD is the balanced default because it uses both groups' variability rather than choosing only the tighter or wider group.
5417367
Two random samples have these summaries. Population A: mean \(70\), MAD \(4\) Population B: mean \(86\), MAD \(5\) Which statement is best supported when the average MAD is used as the common variability unit? a) The populations cannot overlap because their means differ by \(16\). b) The means are about \(3.6\) common MADs apart, suggesting strong separation but not proving zero overlap. c) The means are \(4\) common MADs apart because only Population A's MAD should be used. d) The distributions are nearly identical because their MADs differ by only \(1\).

Hints

- Average the two MADs before measuring the center difference. - Distinguish strong evidence of separation from a claim of impossible overlap. - Check every option against both the center and variability information.

Solution

1. The average MAD is \(\frac{4+5}{2}=4.5\). 2. The mean difference is \(86-70=16\). 3. The standardized separation is \(16\div4.5\approx3.6\) common MADs. 4. This is strong evidence of separated centers and limited likely overlap, but summary statistics cannot prove that no population values overlap. 5. Therefore, statement b is best supported.

Answer

b) The means are about \(3.6\) common MADs apart, suggesting strong separation but not proving zero overlap.
5417407
Two packaging machines are set to fill bags near a target of \(503\,\text{g}\). Machine A: mean \(500\,\text{g}\), MAD \(2\,\text{g}\) Machine B: mean \(506\,\text{g}\), MAD \(3\,\text{g}\) Compare the machines' accuracy and consistency. Then express the difference between their means in average-MAD units.

Hints

- Compare each center with the target separately. - Use the spread measures to judge consistency. - Standardize the distance between the two centers using both spread values.

Solution

1. Each mean is \(3\,\text{g}\) from the target, so the machines are equally accurate by their sample means. 2. Machine A is more consistent because \(2<3\). 3. The mean difference is \(506-500=6\,\text{g}\). 4. The average MAD is \(\frac{2+3}{2}=2.5\,\text{g}\). 5. The means are \(6\div2.5=2.4\) average-MAD units apart.

Answer

The machines are equally accurate by their means. Machine A is more consistent. Their means are \(2.4\) average-MAD units apart.
5417427
Two samples have these summaries. Sample A: mean \(10\), MAD \(3\) Sample B: mean \(18\), MAD \(6\) A student adds the MADs and says, “The mean difference \(8\) is less than \(3+6=9\), so the centers are less than one common variability unit apart.” Check the claim using the average MAD.

Hints

- Decide whether the sum or the average of the two MADs represents one shared unit. - Divide the center difference by that common unit. - Interpret the result without treating either MAD as a boundary around a distribution.

Solution

1. The mean difference is \(18-10=8\). 2. The average MAD is \(\frac{3+6}{2}=4.5\). 3. The standardized separation is \(8\div4.5\approx1.8\) common-MAD units. 4. The sum \(3+6\) is twice the average MAD, not one common variability unit. 5. The student's claim is false. The centers are about \(1.8\) common MADs apart, suggesting a moderate difference with overlap still plausible.

Answer

The claim is false. The average MAD is \(4.5\), and \(8\div4.5\approx1.8\). The centers are about \(1.8\) common-MAD units apart; adding the two MADs does not create one common unit.
5417447
Two samples have means \(18\) and \(30\), with MADs \(3\) and \(5\). A student writes, “The average mean is \(24\), so the means are \(12\div24=0.5\) variability unit apart.” Identify the error and find the correct separation using the average MAD.

Hints

- Decide which quantities describe center and which describe spread. - The measuring unit for overlap should come from variability. - Recalculate the ratio with the appropriate kind of measure.

Solution

1. The student used the average of the means as the variability unit, but variability must be measured with the MADs. 2. The mean difference is \(30-18=12\). 3. The average MAD is \(\frac{3+5}{2}=4\). 4. The correct separation is \(12\div4=3\) average-MAD units.

Answer

The error is using the average mean instead of the average MAD. The correct separation is \(3\) average-MAD units.
5417457
The parallel dot plots show random samples of seedling heights for two varieties. A student says the populations cannot overlap because their sample means are different. Compute the means and MADs, compare the centers in MAD units, and evaluate the statement using the display.
Figure for problem 541745

Hints

- Find each dot plot's balance point and average absolute deviation. - Measure the center difference using the common MAD. - Check the displayed values before making an absolute claim about overlap.

