Free Math Worksheets

- 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 \(5\,^\circ\text{C}\).

- Order the temperatures from least to greatest. - Which value is in the middle of the ordered list? - Add all seven values and divide by \(7\) to find the mean. - Subtract the least value from the greatest to find the range.

1. In order, the seven temperatures are \(-4\,^\circ\text{C}, -2\,^\circ\text{C}, -1\,^\circ\text{C}, 0\,^\circ\text{C}, 1\,^\circ\text{C}, 3\,^\circ\text{C}, 3\,^\circ\text{C}\). The middle value is \(0\,^\circ\text{C}\), so the median is \(0\,^\circ\text{C}\). 2. The mean is \(\frac{3 + 1 + (-2) + (-4) + 0 + (-1) + 3}{7} = \frac{0}{7} = 0\,^\circ\text{C}\). 3. The range is \(3 - (-4) = 7\,^\circ\text{C}\).

a) \(0\,^\circ\text{C}\) b) \(0\,^\circ\text{C}\) c) \(7\,^\circ\text{C}\)

- 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.

- Read each day's value, including its sign. - The range is the greatest value minus the least value. - Add all seven differences, then divide by \(7\).

1. The graph shows the values \(3, -2, 4, -1, -3, 2, 4\). 2. The greatest difference is \(4\,^\circ\text{C}\), and the least difference is \(-3\,^\circ\text{C}\). 3. The range is \(4 - (-3) = 7\,^\circ\text{C}\). 4. The sum is \(3 + (-2) + 4 + (-1) + (-3) + 2 + 4 = 7\). The mean is \(7 \div 7 = 1\,^\circ\text{C}\).

a) Greatest: \(4\,^\circ\text{C}\); least: \(-3\,^\circ\text{C}\) b) \(7\,^\circ\text{C}\) c) \(1\,^\circ\text{C}\)

- 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 \(15\,^\circ\text{C}\).

- Find the highest and lowest temperatures for the range. - Order the temperatures before finding the median. - Which value is in the middle of seven ordered values?

1. The graph shows \(14\,^\circ\text{C}, 16\,^\circ\text{C}, 19\,^\circ\text{C}, 15\,^\circ\text{C}, 12\,^\circ\text{C}, 17\,^\circ\text{C},\) and \(21\,^\circ\text{C}\). 2. The range is \(21 - 12 = 9\,^\circ\text{C}\). 3. In order, the values are \(12, 14, 15, 16, 17, 19, 21\). The middle value is \(16\,^\circ\text{C}\), so the median is \(16\,^\circ\text{C}\).

a) \(9\,^\circ\text{C}\) b) \(16\,^\circ\text{C}\)

- 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.

- Add the frequencies in each graph. - Multiply each rubric score by its frequency to find the total score. - With \(25\) values, identify the score in position \(13\).

1. Class A has \(3 + 7 + 9 + 4 + 2 + 0 = 25\) students. Class B has \(1 + 5 + 11 + 6 + 2 + 0 = 25\) students. 2. Class A's total score is \(1 \times 3 + 2 \times 7 + 3 \times 9 + 4 \times 4 + 5 \times 2 + 6 \times 0 = 70\). Its mean is \(70 \div 25 = 2.8\). 3. Class B's total score is \(1 \times 1 + 2 \times 5 + 3 \times 11 + 4 \times 6 + 5 \times 2 + 6 \times 0 = 78\). Its mean is \(78 \div 25 = 3.12\). Because higher rubric scores indicate stronger performance, Class B has the higher mean. 4. With \(25\) values, the median is position \(13\). In both classes, position \(13\) falls among the scores of \(3\), so both medians are \(3\).

a) Each class has \(25\) students. b) Class A: \(2.8\); Class B: \(3.12\). Class B has the higher mean. c) Both medians are \(3\).

- 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.

- Multiply each time by its frequency before finding the mean. - Add the bar heights to confirm each group's size. - The range is the maximum minus the minimum. - A smaller range indicates less variability.

