Free math worksheets

- Count every dot, including dots stacked at the same x-value. - Read the horizontal-axis label to identify the measured attribute and unit. - Look for the shortest interval containing most of the dots.

1. There are six dots, so \(6\) seedlings were measured. 2. The attribute is seedling height, measured in centimeters. 3. Five of the six observations are at \(9\) or \(10\) centimeters, so the data are concentrated near \(9\) to \(10\) centimeters.

\(6\) seedlings; seedling height in centimeters; most observations are concentrated at \(9\) to \(10\,\text{cm}\).

- Read the frequency of each equal-width interval. - Identify any interval with no observations. - Ask whether one center value would communicate the visible gap.

1. The frequencies are \(2,5,0,4\), totaling \(11\) calls. 2. Seven calls are from \(0\) to less than \(20\) minutes, while four calls are from \(30\) to less than \(40\) minutes. 3. The interval from \(20\) to less than \(30\) minutes is empty, creating a clear gap between the two occupied groups. 4. One typical-value statement would hide the separated groups and the gap, so the shape of the distribution must also be described.

The distribution has one group from \(0\) to less than \(20\) minutes and another from \(30\) to less than \(40\) minutes, with a gap from \(20\) to less than \(30\) minutes. A single “typical” value would hide that two-group structure.

- Read the median line and the two box edges from the scale. - Use the whisker endpoints for the full range. - Separate what a five-number summary shows from details it compresses away.

1. The median commute time is \(18\) minutes. 2. The interquartile range is \(21-16=5\) minutes, so the middle half of commute times spans \(5\) minutes. 3. The range is \(24-12=12\) minutes. 4. A box plot does not reveal exact individual values, frequencies at particular values, or internal gaps and clusters.

The median commute time is \(18\) minutes. The middle \(50\%\) of commute times runs from \(16\) to \(21\) minutes, an IQR of \(5\) minutes, and the full range is \(12\) minutes. The box plot does not show exact individual values or internal gaps/clusters.

- Count every dot, including repeated values in a stack. - Describe the visible group before focusing on the single point far to the right. - With seven observations, the median is the fourth ordered value; use the two extremes for the range.

1. There are \(7\) observations, measured in minutes. 2. Six of the seven times are concentrated from \(4\) through \(6\) minutes. 3. The value \(12\) minutes is separated from that main cluster. 4. In order, the fourth value is \(6\), so the median is \(6\) minutes. 5. The range is \(12-4=8\) minutes.

There are \(7\) completion times, measured in minutes. Six are concentrated from \(4\) to \(6\) minutes, with a separated value at \(12\) minutes. The median is \(6\) minutes and the range is \(8\) minutes.

- Find the mean of each data set separately. - Compare the smallest and largest values in each set. - A more consistent data set has values that are closer together.

1. Class 6A collected \(2.6+1.8+3.2+2.4+2.0=12.0\) pounds, so its mean is \(12.0\div5=2.4\) pounds. 2. Class 6B collected \(1.0+4.8+2.2+0.8+3.2=12.0\) pounds, so its mean is also \(12.0\div5=2.4\) pounds. 3. Class 6A’s values range from \(1.8\) to \(3.2\), while Class 6B’s range from \(0.8\) to \(4.8\). Class 6A’s values are closer together, so its weekly collection was more consistent.

a) Both classes had a mean of \(2.4\) pounds per week. b) Class 6A was more consistent because its weekly amounts varied less.

- How many players are represented in all? - What is the difference between the greatest and least values? - Which value contains the middle player when all \(25\) values are listed in order? - How can you use each frequency to find the total number of baskets?

1. Find the number of players: \(3 + 7 + 8 + 5 + 2 = 25\). 2. Find the total number of baskets: \((4 \times 3) + (6 \times 7) + (10 \times 8) + (12 \times 5) + (15 \times 2) = 224\). The mean is \(224 \div 25 = 8.96\) baskets. 3. The range is the maximum minus the minimum: \(15 - 4 = 11\) baskets. 4. With \(25\) values, the median is the \(13\)th value in order. The cumulative counts are \(3\) players through \(4\) baskets, \(10\) players through \(6\) baskets, and \(18\) players through \(10\) baskets. Therefore, the \(13\)th value is \(10\).

a) The mean is \(8.96\) baskets. b) The range is \(11\) baskets. c) The median is \(10\) baskets.

- Order the temperatures before finding the median. - For an even number of values, average the two middle values. - The range is the maximum minus the minimum. - Decide whether the new record changes the minimum, the maximum, or both.

1. The sum of the \(14\) temperatures is \(285\), so the mean is \(285 \div 14 \approx 20.36\,^\circ\text{C}\). 2. In order, the temperatures are \(18, 18, 18, 19, 19, 19, 20, 20, 21, 21, 22, 22, 23, 25\). The seventh and eighth values are both \(20\), so the median is \(20\,^\circ\text{C}\). 3. The range is \(25 - 18 = 7\,^\circ\text{C}\). 4. With a new maximum of \(30\,^\circ\text{C}\), the range would be \(30 - 18 = 12\,^\circ\text{C}\), an increase of \(5\,^\circ\text{C}\).

a) Mean: \(\approx 20.36\,^\circ\text{C}\); median: \(20\,^\circ\text{C}\) b) \(7\,^\circ\text{C}\) c) The range would increase to \(12\,^\circ\text{C}\), an increase of \(5\,^\circ\text{C}\).

- Find each mean by adding the five temperatures and dividing by \(5\). - Order each set to find the middle value. - Find each range using the maximum and minimum. - Compare all three statistics across the cities.

1. City A has mean \((18 + 22 + 25 + 20 + 15) \div 5 = 100 \div 5 = 20\,^\circ\text{C}\). City B has mean \((16 + 19 + 21 + 24 + 20) \div 5 = 100 \div 5 = 20\,^\circ\text{C}\). The means are equal. 2. City A, in order, is \(15, 18, 20, 22, 25\), so its median is \(20\,^\circ\text{C}\). City B, in order, is \(16, 19, 20, 21, 24\), so its median is also \(20\,^\circ\text{C}\). 3. City A has range \(25 - 15 = 10\,^\circ\text{C}\). City B has range \(24 - 16 = 8\,^\circ\text{C}\). 4. City A has the greater range and greater variation.

a) Both cities have a mean of \(20\,^\circ\text{C}\). b) Both medians are \(20\,^\circ\text{C}\). City A has range \(10\,^\circ\text{C}\), and City B has range \(8\,^\circ\text{C}\). City A has the greater variation.