Solution

1. Variety A has mean \(24\,\text{cm}\) and MAD \(\frac{12}{6}=2\,\text{cm}\). 2. Variety B has mean \(28\,\text{cm}\) and MAD \(\frac{12}{6}=2\,\text{cm}\). 3. The mean difference is \(28-24=4\,\text{cm}\), which is \(4\div2=2\) MADs. 4. The displayed samples share heights of \(25\), \(26\), and \(27\) centimeters. 5. Different means and a two-MAD separation do not prove zero overlap; the student's statement is false.

Answer

The sample means are \(24\,\text{cm}\) and \(28\,\text{cm}\), and both MADs are \(2\,\text{cm}\). The means are \(2\) MADs apart, but the samples overlap at \(25\), \(26\), and \(27\) centimeters. The claim of no population overlap is not justified.
5417467
Two samples have MADs \(4\) and \(8\). Their means are \(2.5\) average-MAD units apart. The lower mean is \(38\). Find the higher mean.

Hints

- First determine the variability unit shared by the comparison. - Convert the number of variability units into a raw center difference. - Use the known center to locate the other one.

Solution

1. The average MAD is \(\frac{4+8}{2}=6\). 2. A separation of \(2.5\) average-MAD units corresponds to a mean difference of \(2.5\cdot6=15\). 3. The higher mean is \(38+15=53\).

Answer

The higher mean is \(53\).
5417477
The parallel box plots show two random samples. Find the numerical overlap of the boxes. What fraction of each box's interval length is covered by that overlap? Explain why these fractions are not percentages of observations.
Figure for problem 541747

Hints

- Read the left and right edges of each box. - Divide the shared interval length by each full box length separately. - Distinguish geometric length on the scale from the number of observations inside an interval.

Solution

1. Sample A's box is \([30,42]\), so its IQR is \(42-30=12\). 2. Sample B's box is \([36,44]\), so its IQR is \(44-36=8\). 3. The boxes overlap on \([36,42]\), which has length \(42-36=6\). 4. The overlap is \(\frac{6}{12}=\frac12\) of Sample A's box length and \(\frac{6}{8}=\frac34\) of Sample B's box length. 5. These fractions compare numerical interval lengths. A box plot does not show how observations are distributed inside each box, so the fractions are not percentages of data points.

Answer

The boxes overlap on \([36,42]\), a length of \(6\). This is \(\frac12\) of Sample A's IQR length and \(\frac34\) of Sample B's IQR length. Those fractions describe interval lengths, not proportions of observations.
5417487
The parallel dot plots show two random samples. Compute each mean and MAD. Compare the centers using the average MAD, and describe what the display shows about variability and overlap.
Figure for problem 541748

Hints

- Find each balance point and average absolute distance from it. - Average the MADs before measuring the center difference. - Compare the horizontal spread of the two dot clusters separately from their centers.

Solution

1. Sample A has mean \(50\) and MAD \(\frac{60}{6}=10\). 2. Sample B has mean \(52\) and MAD \(\frac{24}{6}=4\). 3. The average MAD is \(\frac{10+4}{2}=7\), and the mean difference is \(52-50=2\). 4. The standardized separation is \(2\div7=\frac27\approx0.29\) common MAD. 5. The centers are very close relative to the variability. Sample B is much more tightly clustered, and its displayed range lies inside Sample A's displayed range.

Answer

Sample A: mean \(50\), MAD \(10\) Sample B: mean \(52\), MAD \(4\) The means are about \(0.29\) common MAD apart. The centers are very similar, while Sample B is much less variable and its displayed range is contained within Sample A's.
5417517
The box plots show random samples of completion times for two puzzle teams. Do the boxes overlap? Do the whisker intervals overlap? Find the gap or overlap length in each case.
Figure for problem 541751

Hints

- Treat the box interval and the full whisker interval as different comparisons. - Compare the nearest box edges first. - Then identify the values shared by the two whisker intervals.

