- The data set is already ordered. - Which position is exactly in the middle? - Check that the value keeps the list in order.

1. With seven ordered values, the median is the fourth value. 2. The fourth value is \(x\). 3. Therefore, \(x = 11\).

\(x = 11\)

- Decide whether the numbers represent amounts or only labels. - Ask whether the difference between two numbers has a meaningful interpretation. - Not every list of numbers should be summarized with a mean.

1. Jersey numbers are labels, not measured quantities. 2. Differences between jersey numbers do not represent meaningful amounts. 3. Therefore, the calculated mean of \(15\) does not describe a typical characteristic of the team.

The mean is not useful because jersey numbers are labels rather than measured quantities.

- Distinguish a statement about the whole group from one about each person. - Consider whether the quantity can take non-whole-number values for an individual. - Interpret the mean as an equal share of the total.

1. The mean represents an equal mathematical distribution of \(16\) pets among \(10\) people. 2. Pets are counted in whole numbers, so \(1.6\) cannot be one person's actual number of pets. 3. Therefore, it is incorrect to say that every person has \(1.6\) pets. 4. The correct interpretation is that the group has an average of \(1.6\) pets per person.

The group has an average of \(1.6\) pets per person. This does not mean that every person has \(1.6\) pets.

- Combine the four amounts first. - Redistribute the total equally among the same four students. - Connect that equal-share operation to a measure of center.

1. The total is \(6+8+10+12=36\) cans. 2. Sharing \(36\) cans equally among \(4\) students gives \(36\div4=9\) cans each. 3. The equal-share amount is the mean of the four original values.

\(9\) cans each. The equal-share amount is the mean.

- The middle value cannot be found from an unordered list. - With five values, look for the third one after sorting.

1. Put the numbers in order: \(2,4,5,7,8\). 2. The middle of five numbers is the third, so the median is \(5\).

\(5\) laps.

- Sort the four values first. - With an even number of data values, the median is halfway between the middle two.

1. Order the numbers: \(3,5,7,9\). 2. The two middle numbers are \(5\) and \(7\); their average is \((5+7)\div2=6\).

\(6\) books.

- Add the scores and divide by the number of students. - Determine how much the corrected score changes the total. - Divide the change in the total among all six scores.

1. The original total is \(105+118+92+126+110+103=654\). 2. The original mean is \(654\div6=109\). 3. The corrected score increases the total by \(117-105=12\), so the new total is \(654+12=666\). 4. The new mean is \(666\div6=111\). 5. The mean increased by \(111-109=2\) points.

a) \(109\) points b) The new mean is \(111\) points, an increase of \(2\) points.

- Use the original mean and group size to find the original total height. - Add the new student’s height, then divide by the new group size. - Compare the new value with the old mean.

1. The original total height is \(5\times60=300\) inches. 2. The new total is \(300+66=366\) inches. 3. The new mean is \(366\div6=61\) inches. 4. The new value, \(66\) inches, is greater than the original mean of \(60\) inches, so adding it must increase the mean.

a) \(61\) inches b) The new mean must be greater because the added height is greater than the original mean.

- Add all six amounts and divide by \(6\). - Compare each amount with the mean. - Find half of \(6\) before answering the final question.

1. The total amount is \(11+16+10+18+14+15=84\) ounces. 2. The mean is \(84\div6=14\) ounces. 3. The amounts above the mean are \(16\), \(18\), and \(15\), so \(3\) students collected more than the mean. 4. Half of \(6\) is \(3\), so this is exactly half, not more than half.

a) \(14\) ounces b) \(3\) students; no, that is exactly half of the group.

- Find the total for the first three days. - Find the total for the last two days. - Divide the five-day total by \(5\).

1. From Monday through Wednesday, Maya collects \(3\times4=12\) bottles. 2. On Thursday and Friday, she collects \(2\times9=18\) bottles. 3. The total is \(12+18=30\) bottles. 4. The mean is \(30\div5=6\) bottles per day.

Maya collects \(30\) bottles in all, for a mean of \(6\) bottles per day.

- The halfway point is the mean of the two endpoints. - Add the two values first, then divide the sum by \(2\). - Check that each answer is the same distance from both endpoints.

1. For a), \((12.4+18.6)\div2=31\div2=15.5\). 2. For b), \((5+13)\div2=18\div2=9\). 3. For c), \(\left(\frac{1}{4}+\frac{3}{4}\right)\div2=1\div2=\frac{1}{2}\). 4. For d), \((2.5+7.5)\div2=10\div2=5\).

a) \(15.5\) b) \(9\) c) \(\frac{1}{2}\) d) \(5\)

- Find the mean before measuring distances from it. - For values below the mean, measure how far each is from the mean without using signed subtraction. - Compare the total distance on the two sides of the mean.

1. The sum is \(5+7+8+8+10+10=48\), so the mean is \(48\div6=8\). 2. The values below \(8\) are \(5\) and \(7\). Their distances from \(8\) are \(3\) and \(1\), for a total of \(4\). 3. The values above \(8\) are \(10\) and \(10\). Their distances from \(8\) are \(2\) and \(2\), for a total of \(4\). 4. The total distance below the mean equals the total distance above the mean. The mean acts like a balance point for the data.

1) \(8\) 2) \(4\) 3) \(4\); the total distance below the mean equals the total distance above it.

- How do the day labels match the five bar heights? - Imagine spreading the total of the five highs equally across the days. - Should the mean fall between the week's lowest and highest temperatures?

1. The highs from Monday to Friday are \(54,60,58,52,66\,^{\circ}\text{F}\). 2. Their sum is \(54+60+58+52+66=290\). 3. The mean is \(290\div5=58\,^{\circ}\text{F}\).

a) Monday \(54\,^{\circ}\text{F}\); Tuesday \(60\,^{\circ}\text{F}\); Wednesday \(58\,^{\circ}\text{F}\); Thursday \(52\,^{\circ}\text{F}\); Friday \(66\,^{\circ}\text{F}\). b) \(58\,^{\circ}\text{F}\)

- What does the median represent? - Order the values from least to greatest. - Which value is in the middle of a list with seven values?