- Read each student's value from both graphs. - Add and divide by \(8\) to find each mean. - Use the maximum and minimum for each range. - Explain what the difference in ranges means for the groups.

1. Group A has total \(2 + 4 + 5 + 1 + 6 + 3 + 2 + 9 = 32\) hours, so its mean is \(32 \div 8 = 4\) hours. 2. Group B has total \(3 + 3 + 4 + 4 + 5 + 3 + 5 + 5 = 32\) hours, so its mean is also \(4\) hours. 3. Group A has range \(9 - 1 = 8\) hours. Group B has range \(5 - 3 = 2\) hours. 4. The groups have the same mean, but Group A's values are much more spread out. Group B's values are more consistent.

a) Both groups have a mean of \(4\) hours. b) Group A: \(8\) hours; Group B: \(2\) hours c) Group A has much greater variability, while Group B's exercise times are more consistent.

- Add the bar heights to confirm the number of students. - Multiply each time by its frequency to find the total minutes. - For \(20\) values, locate positions \(10\) and \(11\). - Compare both the means and the medians.

1. Class A has total homework time \(2 \times 15 + 6 \times 30 + 7 \times 45 + 4 \times 60 + 1 \times 75 = 840\) minutes. Its mean is \(840 \div 20 = 42\) minutes. 2. Class B has total homework time \(2 \times 15 + 4 \times 30 + 8 \times 45 + 4 \times 60 + 2 \times 75 = 900\) minutes. Its mean is \(900 \div 20 = 45\) minutes. 3. In Class A, positions \(10\) and \(11\) are both \(45\) minutes, so the median is \(45\) minutes. 4. In Class B, positions \(10\) and \(11\) are also both \(45\) minutes, so the median is \(45\) minutes. 5. Class B's mean is \(45 - 42 = 3\) minutes greater, while the medians are equal.

a) Class A: \(42\) minutes; Class B: \(45\) minutes b) Both medians are \(45\) minutes. c) Class B's mean is \(3\) minutes greater. The medians are the same.

- Add the heights of all the bars. - For \(25\) ordered values, which position contains the median? - “At least \(3\)” includes which bars? - Write the part as a fraction of the total and convert it to a percent.

1. Add the frequencies: \(3 + 6 + 7 + 5 + 3 + 1 = 25\) students. 2. With \(25\) values, the median is the \(13\)th value in order. The first \(3\) values are \(0\), the next \(6\) are \(1\), and positions \(10\) through \(16\) are \(2\). Therefore, the median is \(2\) books. 3. At least \(3\) books includes \(3\), \(4\), and \(5\) books. The number of students is \(5 + 3 + 1 = 9\). The percent is \(\frac{9}{25} \times 100\% = 36\%\).

a) \(25\) students b) \(2\) books c) \(36\%\)

- Read each daily temperature carefully from the graph. - Which two values are needed to find the range? - How do you find the mean of all seven values? - Order the values before finding the median.

1. Read the temperatures from the graph: Monday \(15\,^\circ\text{C}\), Tuesday \(18\,^\circ\text{C}\), Wednesday \(20\,^\circ\text{C}\), Thursday \(17\,^\circ\text{C}\), Friday \(14\,^\circ\text{C}\), Saturday \(13\,^\circ\text{C}\), and Sunday \(15\,^\circ\text{C}\). 2. The range is \(20 - 13 = 7\,^\circ\text{C}\). 3. The mean is \(\frac{15 + 18 + 20 + 17 + 14 + 13 + 15}{7} = \frac{112}{7} = 16\,^\circ\text{C}\). 4. In order, the values are \(13, 14, 15, 15, 17, 18, 20\). The middle value is \(15\,^\circ\text{C}\), so the median is \(15\,^\circ\text{C}\).

a) The range is \(7\,^\circ\text{C}\). b) The mean is \(16\,^\circ\text{C}\). c) The median is \(15\,^\circ\text{C}\).

- Add all bar heights to find the total number of students. - Subtract the least data value from the greatest to find the range. - Which position contains the median of \(21\) ordered values?

1. Add the frequencies: \(4 + 7 + 5 + 3 + 1 + 1 = 21\) students. 2. The least number of pets is \(0\), and the greatest is \(5\). The range is \(5 - 0 = 5\). 3. With \(21\) values, the median is the \(11\)th value in order. Positions \(1\) through \(4\) are \(0\), and positions \(5\) through \(11\) are \(1\). Therefore, the median is \(1\) pet.

a) \(21\) students b) \(5\) pets c) \(1\) pet

- Read the frequency shown by each bar. - Multiply each height by its frequency to find the total height. - For \(20\) values, locate positions \(10\) and \(11\). - Compare both measures of center.

1. Group A has \(2 + 4 + 8 + 4 + 2 = 20\) plants. The total height is \(2 \times 10 + 4 \times 11 + 8 \times 12 + 4 \times 13 + 2 \times 14 = 240\,\text{cm}\), so the mean is \(240 \div 20 = 12\,\text{cm}\). 2. Group B has \(3 + 4 + 3 + 4 + 6 = 20\) plants. The total height is \(3 \times 10 + 4 \times 11 + 3 \times 12 + 4 \times 13 + 6 \times 14 = 246\,\text{cm}\), so the mean is \(246 \div 20 = 12.3\,\text{cm}\). 3. Each group has \(20\) values, so the median is the mean of positions \(10\) and \(11\). In Group A, both middle values are \(12\), so the median is \(12\,\text{cm}\). In Group B, the middle values are \(12\) and \(13\), so the median is \((12 + 13) \div 2 = 12.5\,\text{cm}\). 4. Group B has both the greater mean and the greater median, so Fertilizer B appears to produce taller plants overall.

a) Group A: \(12\,\text{cm}\); Group B: \(12.3\,\text{cm}\) b) Group A: \(12\,\text{cm}\); Group B: \(12.5\,\text{cm}\) c) Fertilizer B appears to produce taller plants overall because Group B has both the greater mean and the greater median.

- Multiply each score by its frequency before finding the mean. - For \(20\) values, use positions \(10\) and \(11\) to find the median. - The range is the maximum minus the minimum.