Solution

1. Team A's box covers \([14,\ 22]\), and Team B's box covers \([24,\ 32]\). 2. The boxes do not overlap; the gap is \(24-22=2\) minutes. 3. Team A's whisker interval is \([10,\ 30]\), and Team B's is \([20,\ 40]\). 4. The whisker intervals overlap from \(20\) to \(30\), a length of \(30-20=10\) minutes. 5. The middle halves are separated even though the wider displayed intervals overlap.

Answer

The boxes have a \(2\)-minute gap. The whisker intervals overlap by \(10\) minutes.
5417537
Two populations have means \(10\) units apart. At first, both samples have MAD \(5\). Later, the first sample still has MAD \(5\), but the second sample has MAD \(15\). Using the average MAD each time, compare the standardized separations. Which situation suggests more overlap?

Hints

- The distance between the centers does not change. - Recalculate the variability unit for each situation. - Compare how large the fixed distance is relative to each unit.

Solution

1. At first, the average MAD is \(\frac{5+5}{2}=5\), so the separation is \(10\div5=2\) units. 2. Later, the average MAD is \(\frac{5+15}{2}=10\), so the separation is \(10\div10=1\) unit. 3. The later situation suggests more overlap because the center difference is smaller relative to the variability.

Answer

The standardized separation changes from \(2\) average-MAD units to \(1\) average-MAD unit. The later situation suggests more overlap.
5417547
Sample A has middle-half interval \([12,\ 20]\). Sample B has \(Q_3=26\) and IQR \(10\). Find Sample B's \(Q_1\). Then find the length of overlap between the two middle-half intervals.

Hints

- Use the relationship between the two quartiles and the middle-half spread. - Write the second interval after finding its missing endpoint. - Identify the portion common to both intervals.

Solution

1. Since the IQR is \(Q_3-Q_1\), \(26-Q_1=10\). 2. Therefore, \(Q_1=16\), so Sample B's middle-half interval is \([16,\ 26]\). 3. The intervals overlap from \(16\) to \(20\). 4. The overlap length is \(20-16=4\).

Answer

Sample B has \(Q_1=16\), and the middle-half intervals overlap by \(4\).
5417557
Random samples of model airplane flight times have these summaries. Design A: mean \(3.8\,\text{s}\), MAD \(0.2\,\text{s}\) Design B: mean \(4.7\,\text{s}\), MAD \(0.4\,\text{s}\) A student says the means are “about one MAD apart.” Use the average MAD to check the statement.

Hints

- Find the distance between the two centers. - Form one variability unit from the two spread measures. - Compare the resulting ratio with the student's estimate.

Solution

1. The mean difference is \(4.7-3.8=0.9\,\text{s}\). 2. The average MAD is \(\frac{0.2+0.4}{2}=0.3\,\text{s}\). 3. The separation is \(0.9\div0.3=3\) average-MAD units. 4. The student's statement is incorrect; the means are three average-MAD units apart.

Answer

The statement is incorrect. The means are \(3\) average-MAD units apart.
5417567
Sample A has mean \(32\) and MAD \(4\). Sample B has mean \(38\) and MAD \(3\). Every value in Sample B increases by \(1\). Find Sample B's new mean and MAD. Then compare the standardized separation before and after the shift using the average MAD.

Hints

- A uniform shift changes a center but not distances from that center. - Use the same average MAD before and after the shift. - Compare the two standardized center differences.

Solution

1. The original average MAD is \(\frac{4+3}{2}=3.5\). 2. The original mean difference is \(38-32=6\), so the original separation is \(6\div3.5\approx1.71\) common MADs. 3. Adding \(1\) to every Sample B value raises its mean to \(39\) and leaves its MAD at \(3\). 4. The new mean difference is \(39-32=7\), so the new separation is \(7\div3.5=2\) common MADs. 5. The uniform shift increases the center separation, so less overlap would generally be expected.

Answer

Sample B's new mean is \(39\), and its MAD remains \(3\). The standardized separation increases from about \(1.71\) to \(2\) common MADs.
5417587
Consider two comparisons. Pair 1: means \(50\) and \(52\), with common MAD \(5\) Pair 2: means \(50\) and \(70\), with common MAD \(2\) A student says Pair 2 should overlap more because its MAD is smaller. Use MAD units to evaluate the claim.

Hints

- Compare each center difference with its own variability. - Do not judge overlap from spread alone. - Focus on the ratio between separation and spread.