1. Order the seven temperatures: \(-5\,^\circ\text{C}, -4\,^\circ\text{C}, -3\,^\circ\text{C}, -1\,^\circ\text{C}, 1\,^\circ\text{C}, 2\,^\circ\text{C}, 4\,^\circ\text{C}\). 2. With seven values, the median is the fourth value in the ordered list. 3. The median is \(-1\,^\circ\text{C}\).

The median low temperature is \(-1\,^\circ\text{C}\).

- Read each bar using the y-axis scale. - Add all five daily values. - Divide the weekly total by \(5\). - Compare the mean with the goal.

1. The graph shows Monday, \(14\); Tuesday, \(18\); Wednesday, \(12\); Thursday, \(22\); and Friday, \(14\) servings. 2. The total is \(14+18+12+22+14=80\) servings. 3. The mean is \(80\div5=16\) servings per day. 4. Since \(16<18\), the goal was not met.

a) Monday: \(14\); Tuesday: \(18\); Wednesday: \(12\); Thursday: \(22\); Friday: \(14\) b) \(16\) servings per day c) No. The mean of \(16\) is less than the goal of \(18\).

- Add all seven observations. - Divide by the number of observations. - Check that the result lies between the least and greatest values.

1. Add the seven visitor counts: \(128 + 143 + 137 + 151 + 146 + 132 + 157 = 994\). 2. Divide by the number of mornings: \(994 \div 7 = 142\).

The mean number of visitors is \(142\) per morning.

- Read each bar value from the vertical axis. - Record exactly six values. - Check whether the mean is reasonable compared with the bar heights.

1. Read the six values from the graph: \(14, 18, 13, 17, 16,\) and \(12\). 2. Their sum is \(14 + 18 + 13 + 17 + 16 + 12 = 90\). 3. The mean is \(90 \div 6 = 15\) parts per station.

The mean is \(15\) defect-free parts per station.

- Keep the decimal points aligned when adding. - Round only after dividing. - Check that the result lies between the least and greatest times.

1. Add the six times: \(11.5 + 12.2 + 10.8 + 13.1 + 12.4 + 12.6 = 72.6\) minutes. 2. Divide by \(6\): \(72.6 \div 6 = 12.1\) minutes. 3. To the nearest tenth, the mean is \(12.1\) minutes.

The mean setup time is \(12.1\) minutes.

- Order the values from least to greatest. - Find the position with the same number of values on each side. - Check that all nine times are included.

1. Order the times: \(6, 7, 8, 9, 10, 11, 12, 14, 15\). 2. With nine values, the median is the fifth value. 3. The median is \(10\) minutes.

The median setup time is \(10\) minutes.

- Order the lengths from least to greatest. - With an even number of values, two values are in the middle. - Find the mean of those two middle values.

1. Order the lengths: \(3.8, 4.2, 4.4, 4.7, 5, 5.1, 5.5, 6\) kilometers. 2. The two middle values are \(4.7\) and \(5\). 3. The median is \((4.7 + 5) \div 2 = 4.85\,\text{km}\).

The median trail length is \(4.85\,\text{km}\).

- Read all seven bar values. - The route order does not matter for the median. - Order the values before finding the middle.

1. Read the values from the graph: \(5, 8, 6, 9, 7, 10,\) and \(6\). 2. Order the values: \(5, 6, 6, 7, 8, 9, 10\). 3. The middle value is \(7\), so the median is \(7\) checkpoints.

The median is \(7\) checkpoints.

- The inserted number does not have to equal the target median. - A seven-value ordered list has one middle position. - Test where each candidate would be placed relative to \(7\), \(9\), and \(12\).

1. After one number is inserted, there are seven values, so the fourth value is the median. 2. Inserting \(5\) gives \(4,5,7,9,12,16,20\). 3. The fourth value is \(9\), so inserting \(5\) gives the required median. 4. Inserting \(10\) makes the fourth value \(10\), while inserting \(14\) or \(25\) makes the fourth value \(12\).

\(5\) must be inserted.

- Are the values already ordered from least to greatest? - The original list position is not necessarily the middle value by size. - Order the values and check that all seven are included.

1. The values are not ordered, so the fourth position in the original list does not determine the median. 2. Order the values: \(12, 14, 15, 16, 18, 19, 21\). 3. The fourth ordered value is \(16\), so the median is \(16\).

Noah must order the data first. The correct median is \(16\).

- Find the middle before and after adding the value. - The number of values changes from odd to even. - The new value belongs at the end of the ordered list.

1. With seven values, the fourth value is the median, so the original median is \(8\). 2. After adding \(100\), the ordered list is \(5, 6, 7, 8, 9, 10, 11, 100\). 3. The middle values are \(8\) and \(9\), so the new median is \((8 + 9) \div 2 = 8.5\).

Original median: \(8\) New median: \(8.5\)

- Insert the additional value in the correct position. - Repeated values are allowed. - With eight values, use the two middle positions.

1. The new ordered list is \(2, 5, 7, 7, 9, 12, 14, 18\). 2. With eight values, the middle values are the fourth and fifth values, \(7\) and \(9\). 3. The new median is \((7 + 9) \div 2 = 8\).

The new median is \(8\).

- Each data set is already ordered. - Use only the two middle positions in each set. - Compare each median with the target value.

1. Each data set has six values, so find the mean of the third and fourth values. 2. For A, the median is \((6 + 8) \div 2 = 7\). 3. For B, the median is \((7 + 11) \div 2 = 9\). 4. For C, the median is \((9 + 10) \div 2 = 9.5\).

Data set B has a median of \(9\).

- Distinguish the line inside the box from the \(\times\). - Use the labeled scale to read both centers. - Compare the two values.

1. The median line is at \(30\) points. 2. The \(\times\) marking the mean is at \(25\) points. 3. Since \(25<30\), the mean is below the median.

a) \(30\) points b) The mean is below the median.

- Put the smaller ranking numbers first. - Find the middle position after ordering the values. - Keep repeated rankings in the list.

1. Order the rankings: \(1, 2, 2, 3, 4, 4, 5, 6, 7, 8, 9\). 2. With eleven values, the median is the sixth value. 3. The sixth value is \(4\).

The median ranking is \(4\).

- Compare the two group sizes first. - Equal group sizes give the two means equal weight. - You can check the result using the groups' total practice hours.

1. The two means represent equal numbers of people, so they have equal weight. 2. The combined mean is \((7.2 + 8.4) \div 2 = 7.8\) hours per month. 3. The same result follows by adding the two groups' total practice hours and dividing by \(30\).