1. Class A has \(1 + 2 + 6 + 8 + 3 = 20\) students. Its total number of points is \(1 \times 1 + 2 \times 2 + 3 \times 6 + 4 \times 8 + 5 \times 3 = 70\), so the mean is \(70 \div 20 = 3.5\) points. 2. Class B has \(3 + 4 + 6 + 4 + 3 = 20\) students. Its total number of points is \(1 \times 3 + 2 \times 4 + 3 \times 6 + 4 \times 4 + 5 \times 3 = 60\), so the mean is \(60 \div 20 = 3\) points. Class A has the higher mean. 3. For Class A, positions \(10\) and \(11\) are both \(4\), so the median is \(4\). For Class B, positions \(10\) and \(11\) are both \(3\), so the median is \(3\). 4. In both classes, the minimum score is \(1\) and the maximum score is \(5\). Each range is \(5 - 1 = 4\) points.

a) Class A: \(3.5\) points; Class B: \(3\) points. Class A has the higher mean. b) Class A: \(4\) points; Class B: \(3\) points c) Both ranges are \(4\) points.

- Multiply each number of books by its frequency before finding the mean. - For \(20\) values, use positions \(10\) and \(11\) to find the median. - Find each range by subtracting the minimum from the maximum.

1. Class A has \(20\) students and a total of \(0 \times 2 + 1 \times 5 + 2 \times 6 + 3 \times 4 + 4 \times 2 + 5 \times 1 = 42\) books. Its mean is \(42 \div 20 = 2.1\) books. 2. Class B has \(20\) students and a total of \(0 \times 1 + 1 \times 3 + 2 \times 8 + 3 \times 5 + 4 \times 3 + 5 \times 0 = 46\) books. Its mean is \(46 \div 20 = 2.3\) books. 3. In both classes, positions \(10\) and \(11\) are \(2\), so both medians are \(2\) books. 4. Class A's range is \(5 - 0 = 5\) books. Class B's range is \(4 - 0 = 4\) books. Class A has the greater range.

a) Class A: \(2.1\) books; Class B: \(2.3\) books b) Both medians are \(2\) books. c) Class A: \(5\) books; Class B: \(4\) books. Class A has the greater range.

- The mode is the value represented by the tallest bar. - The range is the greatest occurring value minus the least occurring value. - Add all bar heights to find the class total. - Add the frequencies for scores of \(6\) or more.

1. The tallest bar for Class A is at \(3\) hits, so its mode is \(3\). The tallest bar for Class B is at \(6\) hits, so its mode is \(6\). 2. Class A's minimum is \(0\) and maximum is \(10\), so its range is \(10 - 0 = 10\). Class B's minimum is \(2\) and maximum is \(9\), so its range is \(9 - 2 = 7\). Class A has the greater range. 3. Class A has \(30\) students, and \(1 + 1 + 1 + 1 = 4\) scored at least \(6\) hits. Class B also has \(30\) students, and \(8 + 6 + 3 + 2 = 19\) scored at least \(6\) hits. Since \(19 > 15\), more than half did so only in Class B.

a) Class A: \(3\) hits; Class B: \(6\) hits b) Class A has the greater range: \(10\), compared with \(7\) for Class B. c) Class B; \(19\) of \(30\) students scored at least \(6\) hits.

- Find the score represented by the tallest bar in each graph. - Add the frequencies at \(9\) and \(10\) points for Group B. - Compare where most of each distribution lies on the horizontal axis.

1. Group A's tallest bar is at \(5\) points, with \(15\) students. Group B's tallest bar is at \(8\) points, with \(16\) students. 2. In Group B, \(10\) students earned \(9\) points and \(6\) earned \(10\) points, for a total of \(10 + 6 = 16\) students. 3. Group A's distribution is centered near \(5\), while Group B's distribution is shifted toward higher scores and centered near \(8\). Group B performed better overall because its scores are generally higher.

a) Group A: \(5\) points; Group B: \(8\) points b) \(16\) students c) Group B performed better overall because its distribution is shifted toward higher scores.

- Compare where most of the bars lie in each distribution. - Values farther right represent higher scores. - The range is the maximum minus the minimum. - A smaller range indicates more compact results.

1. Class B's distribution is shifted farther to the right than Class A's. Class A is centered near \(5\) points, while Class B is centered near \(6.5\) points, so Class B appears to have the higher mean. 2. Class A's minimum is \(1\), and its maximum is \(9\), so its range is \(9 - 1 = 8\) points. Class B's minimum is \(4\), and its maximum is \(9\), so its range is \(9 - 4 = 5\) points. Class B has the more compact results.

1. Class B appears to have the higher mean because its distribution is shifted toward higher scores. 2. Class A: \(8\) points; Class B: \(5\) points. Class B's results are more compact.

- The tallest bar identifies the most common distance. - Check the greatest distance represented in each graph. - “At least \(8\)” includes \(8\), \(9\), and \(10\).

1. Graph A has its tallest bar at \(4\) miles, and no values are greater than \(7\) miles. It matches Blue Ridge Running Club. 2. Graph B has its tallest bar at \(6\) miles, and its values extend to \(10\) miles. It matches Green Valley Running Club. 3. In Graph B, \(6\) members ran \(8\) miles, \(4\) ran \(9\) miles, and \(2\) ran \(10\) miles. Therefore, \(6 + 4 + 2 = 12\) members ran at least \(8\) miles.

a) Graph A: Blue Ridge; Graph B: Green Valley b) \(12\) members

- The range is the greatest occurring score minus the least occurring score. - Compare where the tallest bars occur. - Multiply each score by its frequency to find the total points. - Look for differences that remain even when the means are equal.

1. Class A's scores range from \(4\) to \(8\), so the range is \(8 - 4 = 4\). Class B's scores range from \(2\) to \(10\), so the range is \(10 - 2 = 8\). 2. Class A is concentrated around \(6\) points. Class B has two clusters, one at low scores and one at high scores, with few scores in the middle. 3. Class A's total is \(4 \times 2 + 5 \times 6 + 6 \times 10 + 7 \times 6 + 8 \times 2 = 156\) points for \(26\) students, so its mean is \(156 \div 26 = 6\). 4. Class B's total is \(2 \times 8 + 3 \times 4 + 6 \times 2 + 9 \times 4 + 10 \times 8 = 156\) points for \(26\) students, so its mean is also \(6\). 5. The equal means hide very different variability and shapes. Class A is tightly centered, while Class B is widely spread and split into two clusters.

a) Class A: \(4\); Class B: \(8\) b) Class A is more concentrated around the middle. Class B has clusters at low and high scores. c) Both means are \(6\) points. d) The mean does not show the spread or shape of a distribution.