1. Group 1 has \(10\) students and total screen time \(2 \times 1 + 3 \times 2 + 4 \times 4 + 5 \times 2 + 6 \times 1 = 40\) hours. Its mean is \(40 \div 10 = 4\) hours. 2. Group 2 has \(10\) students and total screen time \(0 \times 2 + 1 \times 1 + 2 \times 1 + 4 \times 2 + 6 \times 1 + 7 \times 1 + 8 \times 2 = 40\) hours. Its mean is also \(40 \div 10 = 4\) hours. 3. Group 1's range is \(6 - 2 = 4\) hours. Group 2's range is \(8 - 0 = 8\) hours. 4. Group 1 has more similar habits because its range is smaller. Group 2 has more varied habits because its values are spread across a wider interval.

a) Both means are \(4\) hours. b) Group 1: \(4\) hours; Group 2: \(8\) hours c) Group 1 has more similar habits; Group 2 has more varied habits.

- 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. - Which bars represent distances of at least \(25\,\text{m}\)? - Think about what happens to a mean when the same amount is added to every value.

1. Add the frequencies: \(2 + 6 + 9 + 5 + 3 = 25\) students. 2. The total distance is \((10 \times 2) + (15 \times 6) + (20 \times 9) + (25 \times 5) + (30 \times 3) = 505\,\text{m}\). The mean is \(505 \div 25 = 20.2\,\text{m}\). 3. The number of students at \(25\,\text{m}\) or farther is \(5 + 3 = 8\). The percent is \(\frac{8}{25} \times 100\% = 32\%\). 4. Adding \(3\,\text{m}\) to every data value adds \(3\,\text{m}\) to the mean. The new mean is \(20.2 + 3 = 23.2\,\text{m}\).

a) \(25\) students b) \(20.2\,\text{m}\) c) \(32\%\) d) 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.

- Check the four statistics separately. - Each half has two middle values. - The range is the maximum minus the minimum.

1. The median of the nine values is \(8\). 2. The lower half is \(2, 4, 5, 7\), so \(Q_1 = (4 + 5) \div 2 = 4.5\). 3. The upper half is \(10, 12, 13, 16\), so \(Q_3 = (12 + 13) \div 2 = 12.5\). 4. The range is \(16 - 2 = 14\). 5. Summary A has all four correct values.

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.

- 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.

- Use the vertical scale to read each bar exactly. - Add the five values and divide by \(5\) to find the mean. - Order the values before finding the median, and subtract the least value from the greatest to find the range. - For the percent increase, use the Year \(3\) value as the original amount. - Add the Year \(6\) harvest before finding the new mean.

1. The graph shows \(45\,\text{kg}\), \(60\,\text{kg}\), \(40\,\text{kg}\), \(80\,\text{kg}\), and \(50\,\text{kg}\). 2. The five-year total is \(45 + 60 + 40 + 80 + 50 = 275\,\text{kg}\), so the mean is \(275 \div 5 = 55\,\text{kg}\). 3. In order, the values are \(40, 45, 50, 60, 80\). The median is \(50\,\text{kg}\), and the range is \(80 - 40 = 40\,\text{kg}\). 4. The increase from Year \(3\) to Year \(4\) is \(80 - 40 = 40\,\text{kg}\). Relative to the Year \(3\) amount, the percent increase is \(\frac{40}{40} \times 100\% = 100\%\). 5. The new total is \(275 + 85 = 360\,\text{kg}\), so the six-year mean is \(360 \div 6 = 60\,\text{kg}\).

a) Year \(1\): \(45\,\text{kg}\); Year \(2\): \(60\,\text{kg}\); Year \(3\): \(40\,\text{kg}\); Year \(4\): \(80\,\text{kg}\); Year \(5\): \(50\,\text{kg}\) b) \(55\,\text{kg}\) c) Median: \(50\,\text{kg}\); range: \(40\,\text{kg}\) d) \(100\%\) e) \(60\,\text{kg}\)

- 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.

- 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 \(Q_3 = 10\).