Solution

1. Pair 1 has mean difference \(52-50=2\), so its separation is \(2\div5=0.4\) MAD unit. 2. Pair 2 has mean difference \(70-50=20\), so its separation is \(20\div2=10\) MAD units. 3. Pair 1 should have much more overlap because its centers are far closer relative to its variability. 4. A smaller MAD alone does not imply more overlap.

Answer

The claim is false. Pair 1 is \(0.4\) MAD unit apart, while Pair 2 is \(10\) MAD units apart, so Pair 1 should overlap more.
5417037
The parallel dot plots show random samples of package-processing times at two centers. Compute each mean and MAD. Then express the difference in means as a multiple of the common MAD and compare the numerical separation with the visible overlap.
Figure for problem 541703

Hints

- Use the balance point of each dot plot to find its mean. - Average the absolute distances from each mean to find the MADs. - Compare the mean difference in MAD units with the values shared by the two displays.

Solution

1. Center A has mean \(14\) minutes and MAD \(\frac{12}{5}=2.4\) minutes. 2. Center B has mean \(20\) minutes and MAD \(\frac{12}{5}=2.4\) minutes. 3. The mean difference is \(20-14=6\) minutes. 4. The separation is \(6\div2.4=2.5\) MADs. 5. The plots share values at \(16\) and \(18\) minutes, but most of Center B's distribution lies to the right of Center A's. The ratio and display both suggest limited, not zero, overlap.

Answer

Center A: mean \(14\) minutes, MAD \(2.4\) minutes Center B: mean \(20\) minutes, MAD \(2.4\) minutes The means are \(2.5\) MADs apart. The plots overlap at \(16\) and \(18\) minutes, but the distributions are noticeably separated overall.
5417107
Three pairs of random samples have equal MADs within each pair. Pair 1: means \(40\) and \(48\), MAD \(4\) Pair 2: means \(70\) and \(77\), MAD \(7\) Pair 3: means \(15\) and \(24\), MAD \(3\) Rank the pairs from greatest likely overlap to least likely overlap.

Hints

- Put each center difference in units of its pair's spread. - Compare the resulting ratios rather than the raw differences. - Smaller ratios indicate more overlap.

Solution

1. Pair 1 has separation \((48-40)\div4=2\) MADs. 2. Pair 2 has separation \((77-70)\div7=1\) MAD. 3. Pair 3 has separation \((24-15)\div3=3\) MADs. 4. Smaller separation means greater likely overlap, so the order is Pair 2, Pair 1, Pair 3.

Answer

Greatest to least likely overlap: Pair 2, Pair 1, Pair 3.
5417177
Every value in a random sample from Population B is \(9\) greater than the corresponding value in a random sample from Population A. Population A has mean \(27\) and MAD \(3\). Find Population B's mean and MAD. Then express the difference in means in MAD units.

Hints

- Think about how a uniform shift affects a center. - Distances from the shifted center stay the same. - Use the unchanged spread to measure the center shift.

Solution

1. Adding \(9\) to every value raises the mean from \(27\) to \(36\). 2. A uniform shift does not change distances from the mean, so Population B's MAD remains \(3\). 3. The mean difference is \(36-27=9\), and \(9\div3=3\) MADs.

Answer

Population B has mean \(36\) and MAD \(3\). The means are \(3\) MADs apart, suggesting little overlap.
5417337
Two random samples have means \(43\) and \(55\). The means are \(1.5\) average-MAD units apart. One sample has MAD \(6\). Find the other sample's MAD.

Hints

- Use the center difference and the stated number of variability units to recover the average variability. - Relate that average to the two individual spread values. - Check that the result produces the stated separation.

Solution

1. The difference between the means is \(55-43=12\). 2. If \(12\) is \(1.5\) average-MAD units, the average MAD is \(12\div1.5=8\). 3. Let the unknown MAD be \(m\). Then \(\frac{6+m}{2}=8\). 4. Solving gives \(6+m=16\), so \(m=10\).

Answer

The other sample's MAD is \(10\).
5417357
Two random samples of daily customer wait times, in minutes, are: Café A: \(11,\ 13,\ 14,\ 15,\ 16,\ 21\) Café B: \(20,\ 21,\ 23,\ 24,\ 25,\ 31\) Find each mean and MAD. Then express the difference between the means in average-MAD units and interpret the result.