- Use each goal total and its frequency to find the total number of goals. - Divide the total goals by the number of games. - Which whole-number goal totals are no more than \(0.5\) away from the mean?

1. The graph shows \(4\) games with \(0\) goals, \(7\) with \(1\), \(5\) with \(2\), \(3\) with \(3\), and \(1\) with \(4\). 2. The total number of goals is \((0 \times 4) + (1 \times 7) + (2 \times 5) + (3 \times 3) + (4 \times 1) = 30\). 3. The mean is \(30 \div 20 = 1.5\) goals per game. 4. Values within \(0.5\) of \(1.5\) are from \(1.0\) through \(2.0\). The whole-number goal counts in that interval are \(1\) and \(2\), which occurred in \(7 + 5 = 12\) games.

a) \(1.5\) goals per game b) \(12\) games

- Add the bar heights to find the number of students. - Count each throw distance as many times as its bar height indicates. - Think about what happens to an average when every value increases by the same amount.

1. Add the frequencies: \(2+6+9+5+3=25\) students. 2. The total distance is \((10\times2)+(15\times6)+(20\times9)+(25\times5)+(30\times3)=505\,\text{m}\). 3. The mean is \(505\div25=20.2\,\text{m}\). 4. Adding \(3\,\text{m}\) to every data value adds \(3\,\text{m}\) to the mean, so the new mean is \(23.2\,\text{m}\).

a) \(25\) students b) \(20.2\,\text{m}\) c) The mean increases by \(3\,\text{m}\), to \(23.2\,\text{m}\).

- Identify where the tallest bars occur in each graph. - A compact distribution has most values near one central region. - A split distribution has two separated peaks and a low middle.

1. Graph 1 has its tallest bars at scores \(3\) and \(4\), and very small frequencies at \(1\) and \(6\). It matches Class A's compact, middle-centered distribution. 2. Graph 2 has two peaks at scores \(2\) and \(5\), with low frequencies at scores \(3\) and \(4\). It matches Class B's split distribution. 3. Graph 3 is nearly uniform across all six scores, so it does not match either description.

Class A matches Graph \(1\). Class B matches Graph \(2\). Graph \(3\) is left over.

- Multiply each age by its frequency before adding. - Check whether actual ages are close to the mean. - Compare whether each distribution is concentrated or spread into separate groups. - Two data sets can have the same mean but very different shapes.

1. The Runners have \(2 + 4 + 4 + 2 = 12\) members. Their total age is \(2 \times 10 + 4 \times 11 + 4 \times 12 + 2 \times 13 = 138\) years, so their mean age is \(138 \div 12 = 11.5\) years. 2. The Climbers have \(6 + 6 = 12\) members. Their total age is \(6 \times 6 + 6 \times 17 = 138\) years, so their mean age is also \(11.5\) years. 3. The Runners' ages are clustered from \(10\) to \(13\) years, close to the mean. The Climbers are split into two groups, age \(6\) and age \(17\), and no member is near the mean. Therefore, equal means can describe very different distributions.

a) Both groups have a mean age of \(11.5\) years. b) The mean does not show how the ages are distributed. The Runners are clustered near the mean, while the Climbers are split between ages \(6\) and \(17\), with no member near \(11.5\).

- Add all bar heights for each class. - Add the frequencies at scores \(5\) and \(6\). - A score with frequency \(0\) is not part of the data set. - The range is the greatest occurring score minus the least occurring score.

1. Class A has \(2 + 3 + 5 + 7 + 9 + 4 = 30\) students. Class B has \(0 + 2 + 8 + 12 + 6 + 2 = 30\) students. 2. Class A has \(9 + 4 = 13\) scores of \(5\) or \(6\). Class B has \(6 + 2 = 8\). Class A has more strong scores. 3. Class A has scores from \(1\) through \(6\), so its range is \(6 - 1 = 5\). Class B has scores from \(2\) through \(6\), so its range is \(6 - 2 = 4\). Class A has the greater range, indicating that its scores are spread across a wider interval.

a) Each class has \(30\) students. b) Class A: \(13\); Class B: \(8\). Class A has more strong scores. c) Class A: \(5\); Class B: \(4\). Class A has the greater range.

- Add the bar heights to find the total number surveyed. - Use each number of sessions and its frequency to find the total number of sessions. - After finding the mean, identify the bars whose values are greater than it.

1. Add the frequencies: \(1 + 4 + 7 + 5 + 4 + 2 + 2 = 25\) athletes. 2. The total number of extra sessions is \((0 \times 1) + (1 \times 4) + (2 \times 7) + (3 \times 5) + (4 \times 4) + (5 \times 2) + (6 \times 2) = 71\). 3. The mean is \(71 \div 25 = 2.84\) sessions per week. 4. Values greater than \(2.84\) are \(3, 4, 5,\) and \(6\). Their frequencies total \(5 + 4 + 2 + 2 = 13\).

a) \(25\) athletes b) \(2.84\) extra practice sessions per week c) \(13\) athletes

- Add the bar heights to find the total number of students. - The most frequently reported time corresponds to the tallest bar. - Multiply each time by its frequency to find the total number of minutes. - Divide the total by the number of students.

1. Add the frequencies: \(4 + 7 + 10 + 6 + 3 = 30\) students. 2. The tallest bar is at \(90\) minutes, with a frequency of \(10\), so the mode is \(90\) minutes. 3. The total usage time is \((4 \times 30) + (7 \times 60) + (10 \times 90) + (6 \times 120) + (3 \times 150) = 2610\) minutes. 4. The mean is \(2610 \div 30 = 87\) minutes.

a) \(30\) students b) \(90\) minutes c) \(87\) minutes

- Locate the middle value of the complete data set. - Divide the remaining values into two equal halves. - Use only the minimum and maximum to find the range.

1. The data are already ordered. The sixth value is the median, \(12\) minutes. 2. The lower half is \(3, 5, 7, 8, 9\), so \(Q_1 = 7\) minutes. 3. The upper half is \(13, 15, 18, 21, 24\), so \(Q_3 = 18\) minutes. 4. The range is \(24 - 3 = 21\) minutes.

\(Q_1 = 7\) minutes, median \(12\) minutes, \(Q_3 = 18\) minutes, and range \(21\) minutes.

- The choices deliberately agree on the median and range, so compute both quartiles. - Each half contains four values and therefore has two middle values. - Compare your two quartiles with all three choices.