Yes. Because the groups are the same size, the combined mean is \(7.8\) hours per month.

- A mean does not have to appear in the data set. - Think of the mean as an equal share of the total. - Check whether the data represent genuine measurements.

1. The mean is \((9 + 10 + 10 + 12) \div 4 = 10.25\,\text{cm}\). 2. A mean does not have to be one of the observed values. 3. It represents the total height distributed equally among the four bridges and lies between the minimum and maximum. 4. Therefore, it is a useful summary.

No. The mean of \(10.25\,\text{cm}\) is a useful summary even though no bridge has exactly that height.

- Read the median, mean, and outliers separately. - Compare the measures with the interval containing the box and whiskers. - Focus on what “typical” means in this context.

1. The median is \(8\) minutes, and the mean is \(10\) minutes. 2. Most displayed values lie between \(4\) and \(12\) minutes. 3. The high outliers at \(30\) and \(36\) minutes pull the mean upward. 4. Therefore, the median better describes a typical group.

The median of \(8\) minutes is more representative because the high outliers pull the mean up to \(10\) minutes.

- Compare the stated mean with the middle line of the box. - Compare the lengths on the two sides of the median. - What information about individual scores is missing from a five-number summary?

1. The five-number summary is balanced around the median of \(30\) points. 2. The stated mean is also \(30\) points, so it is a reasonable description of the center from the available information. 3. The box plot does not show all individual scores. A balanced five-number summary does not prove that the individual scores are distributed symmetrically.

Yes. The mean and median are both \(30\) points, and the box-plot summary is balanced. No, the box plot does not show enough detail to prove that the individual scores form a symmetric distribution.

- Decide whether the numbers would be measurements or only labels. - Consider whether changing the codes would change the result. - Use summaries that do not depend on arbitrary coding.

1. The numbers \(1\) through \(4\) are arbitrary codes for categories. 2. The distances between the codes do not represent measurable differences between transportation methods. 3. A different coding system would produce a different mean. 4. Therefore, a mean is not meaningful. Percentages or the most common category are appropriate summaries.

A mean is not meaningful because the numbers are only category codes. The transportation methods should be described using their percentages or the most common category.

- Imagine combining all the containers first. - Divide the total equally among the teams. - The question asks about equal sharing, not only a typical position.

1. There are \(3 + 5 + 7 + 9 + 11 = 35\) containers altogether. 2. An equal distribution gives \(35 \div 5 = 7\) containers per team. 3. The arithmetic mean represents this equal-share value. 4. The median also happens to be \(7\), but the equal-sharing interpretation comes from the mean.

Each team receives \(7\) containers. The arithmetic mean represents this equal distribution.

- Recall how to find the median for an even number of values. - Count how many values lie on each side. - A summary statistic does not have to appear in the data set.

1. With four values, the median is the mean of the two middle values, \(4\) and \(8\). 2. The median is \((4 + 8) \div 2 = 6\). 3. Two data values lie on each side of \(6\). 4. The median still divides the ordered data into two equal halves even though it is not an observed value.

The median of \(6\) is correct and meaningful because it divides the ordered data into two equal halves.

- Read all seven values directly from the dotplot before finding the mean. - For the trimmed mean, remove exactly one smallest observation and one largest observation. - Compare how the large value on the right affects the average before and after trimming.

1. The displayed values are \(5,10,15,20,20,25,45\). 2. Their total is \(5+10+15+20+20+25+45=140\), so the ordinary mean is \(\frac{140}{7}=20\). 3. After removing the smallest value \(5\) and the largest value \(45\), the remaining values are \(10,15,20,20,25\). 4. Their total is \(90\), so the trimmed mean is \(\frac{90}{5}=18\). 5. The ordinary mean is larger because the high value \(45\) pulls the mean upward. Trimming removes that influence.

The ordinary mean is \(20\), and the trimmed mean is \(18\). The ordinary mean is larger because the high value \(45\) raises it.

- Look for pairs of marked values that are equally far to the left and right of one value. - Add the four values and divide by \(4\) to confirm the balance point. - Check whether the balance point is one of the marked data values.

1. The marked values are \(2,4,8,10\). 2. The pair \(2,10\) is balanced around \(6\), and the pair \(4,8\) is also balanced around \(6\). 3. The mean is \((2+4+8+10)\div4=24\div4=6\). 4. There is no observed value at \(6\), so a mean can be a balance point without being one of the data values.

Balance point and mean: \(6\) No. The mean does not have to be an observed data value.

- Add the three reading times, then divide by the number of students. - Check whether your result appears in the original list. - What does a mean describe: one student or the group?

1. The three reading times total \(2+2+8=12\) minutes. 2. The mean is \(12\div3=4\) minutes. 3. None of the three students read for \(4\) minutes. A mean describes the group and does not have to match one person's value.

The mean is \(4\,\text{min}\). Yes. The mean can be a value nobody actually recorded.

- Find each group's total before calculating the mean. - Reorder the second list before finding the middle value. - A much larger last value may affect the mean and median differently.

1. The first list has sum \(20\), so its mean is \(20\div5=4\); its median is \(4\). 2. The second list, \(3,4,4,5,14\), has sum \(30\), so its mean is \(30\div5=6\); its median is still \(4\). 3. The mean increases by \(2\) minutes, but the median does not change.

The mean rises from \(4\) to \(6\) minutes; the median stays at \(4\) minutes.

- Work backward from the mean to the total of all four distances. - Find how much the three stated distances account for. - The missing jump completes that total.

1. A mean of \(9\) over four jumps means the total distance is \(4\times9=36\) feet. 2. The known distances total \(6+8+12=26\) feet. 3. The missing distance is \(36-26=10\) feet.

\(10\,\text{ft}\).

- Use the target mean to find the required total for five games. - Add the first four scores. - Subtract the current total from the required total.

1. A mean of \(46\) points over \(5\) games requires a total of \(46\times5=230\) points. 2. The first four games total \(42+38+55+45=180\) points. 3. The fifth-game score must be \(230-180=50\) points.

The team must score \(50\) points in the fifth game.

- Multiply each class size by its mean to find the total distance. - A group mean does not identify individual values. - A larger group can have a greater total even with a smaller mean.