Hints

- Find each center before measuring absolute distances from it. - Average the two MADs to create the common variability unit. - Divide the mean difference by that unit, then interpret the size of the result cautiously.

Solution

1. Café A's mean is \(\frac{11+13+14+15+16+21}{6}=15\). 2. Its absolute deviations are \(4,\ 2,\ 1,\ 0,\ 1,\ 6\), so its MAD is \(\frac{14}{6}=\frac{7}{3}\). 3. Café B's mean is \(\frac{20+21+23+24+25+31}{6}=24\). 4. Its absolute deviations are \(4,\ 3,\ 1,\ 0,\ 1,\ 7\), so its MAD is \(\frac{16}{6}=\frac{8}{3}\). 5. The average MAD is \(\frac{\frac{7}{3}+\frac{8}{3}}{2}=2.5\), and the mean difference is \(24-15=9\). 6. The standardized separation is \(9\div2.5=3.6\) average-MAD units, suggesting a strong shift and limited likely overlap.

Answer

Café A: mean \(15\), MAD \(\frac{7}{3}\) Café B: mean \(24\), MAD \(\frac{8}{3}\) The means are \(3.6\) average-MAD units apart, suggesting a strong difference in typical wait time and limited likely overlap.
5417387
The parallel dot plots show random samples from two populations. Compute each mean and MAD. Then use the average MAD to measure the center difference and compare the result with the visible sample overlap.
Figure for problem 541738

Hints

- Find each dot plot's mean and average absolute distance from that mean. - Average the two MADs before standardizing the center difference. - Compare the numerical separation with values occupied by both samples.

Solution

1. Population P has mean \(108\) and MAD \(\frac{36}{6}=6\). 2. Population Q has mean \(120\) and MAD \(\frac{24}{6}=4\). 3. The average MAD is \(\frac{6+4}{2}=5\), and the mean difference is \(120-108=12\). 4. The means are \(12\div5=2.4\) common MADs apart. 5. The samples share displayed values at \(114\) and \(118\), so the visual overlap is limited but not zero.

Answer

Population P: mean \(108\), MAD \(6\) Population Q: mean \(120\), MAD \(4\) The means are \(2.4\) common MADs apart. The samples overlap at \(114\) and \(118\), showing noticeable separation with some overlap.
5417397
The parallel dot plots show random samples of the number of minutes students spent reading. Find each median and IQR. Compare the middle-half intervals and the visible overlap of the full samples.
Figure for problem 541739

Hints

- Count the ordered dots to locate each median and the medians of the lower and upper halves. - Use the quartiles to compare the middle-half intervals. - Compare that result with the complete horizontal regions occupied by both dot plots.

Solution

1. Group A has median \(\frac{7+7}{2}=7\), \(Q_1=\frac{5+6}{2}=5.5\), and \(Q_3=\frac{8+9}{2}=8.5\), so its IQR is \(8.5-5.5=3\). 2. Group B has median \(\frac{10+10}{2}=10\), \(Q_1=\frac{8+9}{2}=8.5\), and \(Q_3=\frac{11+12}{2}=11.5\), so its IQR is \(11.5-8.5=3\). 3. The middle-half intervals \([5.5,8.5]\) and \([8.5,11.5]\) meet only at \(8.5\). 4. The full samples overlap visibly from \(7\) to \(10\) minutes. 5. The samples have equal middle-half variability, different centers, and more full-distribution overlap than the quartile intervals alone show.

Answer

Group A: median \(7\), IQR \(3\) Group B: median \(10\), IQR \(3\) The middle-half intervals meet only at \(8.5\), while the full displayed samples overlap from \(7\) to \(10\) minutes.
5417417
The box plots show random samples of event setup times for two crews. Compare the IQRs and the overlap of the boxes. Then compare the overall data ranges, including the marked outlier.
Figure for problem 541741

Hints

- Use the edges of each box for the middle-half spread. - Find the interval common to both boxes. - For the overall range, include any point shown beyond a whisker.