1. All three choices use the same median and range, so those values cannot distinguish the choices. 2. The lower half is \(2,4,5,7\), so \(Q_1=(4+5)\div2=4.5\). 3. The upper half is \(10,12,13,16\), so \(Q_3=(12+13)\div2=12.5\). 4. Therefore Summary A is correct.

Summary A is correct.

- The bar heights are frequencies. - Identify the relevant positions in the ordered data. - Use cumulative frequencies instead of writing every value.

1. The frequencies total \(2 + 4 + 8 + 6 + 4 = 24\) values. 2. For the lower half, positions \(6\) and \(7\) are \(1\) and \(2\), so \(Q_1 = (1 + 2) \div 2 = 1.5\). 3. Positions \(12\) and \(13\) are both \(2\), so the median is \(2\). 4. For the upper half, positions \(18\) and \(19\) are both \(3\), so \(Q_3 = 3\).

\(Q_1 = 1.5\), median \(2\), and \(Q_3 = 3\).

- Read all nine points from the graph. - Order the values to find the median. - Use the minimum and maximum to find the range.

1. The values are \(18, 22, 20, 25, 19, 24, 21, 23, 26\). 2. In order, they are \(18, 19, 20, 21, 22, 23, 24, 25, 26\). 3. The median is the fifth value, \(22\). 4. The range is \(26 - 18 = 8\).

The median is \(22\) artifacts, and the range is \(8\) artifacts.

- Divide the even number of values into two equal halves. - Each half also has two middle values. - Use the minimum and maximum for the range.

1. The median is \((11 + 12) \div 2 = 11.5\). 2. The lower half is \(4, 6, 7, 8, 9, 11\), so \(Q_1 = (7 + 8) \div 2 = 7.5\). 3. The upper half is \(12, 14, 15, 18, 20, 24\), so \(Q_3 = (15 + 18) \div 2 = 16.5\). 4. The range is \(24 - 4 = 20\).

\(Q_1 = 7.5\), median \(11.5\), \(Q_3 = 16.5\), and range \(20\).

- Consider the order of the values, not only their mean. - Look for a trend over time. - A summary of past values is not automatically a forecast.

1. The number of visits increases steadily from \(100\) in week \(1\) to \(170\) in week \(8\). 2. The mean of \(135\) describes the center of all eight previous values. 3. It does not account for the clear upward trend. 4. Therefore, \(135\) is not a good prediction for week \(9\); a value above \(170\) better follows the pattern.

No. The upward trend means that \(135\) visits would likely underestimate week \(9\).

- Look for a point where the level changes sharply. - Compare the overall mean with each phase. - Consider whether separate summaries would preserve more information.

1. On days \(1\) through \(4\), the workshop processed from \(7\) to \(9\) packages per day. 2. On days \(5\) through \(8\), it processed from \(19\) to \(21\) packages per day. 3. The overall mean of \(14\) lies between the two phases and did not occur on any day. 4. It does not describe either process well. Separate means for the two phases would be more informative.

The overall mean does not describe either phase well. The old and new processes should be summarized separately.

- Read the frequencies of the ratings that occur. - Compare the mean with the actual data values. - Consider whether the ratings cluster near the mean or at the extremes.

1. There are \(12\) one-star ratings and \(12\) five-star ratings. 2. The mean is \((12 \times 1 + 12 \times 5) \div 24 = 3\). 3. However, no audience member gave a rating of \(2\), \(3\), or \(4\) stars. 4. The mean hides the complete split between the two extremes, so the mean alone does not describe the distribution well.

The mean of \(3\) stars is mathematically correct, but it does not describe the polarized reactions well. The two clusters at \(1\) star and \(5\) stars should also be reported.

- An unusual value can be genuine or erroneous. - Look for a factual reason, not a preferred result. - Data should not be altered only to make a statistic look better.

1. The value \(48\) is far from the other values and strongly affects the mean. 2. However, being unusual does not prove that a value is a measurement error. 3. The value should be removed only for a valid reason, such as a documented error in measurement or recording. 4. Without such evidence, the value and its effect on the results must be reported.

Removing \(48\) is not justified without evidence of an error. The outlier and its effect on the mean should be reported transparently.

- Group the values by the situation in which they were measured. - Look for two clearly separated clusters. - An overall mean can hide differences between groups.

1. The morning values \(4, 5, 3\) have a mean of \(4\,^\circ\text{C}\). 2. The afternoon values \(18, 19, 17\) have a mean of \(18\,^\circ\text{C}\). 3. The overall mean of \(11\,^\circ\text{C}\) lies between two clearly separated time-of-day groups. 4. Separate means for morning and afternoon are more informative.

The mean of \(11\,^\circ\text{C}\) combines two different time-of-day groups. The separate means, \(4\,^\circ\text{C}\) in the morning and \(18\,^\circ\text{C}\) in the afternoon, are more useful.

- Read the median, box edges, mean markers, and outlier separately. - Use the specific measure requested in each part. - Consider how one unusually large value affects the mean.

1. Studio North has a median of \(4\) minutes, and Studio South has a median of \(5\) minutes. Studio North is typically faster according to the median. 2. Studio North's box has length \(6 - 3 = 3\) minutes. Studio South's box also has length \(6 - 3 = 3\) minutes. Their middle \(50\%\) have equal variability according to the IQR. 3. Studio North has a high outlier at \(14\) minutes. This value raises North's mean to \(5\) minutes, which is greater than South's mean of \(4.5\) minutes.

a) Studio North is typically faster: median \(4\) minutes versus \(5\) minutes for Studio South. b) Both IQRs are \(3\) minutes. c) North's \(14\)-minute outlier raises its mean to \(5\) minutes, above South's mean of \(4.5\) minutes.

- Calculate each measure separately for each route. - Equal medians do not require equal means. - A smaller range indicates more consistent values.

1. Route A has a sum of \(126\), so its mean is \(126 \div 7 = 18\). Its median is \(18\), and its range is \(22 - 14 = 8\). 2. Route B has a sum of \(133\), so its mean is \(133 \div 7 = 19\). Its median is \(18\), and its range is \(31 - 10 = 21\). 3. The routes have the same median, but Route B has the greater mean. 4. Route A has more consistent daily counts because its range is smaller.

a) Route A: mean \(18\), median \(18\); Route B: mean \(19\), median \(18\) b) Route A, because its range is \(8\), compared with \(21\) for Route B.

- Treat each bar height as a frequency. - Compare the measures of center first, then compare the shapes. - Notice how strongly each distribution is concentrated near the middle.