1. Class 6A’s total distance is \(20\times4.5=90\) miles. 2. Class 6B’s total distance is \(25\times4.0=100\) miles. 3. Statement a is false because \(90<100\). 4. Statement b is false. A mean summarizes all values but does not determine any individual student’s distance. Individual distances in the two classes may overlap.

a) False. Class 6A ran \(90\) miles in total, while Class 6B ran \(100\) miles. b) False. The means do not show how far each individual student ran.

- Multiply the mean by the number of players to find the required total. - Add the four known scores. - Subtract the known total from each required total.

1. A mean of \(12\) for \(5\) players requires a total of \(5\times12=60\) points. 2. The four known scores total \(8+15+10+14=47\) points. 3. The fifth player scored \(60-47=13\) points. 4. A mean of \(15\) would require a total of \(5\times15=75\) points. 5. The fifth player would need \(75-47=28\) points.

a) \(13\) points b) \(28\) points

- Add the five scores and divide by \(5\). - Identify the greatest and least scores before finding their distances from the mean. - Use the target mean to find the required six-quiz total.

1. The current total is \(8+12+15+9+11=55\), so the mean is \(55\div5=11\). 2. The highest score is \(15\), so its distance from the mean is \(15-11=4\). 3. The lowest score is \(8\), so its distance from the mean is \(11-8=3\). 4. A mean of \(12\) after \(6\) quizzes requires a total of \(12\times6=72\) points. 5. The required sixth score is \(72-55=17\).

a) \(11\) b) Distance to the highest score: \(4\); distance to the lowest score: \(3\) c) \(17\) points

- Add each runner’s five times separately. - Divide each total by \(5\). - In a race, a smaller time is faster.

1. Jordan’s total time is \(14.5+13.8+14.2+15.1+14.4=72.0\,\text{s}\). 2. Jordan’s mean is \(72.0\div5=14.4\,\text{s}\). 3. Maya’s total time is \(13.9+14.1+14.0+14.3+14.7=71.0\,\text{s}\). 4. Maya’s mean is \(71.0\div5=14.2\,\text{s}\). 5. Since \(14.2<14.4\), Maya was faster on average.

Jordan’s mean time was \(14.4\,\text{s}\). Maya’s mean time was \(14.2\,\text{s}\), so Maya was faster on average.

- Multiply each score by the number of students who earned it. - Add those products and divide by \(25\). - Determine how the correction changes the total score.

1. The original total is \(3\times1+7\times2+10\times3+4\times4+1\times5=68\) points. 2. The original mean is \(68\div25=2.72\) points. 3. Correcting a \(1\) to a \(5\) increases the total by \(5-1=4\), so the corrected total is \(72\). 4. The corrected mean is \(72\div25=2.88\) points. 5. The mean increases by \(2.88-2.72=0.16\) point.

a) \(2.72\) points b) The corrected mean is \(2.88\) points, an increase of \(0.16\) point.

- Multiply each number of books by its frequency. - Add the products to find the total number of books. - Divide by the \(25\) students, then compare with \(1.5\).

1. The total number of books is \(4\times0+10\times1+6\times2+5\times3=37\). 2. The mean is \(37\div25=1.48\) books per student. 3. Since \(1.48<1.5\), the mean is less than \(1.5\).

The mean is \(1.48\) books per student, which is less than \(1.5\).

- The two class sizes are different, so do not simply average \(\$10\) and \(\$15\). - Find the total collected by each class. - Divide the combined total by the combined number of participants.

1. Class 6A collected \(24\times\$10=\$240\). 2. Class 6B collected \(26\times\$15=\$390\). 3. Together, the classes collected \(\$240+\$390=\$630\) from \(24+26=50\) participants. 4. The combined mean is \(\$630\div50=\$12.60\) per participant.

The combined mean is \(\$12.60\) per participant.

- Multiply the target mean by \(3\) to find the required total. - Subtract the contents of the first two barrels. - A mean can be interpreted as an equal share.

1. A mean of \(36\) gallons for \(3\) barrels requires a total of \(3\times36=108\) gallons. 2. The first two barrels contain \(30+40=70\) gallons. 3. The third barrel must contain \(108-70=38\) gallons. 4. The mean is the equal-share amount, so distributing the total evenly would put \(36\) gallons in each barrel.

a) \(38\) gallons b) \(36\) gallons in each barrel

- Find how many pages have already been read. - Subtract from the total number of pages. - Divide the remaining pages by the remaining days.

1. Jordan reads \(4\times15=60\) pages during the first four days. 2. The number of pages remaining is \(123-60=63\). 3. There are \(7-4=3\) days remaining. 4. The required mean is \(63\div3=21\) pages per day.

Jordan must read a mean of \(21\) pages per day during the remaining \(3\) days.

- What total is needed when you know the desired mean and the number of values? - Compare the needed total for six days with the total of the first five temperatures.

1. For a), add the five temperatures: \(14 + 17 + 13 + 16 + 15 = 75\). Then divide by \(5\): \(75 \div 5 = 15\). The mean temperature is \(15\,^\circ\text{C}\). 2. For b), a mean of \(16\,^\circ\text{C}\) over \(6\) days requires a total of \(6 \times 16 = 96\). 3. The first five temperatures total \(75\), so the sixth temperature must be \(96 - 75 = 21\,^\circ\text{C}\).

a) \(15\,^\circ\text{C}\) b) \(21\,^\circ\text{C}\)

- How can you find the total when you know the mean and the number of values? - For part c), compare the fifth-game score with the target mean of \(18\). - What happens to a mean when a new value equals the mean? What happens when it is smaller?

1. For a), multiply the mean by the number of games: \(4 \times 18 = 72\) points. 2. For b), the new total is \(72 + 28 = 100\) points. The new mean is \(100 \div 5 = 20\) points per game. 3. For c), scoring \(18\) points in the fifth game would have kept the mean at \(18\). Instead, she scored \(10\) extra points because \(28 - 18 = 10\). 4. To balance those extra points, she must score \(10\) fewer than \(18\) in the sixth game: \(18 - 10 = 8\) points. Check: \(100 + 8 = 108\), and \(108 \div 6 = 18\).

a) \(72\) points b) \(20\) points per game c) \(8\) points. Her fifth-game score was \(10\) points above the target mean, so her sixth-game score must be \(10\) points below it.