Solution

1. Crew A has IQR \(32-24=8\) minutes. 2. Crew B has IQR \(34-26=8\) minutes, so the IQRs are equal. 3. The boxes overlap from \(26\) to \(32\), a length of \(32-26=6\) minutes. 4. Including the outlier at \(50\), Crew A's overall range is \(50-20=30\) minutes. 5. Crew B's overall range is \(38-22=16\) minutes.

Answer

Both IQRs are \(8\) minutes. The boxes overlap by \(6\) minutes. Including the outlier, Crew A's range is \(30\) minutes and Crew B's range is \(16\) minutes.
5417437
Three pairs of random samples have these summaries. Pair A: mean difference \(6\), MADs \(2\) and \(2\) Pair B: mean difference \(9\), MADs \(6\) and \(6\) Pair C: mean difference \(8\), MADs \(3\) and \(5\) Use the average MAD for each pair. Rank the pairs from greatest expected overlap to least expected overlap.

Hints

- Build a separate variability unit for each pair. - Compare each center difference with its own unit. - Order the pairs by standardized separation, not by raw difference.

Solution

1. Pair A has separation \(6\div2=3\) average-MAD units. 2. Pair B has separation \(9\div6=1.5\) average-MAD units. 3. Pair C has average MAD \(\frac{3+5}{2}=4\), so its separation is \(8\div4=2\) units. 4. Smaller standardized separation suggests greater overlap. 5. The order from greatest to least expected overlap is Pair B, Pair C, Pair A.

Answer

Greatest to least expected overlap: Pair B, Pair C, Pair A.
5417507
The parallel dot plots show two random samples. Find each mean and MAD. Then measure the mean difference using the average MAD and compare the result with the visible sample overlap.
Figure for problem 541750

Hints

- Compute each mean before averaging absolute deviations. - Average the two MADs to create the common variability unit. - Compare the standardized separation with the horizontal values occupied by both samples.

Solution

1. Sample A has mean \(10\). Its absolute deviations sum to \(2+1+0+1+2=6\), so its MAD is \(6\div5=1.2\). 2. Sample B has mean \(16\). Its absolute deviations sum to \(4+2+0+2+4=12\), so its MAD is \(12\div5=2.4\). 3. The average MAD is \(\frac{1.2+2.4}{2}=1.8\), and the mean difference is \(16-10=6\). 4. The standardized separation is \(6\div1.8=\frac{10}{3}\approx3.3\) common MADs. 5. The displayed samples meet at \(12\), so the strong center separation does not mean the sample ranges are separated by a gap.

Answer

Sample A: mean \(10\), MAD \(1.2\) Sample B: mean \(16\), MAD \(2.4\) The means are about \(3.3\) common MADs apart. The displayed samples meet at \(12\), showing strong separation with a small amount of sample overlap.
5417527
The parallel dot plots show random samples of temperature changes, in degrees Fahrenheit, from two regions. Compute each mean and MAD. Express the mean difference in average-MAD units, and compare that result with the visible overlap.
Figure for problem 541752

Hints

- Account carefully for negative values when finding each mean. - Average the two MADs before measuring the center difference. - Compare the numerical separation with values and regions occupied by both dot plots.

Solution

1. Region A has mean \(-4\,^{\circ}\text{F}\) and MAD \(\frac{18}{6}=3\,^{\circ}\text{F}\). 2. Region B has mean \(5\,^{\circ}\text{F}\) and MAD \(\frac{36}{6}=6\,^{\circ}\text{F}\). 3. The mean difference is \(5-(-4)=9\,^{\circ}\text{F}\). 4. The average MAD is \(\frac{3+6}{2}=4.5\,^{\circ}\text{F}\), so the separation is \(9\div4.5=2\) common MADs. 5. The samples share values at \(-5\,^{\circ}\text{F}\) and \(-1\,^{\circ}\text{F}\), and their displayed ranges overlap from \(-5\,^{\circ}\text{F}\) to \(1\,^{\circ}\text{F}\).

Answer

Region A: mean \(-4\,^{\circ}\text{F}\), MAD \(3\,^{\circ}\text{F}\) Region B: mean \(5\,^{\circ}\text{F}\), MAD \(6\,^{\circ}\text{F}\) The means are \(2\) common MADs apart. The samples still overlap visibly, including shared values at \(-5\,^{\circ}\text{F}\) and \(-1\,^{\circ}\text{F}\).

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