1. The Blue Room data include \(26\) teams and a total of \(0 \times 2 + 1 \times 6 + 2 \times 10 + 3 \times 6 + 4 \times 2 = 52\) hints. The mean is \(52 \div 26 = 2\) hints, and the median is \(2\) hints. 2. The Orange Room data also include \(26\) teams and a total of \(0 \times 5 + 1 \times 5 + 2 \times 6 + 3 \times 5 + 4 \times 5 = 52\) hints. The mean is \(2\) hints, and the median is \(2\) hints. 3. Although both rooms have the same mean and median, their distributions differ. Blue Room values are strongly concentrated at \(2\), while Orange Room values are spread more evenly from \(0\) through \(4\).

Both data sets have a mean and median of \(2\) hints. The Blue Room distribution is strongly concentrated at \(2\), while the Orange Room distribution is spread more evenly from \(0\) to \(4\).

- Read all eight values for each team. - Equal measures of center do not imply equal consistency. - Because the data are ordered by trial, also look for a trend over time.

1. Team Green's values have a sum of \(64\), so the mean is \(64 \div 8 = 8\) errors. Its median is \((8 + 8) \div 2 = 8\), and its range is \(10 - 6 = 4\). 2. Team Purple's values also have a sum of \(64\), so its mean is \(8\) errors. Its median is \((7 + 9) \div 2 = 8\), and its range is \(15 - 1 = 14\). 3. The teams have equal means and medians, but Team Green is more consistent because its range is much smaller. 4. Team Purple's errors increase by \(2\) in every successive trial. The measures of center do not show this time trend.

a) Both teams have a mean and median of \(8\) errors. b) Team Green, with a range of \(4\) rather than \(14\) c) Team Purple's errors increase steadily by \(2\) per trial.

- Verify the advertised calculation first. - Identify experiences that the mean does not show. - A true statement about an average can still be incomplete.

1. The mean wait time is \((18 \times 2 + 2 \times 20) \div 20 = 3.8\) minutes. 2. The statement “under \(4\) minutes” is mathematically correct. 3. However, two people waited \(20\) minutes each, and this experience is hidden by the mean. 4. The median is \(2\) minutes, but it also does not show the long waits. The distribution or the percent of long waits should be reported as well.

The advertisement is mathematically true because the mean is \(3.8\) minutes, but it is incomplete: \(10\%\) of the people waited \(20\) minutes.

- Begin with how many observations were collected and what each value measures. - Describe the overall pattern before choosing numerical summaries. - For a symmetric distribution without outliers, consider a mean-based center-and-spread pair.

1. There are \(8\) observations. The attribute is the number of hints required per puzzle station, measured in hints. 2. The values occur once each at consecutive integers from \(2\) through \(9\), so the distribution is flat and symmetric. 3. The mean and median are both \(5.5\) hints. 4. The absolute distances from the mean total \(16\), so the mean absolute deviation is \(16\div8=2\) hints. 5. There are no gaps and no outliers; the endpoints continue the even pattern. 6. Because the distribution is symmetric and has no outliers, the mean and mean absolute deviation are appropriate measures of center and variability.

The data contain \(8\) observations of hints required per puzzle station. The distribution is flat and symmetric from \(2\) to \(9\) hints, with no gaps or outliers. Its mean is \(5.5\) hints and its mean absolute deviation is \(2\) hints. The mean and mean absolute deviation are appropriate because the distribution is symmetric and has no outliers.

- Include context, number of observations, and units before computing statistics. - Check whether values are arranged similarly on both sides of a center. - Interpret MAD as an average distance from the mean, not as the full range.

1. There are \(5\) puzzle-solving times measured in minutes. 2. The values are evenly spaced and symmetric around \(6\) minutes, with no separated observation. 3. The mean is \((2+4+6+8+10)\div5=30\div5=6\) minutes. 4. The distances from the mean are \(4,2,0,2,4\) minutes. 5. The MAD is \((4+2+0+2+4)\div5=12\div5=2.4\) minutes. 6. A student's solving time was an average distance of \(2.4\) minutes from the mean solving time of \(6\) minutes.

There are \(5\) solving times, measured in minutes. The distribution is symmetric and evenly spread around \(6\) minutes with no separated value. The mean is \(6\) minutes and the MAD is \(2.4\) minutes, meaning the times are an average distance of \(2.4\) minutes from the mean.

- Look for an interval with zero frequency in the histogram. - Identify which five numerical landmarks a box plot is built to display. - Ask whether either summary lists every observation separately.

1. The histogram has frequency \(0\) from \(10\) to less than \(15\), so it clearly shows a gap separating lower and higher groups of observations. 2. The box plot gives the median \(7.5\), quartiles \(4.5\) and \(15.5\), and the extreme values \(2\) and \(18\) directly, so center and quartile spread are easier to read there. 3. Neither summary determines every exact raw data value; many different data sets can share those summaries.

The histogram makes the gap from \(10\) to less than \(15\) obvious. The box plot makes the median, quartiles, and overall range easier to read. Neither representation determines every exact individual data value.

- Match the communication goal to what each representation preserves or groups. - Tally repeated values directly from the raw list for the construction data. - After ordering the \(12\) values, use the two middle positions for the median and look for consecutive missing values.

1. A dot plot is the best choice because it preserves each exact observed value and shows repeated values directly. A \(3\)-unit histogram groups exact values, and a box plot compresses them into five landmarks. 2. The dot-stack heights are: \(2\) has \(2\) dots; \(3\) has \(1\); \(4\) has \(2\); \(5\) has \(3\); \(9\) has \(1\); \(10\) has \(2\); \(11\) has \(1\). 3. With \(12\) observations, the median is the average of the sixth and seventh values. Both are \(5\), so the median is \(5\). 4. There are no observations at \(6,7,8\), creating a visible gap between the lower group through \(5\) and the upper group beginning at \(9\).

Best display: dot plot, because it preserves exact values and repeats. Stack heights: value \(2\): \(2\) dots; \(3\): \(1\); \(4\): \(2\); \(5\): \(3\); \(9\): \(1\); \(10\): \(2\); \(11\): \(1\). Median: \(5\). Visible gap: no observations at \(6,7,8\).

- Work only with the lower half of the ordered data set. - The lower half has an even number of values. - Check that your result lies between its neighboring values.