- 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.

- Use the frequencies to find the total number of books. - Consider how one unusually high value affects the mean. - Locate the middle value using cumulative frequencies. - Think about whether increasing only the largest value would change the median.

1. There are \(5 + 8 + 6 + 4 + 1 + 1 = 25\) students. 2. The total number of books is \(0 \times 5 + 1 \times 8 + 2 \times 6 + 3 \times 4 + 4 \times 1 + 12 \times 1 = 48\). The mean is \(48 \div 25 = 1.92\) books. 3. With \(25\) values, the median is the value in position \(13\). Positions \(1\) through \(5\) are \(0\), and positions \(6\) through \(13\) are \(1\), so the median is \(1\) book. 4. The median better represents a typical student because the unusually high value of \(12\) pulls the mean upward.

a) Mean: \(1.92\) books; median: \(1\) book b) The median better represents a typical student because the value \(12\) has a strong effect on the mean.

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

- Treat each four-day total as being distributed evenly across the four days to find its mean. - Find the two means before comparing them. - Keep the units as dollars per day while interpreting the result.

1. The mean daily revenue is \(\$144\div4=\$36\). 2. The mean daily ingredient cost is \(\$52\div4=\$13\). 3. The difference between the two means is \(\$36-\$13=\$23\).

The mean daily revenue is \(\$23\) greater than the mean daily ingredient cost.

- Read each temperature from the graph. - Add the five values and divide by \(5\). - Compare each daily temperature with the mean.

1. The graph shows Monday, \(58^\circ\text{F}\); Tuesday, \(64^\circ\text{F}\); Wednesday, \(56^\circ\text{F}\); Thursday, \(60^\circ\text{F}\); and Friday, \(62^\circ\text{F}\). 2. The mean is \(\frac{58+64+56+60+62}{5}=\frac{300}{5}=60^\circ\text{F}\). 3. Tuesday and Friday are above \(60^\circ\text{F}\), so \(2\) days were above the mean.

a) \(60^\circ\text{F}\) b) \(2\) days

- How can you use the bar heights to confirm the number of students? - How can you find the total amount represented by each bar? - Which positions are in the middle of an ordered data set with \(20\) values? - What do you do with the two middle values?

1. The bar heights represent \(3 + 5 + 6 + 4 + 2 = 20\) students. 2. Find the total allowance represented by the graph: \((3 \times \$10) + (5 \times \$15) + (6 \times \$20) + (4 \times \$25) + (2 \times \$30) = \$385\). 3. The mean is \(\$385 \div 20 = \$19.25\). 4. With \(20\) values, the median is the mean of the \(10\)th and \(11\)th values in order. Both values are \(\$20\), so the median is \(\$20\).

a) The mean weekly allowance is \(\$19.25\). b) The median weekly allowance is \(\$20\).

- Add all bar heights to find the number of students. - Multiply each number of books by its frequency. - Divide the total number of books by the number of students. - For part c), multiply the mean by the class size.

1. The total number of students is \(2+5+6+5+2=20\). 2. The total number of books is \(1\times2+2\times5+3\times6+4\times5+5\times2=60\). 3. The mean is \(60\div20=3\) books per student. 4. The other class read \(3.5\times20=70\) books in all.

a) \(20\) students b) \(3\) books per student c) \(70\) books

- Read each daily value from the graph. - Add all seven values and divide by \(7\). - Think about how weekends, weather, and special events can affect attendance.

1. The graph shows Monday, \(40\); Tuesday, \(60\); Wednesday, \(110\); Thursday, \(50\); Friday, \(140\); Saturday, \(480\); and Sunday, \(520\) visitors. 2. The weekly total is \(40+60+110+50+140+480+520=1400\). 3. The mean is \(1400\div7=200\) visitors per day. 4. Possible reasons include greater free time on weekends, differences in weather, special events, or organized group trips on particular days.

a) \(200\) visitors per day b) Possible reasons include higher weekend attendance and changes in weather or scheduled group events.

- Read each weekday value carefully from the graph. - How do you find the mean of several values? - What total is needed for six days to have a mean of \(6.5\) hours? - Compare that total with the total for the first five days.

1. Read the five values from the graph: Monday \(6\) hours, Tuesday \(8\) hours, Wednesday \(4\) hours, Thursday \(9\) hours, and Friday \(3\) hours. 2. Their total is \(6 + 8 + 4 + 9 + 3 = 30\) hours, so the mean is \(30 \div 5 = 6\) hours. 3. A mean of \(6.5\) hours over \(6\) days requires a total of \(6 \times 6.5 = 39\) hours. 4. Saturday must account for \(39 - 30 = 9\) hours.

a) The mean is \(6\) hours per day. b) Saturday had \(9\) hours of sunshine.

- Read one value for each student from the graph. - Order the values before finding the median. - Which position is in the middle of nine values? - Add all nine values and divide by \(9\) to find the mean.

1. Read the values from the graph: \(3, 6, 2, 5, 1, 7, 4, 6,\) and \(2\). 2. In order, the values are \(1, 2, 2, 3, 4, 5, 6, 6, 7\). With nine values, the fifth value is the median, so the median is \(4\) books. 3. The total is \(3 + 6 + 2 + 5 + 1 + 7 + 4 + 6 + 2 = 36\). The mean is \(36 \div 9 = 4\) books.

a) \(4\) books b) \(4\) books

- Read each value from the bar graph. - Add the six values and divide by \(6\). - What happens to the total when one value decreases by \(\$3\) and another increases by \(\$3\)?

1. Read the amounts from the graph: Anna \(\$8\), Ben \(\$12\), Clara \(\$15\), David \(\$6\), Emily \(\$10\), and Felix \(\$9\). 2. The total is \(8 + 12 + 15 + 6 + 10 + 9 = 60\), so the mean is \(\$60 \div 6 = \$10\). 3. If Clara receives \(\$3\) less and David receives \(\$3\) more, the changes add to \(-\$3 + \$3 = \$0\). The total and the number of people stay the same, so the mean remains \(\$10\). Felix is not correct.

a) The mean weekly allowance is \(\$10\). b) No. The two changes cancel, so the total remains \(\$60\) and the mean remains \(\$10\).

- Add the frequencies for all six bars. - For an even number of values, which two positions determine the median? - Use cumulative counts to locate the \(12\)th and \(13\)th values.

1. Add the frequencies: \(3 + 5 + 4 + 6 + 4 + 2 = 24\) students. 2. With \(24\) values, the median is the mean of the \(12\)th and \(13\)th values in order. 3. The cumulative frequencies show that the \(12\)th value is \(3\) hours and the \(13\)th value is \(4\) hours. The median is \(\frac{3 + 4}{2} = 3.5\) hours.

a) \(24\) students b) \(3.5\) hours

- Read the Game 4 bar using the y-axis. - Subtract the smallest score from the greatest score. - Add all five scores and divide by \(5\).

1. The graph shows scores of \(12\), \(18\), \(15\), \(20\), and \(10\) points. Jordan scored \(20\) points in Game 4. 2. The highest score is \(20\), and the lowest is \(10\), so the difference is \(20-10=10\) points. 3. The mean is \((12+18+15+20+10)\div5=75\div5=15\) points per game.

a) \(20\) points b) \(10\) points c) \(15\) points per game

- Compare the tallest bars using the rainfall scale. - Does the total for six months describe one month or all of them together? - Imagine six equal monthly amounts with the same combined rainfall.

1. May has the most rainfall at \(3.2\,\text{in}\). 2. The sum is \(2.4+1.8+2.2+2.8+3.2+2.0=14.4\,\text{in}\). 3. The mean is \(14.4\div6=2.4\,\text{in}\) per month.

1. May, with \(3.2\,\text{in}\). 2. \(14.4\,\text{in}\). 3. \(2.4\,\text{in}\) per month.

- Add the bar heights to find the number of students. - Multiply each error count by its frequency. - Divide the total errors by the number of students. - “Fewer than \(3\)” includes \(1\) and \(2\).

1. The graph shows \(4\) students with \(1\) error, \(9\) with \(2\), \(7\) with \(3\), \(3\) with \(4\), and \(2\) with \(5\). 2. The number of students is \(4+9+7+3+2=25\). 3. The total number of errors is \(4\times1+9\times2+7\times3+3\times4+2\times5=65\). 4. The mean is \(65\div25=2.6\) errors. 5. Fewer than \(3\) errors means \(1\) or \(2\) errors, so \(4+9=13\) students.

a) \(25\) students b) \(2.6\) errors c) \(13\) students

- Add all seven values and divide by \(7\) to find the mean. - Put the seven attendance values in order to find the middle value. - Consider how unusually large values affect the mean.

1. The graph shows Monday, \(1500\); Tuesday, \(2000\); Wednesday, \(500\); Thursday, \(1500\); Friday, \(2500\); Saturday, \(4500\); and Sunday, \(5000\) visitors. 2. The weekly total is \(1500+2000+500+1500+2500+4500+5000=17{,}500\). 3. The mean is \(17{,}500\div7=2500\) visitors per day. 4. In order, the values are \(500, 1500, 1500, 2000, 2500, 4500, 5000\). The middle value is \(2000\), so the median is \(2000\) visitors. 5. The mean is greater than the median because the two large weekend values raise the mean.

a) Mean: \(2500\) visitors per day b) Median: \(2000\) visitors per day c) The mean is greater because the high Saturday and Sunday attendance pulls it upward.

- A frequency tells how many times each score occurs. - Use cumulative frequencies to locate the middle score. - Multiply each score by its frequency before finding the mean.

1. Add the frequencies: \(1 + 2 + 4 + 4 + 3 + 1 = 15\) students. 2. With \(15\) scores, the median is the \(8\)th score in order. The cumulative frequency through \(6\) points is \(7\), and through \(7\) points it is \(11\), so the \(8\)th score is \(7\). 3. The total number of points is \((4 \times 1) + (5 \times 2) + (6 \times 4) + (7 \times 4) + (8 \times 3) + (9 \times 1) = 99\). The mean is \(99 \div 15 = 6.6\) points.

a) \(15\) students b) \(7\) points c) \(6.6\) points

- Record the frequency for each possible number of bicycles. - How can you use each value and its frequency to find the total number of bicycles? - Divide the total number of bicycles by the number of households.

1. The graph shows \(4\) households with \(0\) bicycles, \(8\) with \(1\), \(12\) with \(2\), \(10\) with \(3\), \(4\) with \(4\), and \(2\) with \(5\). 2. The total number of bicycles is \((4 \times 0) + (8 \times 1) + (12 \times 2) + (10 \times 3) + (4 \times 4) + (2 \times 5) = 88\). 3. The mean is \(88 \div 40 = 2.2\) bicycles per household.

The mean is \(2.2\) bicycles per household.

- Add all bar heights to find the total number of students. - Which bars represent more than one sibling? - To find the mean, first determine the total number of siblings represented. - Divide that total by the number of students.

1. Add the bar heights to find the number of students: \(2 + 6 + 9 + 6 + 2 = 25\). 2. More than one sibling means \(2\), \(3\), or \(4\) siblings. The number of students is \(9 + 6 + 2 = 17\). 3. The total number of siblings represented is \((2 \times 0) + (6 \times 1) + (9 \times 2) + (6 \times 3) + (2 \times 4) = 50\). 4. The mean is \(50 \div 25 = 2\) siblings per student.

a) \(25\) students b) \(17\) students c) \(2\) siblings per student

- 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 \(23.2\,\text{m}\).

- 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

- Read all five scores from each graph. - Add the scores in each group and divide each sum by \(5\). - Compare the two means.

1. Group A has scores \(2, 4, 5, 5,\) and \(9\). Their sum is \(25\), so the mean is \(25 \div 5 = 5\). 2. Group B has scores \(3, 4, 6, 7,\) and \(10\). Their sum is \(30\), so the mean is \(30 \div 5 = 6\). 3. Since \(6 > 5\), Group B has the higher mean score.

Group B has the higher mean score. Group A’s mean is \(5\) points, and Group B’s mean is \(6\) points.

- Some values occur more than once. - Multiply each stroke rate by its frequency. - Confirm the total number of measurements before dividing.

1. The number of measurements is \(3 + 5 + 6 + 4 + 2 = 20\). 2. The total of all recorded stroke rates is \((18 \times 3) + (20 \times 5) + (22 \times 6) + (24 \times 4) + (26 \times 2) = 434\). 3. The mean is \(434 \div 20 = 21.7\) strokes per minute.