1. The overall median, \(10\), is excluded when the halves are formed. 2. The lower half is \(2, 4, x, 8\). 3. Its median is \((4 + x) \div 2\). Set \((4 + x) \div 2 = 5\). 4. Then \(4 + x = 10\), so \(x = 6\). 5. The value \(6\) lies between \(4\) and \(8\), so the list remains ordered.

\(x = 6\)

- Fix the positions of the minimum, median, and maximum first. - Each quartile can be the mean of two neighboring values. - More than one data set can satisfy the conditions.

1. Place the minimum, median, and maximum in positions \(1\), \(5\), and \(9\): \(1\), \(10\), and \(20\). 2. For \(Q_1 = 5\), the second and third values must have a mean of \(5\). Choose \(4\) and \(6\). 3. For \(Q_3 = 15\), the seventh and eighth values must have a mean of \(15\). Choose \(14\) and \(16\). 4. Choose \(8\) and \(12\) for the remaining positions. One valid data set is \(1, 4, 6, 8, 10, 12, 14, 16, 20\).

One possible data set is \(1, 4, 6, 8, 10, 12, 14, 16, 20\).

- Find the four given statistics for each data set. - A box plot does not show every individual data value. - More than one data set can have the same five-number summary.

1. Data set A has \(Q_1 = (3 + 5) \div 2 = 4\) and \(Q_3 = (13 + 15) \div 2 = 14\), so it does not match. 2. Data set B has \(Q_1 = 5\), median \(9\), \(Q_3 = 13\), and range \(17 - 1 = 16\). 3. Data set C also has \(Q_1 = 5\), median \(9\), \(Q_3 = 13\), and range \(16\). 4. Therefore, both B and C could produce the given box plot.

Data sets B and C both match. A box plot does not determine every value in the original data set.

- Use the quartile convention stated in the problem. - Compare the main group of observations with any separated value. - Decide which measures are less affected by a distant value. - State the pattern using the time unit and volunteer context.

1. There are \(11\) labeling times measured in minutes. 2. The median is the sixth value, \(18\) minutes. 3. The lower-half median is \(Q_1=16\) minutes, and the upper-half median is \(Q_3=20\) minutes. 4. The interquartile range is \(20-16=4\) minutes. 5. Most times lie from \(15\) to \(22\) minutes, while \(2\) minutes is a clear low outlier, so the distribution has a long left tail. 6. The median and interquartile range are the better pair because the low outlier would pull the mean downward and increase a mean-based measure of spread.

The distribution has \(11\) volunteer labeling times measured in minutes. Its median is \(18\) minutes and its interquartile range is \(4\) minutes. Most values are between \(15\) and \(22\) minutes, with a low outlier at \(2\) minutes, so the distribution has a long left tail. The median and interquartile range are the better summaries.

- Begin with the number of observations and what each value measures. - Identify the main cluster, any gaps, and any value separated from the rest. - For a skewed distribution with an outlier, choose resistant measures of center and variability.

1. There are \(12\) observations. The attribute is cave-tour group size, measured in visitors. 2. Most group sizes are clustered from \(8\) through \(13\) visitors, with no gaps in that interval. The value \(25\) is a high outlier and creates a long right tail. 3. The median is \((10+11)\div2=10.5\) visitors. 4. The first quartile is \((9+10)\div2=9.5\), and the third quartile is \((12+12)\div2=12\), so the interquartile range is \(12-9.5=2.5\) visitors. 5. The mean is \(\frac{140}{12}\approx11.67\) visitors, which is above the median because the high outlier pulls it upward. 6. The median and interquartile range are appropriate because they are resistant to the outlier and right skew.

The distribution contains \(12\) cave-tour group sizes, measured in visitors. Most values cluster from \(8\) through \(13\) visitors with no gaps there, while \(25\) is a high outlier that makes the distribution right-skewed. The median is \(10.5\) visitors and the interquartile range is \(2.5\) visitors. These measures are appropriate because they are resistant to the outlier.

- Begin with the sample size and what each score measures. - Compare frequencies at equal distances from the center, then check for missing values inside the range. - Choose a center-and-spread pair that matches the distribution’s symmetry and balanced extremes.

1. There are \(11\) observations. The attribute is balance-test score, measured in points. 2. Matching scores the same distance below and above \(5\) have matching frequencies, so the distribution is symmetric and mound-shaped around \(5\). 3. The mean and median are both \(5\) points. 4. The absolute distances from \(5\) total \(16\), so the mean absolute deviation is \(\frac{16}{11}\approx1.45\) points. 5. There are gaps at \(2\) and \(8\). The extreme scores \(1\) and \(9\) occur symmetrically, so neither creates one-sided skew; together they make the range \(8\) points. 6. The mean and mean absolute deviation are appropriate because the distribution is symmetric and its extreme values are balanced on both sides.

The data contain \(11\) balance-test scores, measured in points. The distribution is symmetric and mound-shaped around \(5\) points, with gaps at \(2\) and \(8\). Its mean is \(5\) points and its mean absolute deviation is \(\frac{16}{11}\approx1.45\) points. The extreme scores \(1\) and \(9\) are balanced on opposite sides and widen the range to \(8\) without creating skew. The mean and mean absolute deviation are appropriate because of this symmetry.

- Begin with how many days were observed and what each value measures. - Look for empty values or a separated observation in the ordered data. - Match a resistant measure of center with a resistant measure of variability when the distribution has a long tail.

1. There are \(10\) observations. The attribute is the number of rescue calls per day, measured in calls. 2. The median is \((5+6)\div2=5.5\) calls, and the mean is \(61\div10=6.1\) calls, so “about \(6\)” is a reasonable but incomplete center statement. 3. The first quartile is \(Q_1=4\), and the third quartile is \(Q_3=7\), so the interquartile range is \(7-4=3\) calls. 4. Most days had \(3\) through \(7\) calls. There is a gap at \(8\), another gap from \(10\) through \(11\), and \(12\) is a separated high value, giving the distribution a right tail. 5. The median and interquartile range are appropriate because they are resistant to the separated high value and right-skewed shape.

The data contain \(10\) daily rescue-call counts, measured in calls. Most days had \(3\) through \(7\) calls, with gaps at \(8,10,11\); \(12\) calls is a separated high value, so the distribution has a right tail. The median is \(5.5\) calls and the interquartile range is \(3\) calls. These measures are appropriate because they are resistant to the high value. The statement “about \(6\) calls” gives a reasonable center but omits variability and shape.