The mean stroke rate is \(21.7\) strokes per minute.

- What total length corresponds to the given mean for six recordings? - Add the five known lengths. - Subtract the known total from the required total.

1. The required total length is \(7.8 \times 6 = 46.8\) minutes. 2. The five known recordings total \(7.4 + 8.1 + 6.9 + 7.8 + 8.3 = 38.5\) minutes. 3. The missing length is \(46.8 - 38.5 = 8.3\) minutes.

The sixth recording must be \(8.3\) minutes long.

- The original mean represents four scores. - Find the original total first. - After adding the fifth score, divide by the new number of scores.

1. The total for the first four runs is \(30.5 \times 4 = 122\) points. 2. The new total is \(122 + 38 = 160\) points. 3. The new mean is \(160 \div 5 = 32\) points.

The new mean score is \(32\) points.

- Use the original mean to find the original total. - Only one value in the total needs to be replaced. - The number of measurements does not change.

1. The incorrect total was \(46 \times 6 = 276\) minutes. 2. Correct the total by replacing \(52\) with \(46\): \(276 - 52 + 46 = 270\) minutes. 3. The corrected mean is \(270 \div 6 = 45\) minutes.

The corrected mean is \(45\) minutes.

- The two groups are different sizes. - Find the total amount collected by each group. - Divide the combined total by the combined number of people.

1. The first group collected \(12 \times 6.5 = 78\,\text{kg}\). 2. The second group collected \(18 \times 8 = 144\,\text{kg}\). 3. Together, the groups collected \(78 + 144 = 222\,\text{kg}\), and there were \(12 + 18 = 30\) people. 4. The combined mean is \(222 \div 30 = 7.4\,\text{kg}\) per person.

The combined mean is \(7.4\,\text{kg}\) per person.

- Use the grid lines to read each marked point. - The order of the measurements does not affect the mean. - Keep track of the decimal places.

1. Read the eight values from the graph: \(3.2\,\text{m}, 3.6\,\text{m}, 3.4\,\text{m}, 3.9\,\text{m}, 3.7\,\text{m}, 3.5\,\text{m}, 3.8\,\text{m},\) and \(3.3\,\text{m}\). 2. Their sum is \(28.4\,\text{m}\). 3. The mean is \(28.4 \div 8 = 3.55\,\text{m}\).

The mean water level is \(3.55\,\text{m}\).

- Read the frequency for each rating. - Multiply each rating by its frequency. - Divide the total points by the total number of ratings.

1. The frequencies are \(3\) ratings of \(1\), \(7\) ratings of \(2\), \(6\) ratings of \(3\), and \(4\) ratings of \(4\). 2. The total number of points is \(3 \times 1 + 7 \times 2 + 6 \times 3 + 4 \times 4 = 51\). 3. The mean is \(51 \div 20 = 2.55\).

The mean rating is \(2.55\) points.

- The first row gives the values, and the second row gives their frequencies. - Find both the total number of students and the total number of laps. - A mean does not have to be a whole number.

1. The number of students is \(4 + 7 + 5 + 3 + 1 = 20\). 2. The total number of laps is \((2 \times 4) + (3 \times 7) + (4 \times 5) + (5 \times 3) + (6 \times 1) = 70\). 3. The mean is \(70 \div 20 = 3.5\) laps.

The mean is \(3.5\) laps.

- What does the second row of the table represent? - Does Mara’s calculation include every student? - Imagine writing each number of visits as many times as its frequency.

1. Mara counted each number of visits only once and ignored the frequencies. 2. The table represents \(8 + 6 + 4 + 2 = 20\) students. 3. The total number of visits is \((1 \times 8) + (2 \times 6) + (3 \times 4) + (4 \times 2) = 40\). 4. The correct mean is \(40 \div 20 = 2\) visits.

Mara ignored the frequencies. The correct mean is \(2\) visits.

- Compare the totals before and after the ninth round. - The new mean represents one more round. - The difference between the two totals is the ninth-round result.

1. Her total after eight rounds was \(15 \times 8 = 120\) hits. 2. Her total after nine rounds was \(16 \times 9 = 144\) hits. 3. The ninth-round score was \(144 - 120 = 24\) hits.

She made \(24\) hits in the ninth round.

- Add the frequencies first. - Identify the two middle positions. - Use cumulative frequencies; you do not need to write every value.

1. The table represents \(3 + 6 + 8 + 5 + 2 = 24\) robots. 2. The median is the mean of the \(12\)th and \(13\)th values in order. 3. The cumulative frequency through \(1\) failed attempt is \(9\), and through \(2\) failed attempts it is \(17\). Therefore, both middle values are \(2\).

The median is \(2\) failed attempts.

- Which position is in the middle of \(21\) values? - Add the frequencies from left to right. - Keep the unit in your answer.

1. The table represents \(2 + 5 + 7 + 4 + 3 = 21\) rehearsals. 2. The median is the \(11\)th value in order. 3. The cumulative frequency through \(2\) hours is \(7\), and through \(2.5\) hours it is \(14\). Therefore, the \(11\)th value is \(2.5\) hours.

The median rehearsal length is \(2.5\) hours.

- Convert the pie slices into frequencies. - Order the categories by number of transfers. - Use cumulative frequencies to locate the two middle trips.

1. There are \(24\) values, so the median is determined by positions \(12\) and \(13\). 2. The first \(6\) values are \(0\) transfers, and the next \(10\) values are \(1\) transfer. 3. Both middle positions are in the \(1\)-transfer category, so the median is \(1\) transfer.

The median is \(1\) transfer.

- Read the frequency for each star rating. - Place the ratings in order using their frequencies. - For \(20\) values, identify positions \(10\) and \(11\).

1. There are \(4\) one-star ratings, \(7\) two-star ratings, \(5\) three-star ratings, and \(4\) four-star ratings. 2. With \(20\) data values, the median is the mean of the values in positions \(10\) and \(11\) when the ratings are ordered. 3. Positions \(1\) through \(4\) are one-star ratings, and positions \(5\) through \(11\) are two-star ratings. Thus, both middle values are \(2\). 4. The median is \((2 + 2) \div 2 = 2\).

The median rating is \(2\) stars.

- Read all nine marked points. - The measurement order does not affect the median. - Order the depths and find the middle value.