- Use the stated quartile convention before comparing the middle halves. - Inspect the full ordered values for gaps and separated observations. - Use measures that fit the shape instead of relying on the same pair for every distribution.

1. Each set contains \(9\) observations of practice laps completed, measured in laps. 2. Both sets have median \(5\). In both sets, \(Q_1=(3+4)\div2=3.5\) and \(Q_3=(6+7)\div2=6.5\), so both interquartile ranges are \(3\). 3. Set A is symmetric around \(5\), has no gaps from \(2\) through \(8\), and has no separated value. Its mean is \(5\), and its mean absolute deviation is \(12\div9=\frac{4}{3}\) laps. 4. Set B has a long right tail, a gap from \(8\) through \(19\), and a clearly separated high value at \(20\). 5. Set B's median \(5\) and interquartile range \(3\) are appropriate because they resist that high value. Its range is \(20-3=17\), compared with Set A's range \(8-2=6\). 6. The shared median and interquartile range describe the same center and middle-half spread but hide the very different tail shape, gap, separated value, and full range.

Each distribution contains \(9\) practice-lap counts, measured in laps. Both have median \(5\) and interquartile range \(3\). Set A is symmetric with no gaps from \(2\) through \(8\); its mean is \(5\), its mean absolute deviation is \(\frac{4}{3}\) laps, and its range is \(6\). Set B has a long right tail, a gap from \(8\) through \(19\), and a separated high value at \(20\); its range is \(17\). The shared median and interquartile range do not reveal those differences in tail shape and overall spread.

- Begin with sample size, attribute, and units. - Locate the quartiles, then inspect whether the resulting center lies where observations actually occur. - When a distribution has separated clusters, report the clusters and gap in addition to numerical summaries.

1. The data contain \(10\) observations of one-way student commute time, measured in minutes. 2. The median is \((7+18)\div2=12.5\) minutes. 3. The first quartile is \(Q_1=6\), and the third quartile is \(Q_3=19\), so the interquartile range is \(19-6=13\) minutes. 4. The range is \(20-5=15\) minutes. 5. The distribution has two clusters, from \(5\) through \(7\) minutes and from \(18\) through \(20\) minutes, with a gap from \(8\) through \(17\) minutes. There are no isolated outliers beyond the two clusters. 6. The median \(12.5\) lies in the gap, and the interquartile range spans both clusters. Therefore, those numerical summaries are valid but do not represent either cluster well; the two clusters and gap must be reported.

The data contain \(10\) one-way student commute times, measured in minutes. The median is \(12.5\) minutes, the interquartile range is \(13\) minutes, and the range is \(15\) minutes. The distribution has two clusters, from \(5\) through \(7\) minutes and from \(18\) through \(20\) minutes, separated by a gap from \(8\) through \(17\) minutes, with no isolated outliers. Because the median lies in the gap and the interquartile range spans both clusters, one center-and-spread pair is incomplete without the shape description.

- Locate the positions of \(Q_1\), the median, and \(Q_3\) in a list of \(11\) values. - Determine the total needed for a mean of \(20\). - Fill the required positions, then choose remaining values without breaking the order. - Use the last values to reach the required total.

1. With \(11\) ordered values, the median is the sixth value. The first quartile is the third value, and the third quartile is the ninth value. Therefore, positions \(3\), \(6\), and \(9\) must contain \(5\), \(10\), and \(15\), respectively. 2. A mean of \(20\) requires a total of \(11 \times 20 = 220\). 3. Choose the first nine values as \(5, 5, 5, 10, 10, 10, 15, 15, 15\). Their sum is \(90\), so the final two values must have a sum of \(220 - 90 = 130\). 4. Choosing \(65\) and \(65\) keeps the list sorted. The completed list is \(5, 5, 5, 10, 10, 10, 15, 15, 15, 65, 65\). 5. Its mean is \(220 \div 11 = 20\), and its quartiles and median have the required values.

One possible list is \(5, 5, 5, 10, 10, 10, 15, 15, 15, 65, 65\). Its mean is \((5 + 5 + 5 + 10 + 10 + 10 + 15 + 15 + 15 + 65 + 65) \div 11 = 220 \div 11 = 20\).

- Identify the median position in a sorted list of \(25\) values. - Identify the two positions that determine the median of the upper half. - Compare values in later positions with the overall median. - Decide when the mean of two values that are each at least \(\$20\) can equal \(\$20\).

1. In a sorted list of \(25\) values, the median is the value in position \(13\). Therefore, the value in position \(13\) is \(\$20\). 2. The upper half consists of positions \(14\) through \(25\), so \(Q_3\) is the mean of the values in positions \(19\) and \(20\). 3. Because the list is sorted, the values in positions \(19\) and \(20\) are each at least \(\$20\). Their mean cannot be \(\$15\), so the claim is impossible. 4. For their mean to equal \(\$20\), both values must equal \(\$20\). The sorted order then requires every value from position \(13\) through position \(20\) to equal \(\$20\).

a) The claim is impossible. The values in positions \(19\) and \(20\), which determine \(Q_3\), cannot be less than the median value of \(\$20\). b) Every value from position \(13\) through position \(20\) must be \(\$20\). Then the median is \(\$20\) and \(Q_3 = (\$20 + \$20) \div 2 = \$20\).

- Calculate the sum and identify the middle value. - Consider how replacing one very large value affects each measure. - To keep the mean unchanged, increases and decreases must balance. - Find \(Q_3\) as the median of the upper half.

1. The sum of the original values is \(56\), so the mean is \(56 \div 7 = 8\). The median is the fourth value, \(6\). 2. Replacing \(21\) with \(8\) gives \(4, 5, 5, 6, 7, 8, 8\), with sum \(43\). Its mean is \(43 \div 7 \approx 6.14\), while its median remains \(6\). The mean changes by about \(1.86\); the median does not change. 3. In the original data set, \(Q_3\) is the median of \(7, 8, 21\), so \(Q_3 = 8\). 4. Increase \(8\) to \(10\) and decrease \(21\) to \(19\). The new ordered data set is \(4, 5, 5, 6, 7, 10, 19\). Its sum is still \(56\), its median is still \(6\), and its new third quartile is \(10\).

a) Mean: \(8\); median: \(6\) b) The mean changes more. It decreases from \(8\) to \(\approx 6.14\), while the median remains \(6\). c) Yes. One example is \(4, 5, 5, 6, 7, 10, 19\). Its mean is \(8\), its median is \(6\), and \(Q_3 = 10\).