1. Read the nine depths from the graph: \(42\,\text{cm}, 38\,\text{cm}, 45\,\text{cm}, 41\,\text{cm}, 39\,\text{cm}, 44\,\text{cm}, 40\,\text{cm}, 43\,\text{cm},\) and \(37\,\text{cm}\). 2. Order the values: \(37, 38, 39, 40, 41, 42, 43, 44, 45\) centimeters. 3. The fifth value is \(41\,\text{cm}\), so the median is \(41\,\text{cm}\).

The median depth is \(41\,\text{cm}\).

- Use cumulative frequencies. - With an even number of observations, identify two middle positions. - Determine which value contains both positions.

1. With \(40\) observations, the median is determined by positions \(20\) and \(21\). 2. The cumulative frequency through \(3\) tools is \(14\). 3. The next \(12\) observations have a value of \(4\), so both middle positions are \(4\).

The median is \(4\) tools checked out.

- Identify which two positions determine the median of eight observations. - Work backward from their average to find the unknown middle value. - Check that your number fits between its neighbors.

1. Eight ordered values have a median halfway between their fourth and fifth values, so \(\frac{x+14}{2}=12\). 2. The two middle values must sum to \(24\), giving \(x=24-14=10\). 3. Since \(8<10<14\), the list remains in increasing order.

\(x=10\)

- Place the median in the middle position first. - Put the same number of values on each side. - Check that all seven integers are different.

1. With seven ordered values, the fourth value must be \(20\). 2. Choose three different integers less than \(20\) and three different integers greater than \(20\), while keeping \(4\) as the least value. 3. One possible data set is \(4, 11, 17, 20, 23, 28, 35\).

One possible data set is \(4, 11, 17, 20, 23, 28, 35\).

- The bar heights show how many teams earned each score. - Add the heights before locating the middle positions. - Keep a running total from the smallest checkpoint count.

1. There are \(2+5+9+6+2=24\) teams. 2. The median uses the twelfth and thirteenth observations. 3. The first two categories cover \(7\) teams, and through \(3\) checkpoints they cover \(16\). Both middle teams completed \(3\) checkpoints.

\(3\) checkpoints

- Treat the two data sets separately. - Notice that the number of values changes from even to odd. - Insert the missing value in the correct position.

1. The incomplete data set has six values. Its middle values are \(9\) and \(11\), so its median is \((9 + 11) \div 2 = 10\). 2. The complete ordered data set is \(4, 6, 8, 9, 11, 13, 15\). 3. With seven values, the median is the fourth value, \(9\).

a) The incomplete median is \(10\). b) The correct median is \(9\).

- Compare where most of the values lie with each measure. - Look for a value far from the others. - A useful typical value should represent most observations.

1. The mean is \((12 + 14 + 15 + 13 + 16 + 56) \div 6 = 126 \div 6 = 21\) minutes. 2. In order, the times are \(12, 13, 14, 15, 16, 56\). The median is \((14 + 15) \div 2 = 14.5\) minutes. 3. The value \(56\) is much greater than the other times and pulls the mean upward. 4. The median better describes a typical repair time.

Mean: \(21\) minutes; median: \(14.5\) minutes. The median better describes a typical repair.

- Consider why one value is much lower than the others. - Compare both measures with the interval containing most values. - The purpose of the summary determines which measure is more useful.

1. The mean is \((0 + 28 + 30 + 31 + 32 + 33 + 35) \div 7 = 189 \div 7 = \$27\). 2. The median of the ordered data is \(\$31\). 3. The unusual value \(\$0\) pulls the mean downward. 4. The median is closer to most of the amounts and better describes the usual earnings.

Mean: \(\$27\); median: \(\$31\). The median better describes the groups' usual earnings.

- What do the separate circles stand for in a box plot? - Compare where the mean and median fall on the horizontal scale. - Which value better matches the middle half of the scores?

1. The two low outliers are at \(2.5\) and \(5\) points. 2. The mean marker is at \(17.5\) points, below the median line at \(20\) points. The low values pull the mean down. 3. The central half spans from \(15\) to \(25\), so the median of \(20\) points is a more useful usual score in this setting.

The low outliers pull the mean down to \(17.5\) points. The median, \(20\) points, better describes a usual score.

- The y-axis counts groups, while the x-axis gives minutes. - Find the total and work with each bar's frequency. - Look at how far the rare long waits lie from the main group.

1. The total number of groups is \(8+10+6+2=26\). 2. The weighted total time is \(2\times8+3\times10+4\times6+17\times2=104\) minutes, so the mean is \(104\div26=4\) minutes. 3. The thirteenth and fourteenth values both equal \(3\), so the median is \(3\) minutes. 4. The two waits of \(17\) minutes raise the mean; the median better reflects the typical short wait.

Mean: \(4\) minutes; median: \(3\) minutes. The median better describes a usual wait.

- An order does not automatically create measurable equal distances. - Identify the extra assumption introduced by the numerical codes. - State the conclusion cautiously rather than treating the mean as exact measurement.

1. The responses have an order, but the distances between the categories are not measured and may not be equal. 2. Coding the responses as \(1, 2, 3\) assumes equal spacing between the levels. 3. The resulting mean depends on that assumption. 4. The category percentages and the median category are safer to interpret.

Mei must state that the mean depends on an artificial coding system and the assumption that the response levels are equally spaced.

- Consider the effect of the special event. - Compare both measures with the normal Saturdays. - The situation being planned determines which values are relevant.

1. The mean is \((92 + 105 + 98 + 101 + 96 + 455 + 103) \div 7 = 1050 \div 7 = 150\) visitors. 2. In order, the values are \(92, 96, 98, 101, 103, 105, 455\), so the median is \(101\) visitors. 3. The festival attendance of \(455\) is not typical of a normal Saturday and greatly increases the mean. 4. The median is close to the six normal Saturdays and is better for planning.

The median of \(101\) visitors is better for planning a normal Saturday than the mean of \(150\) visitors.

- Count how many days each average represents. - Find the total rentals for the week. - Divide the weekly total by \(7\).

1. The weekday mean applies to \(5\) days, while the weekend mean applies to \(2\) days. 2. The total number of weekly rentals is \(5 \times 14 + 2 \times 28 = 126\). 3. The daily average is \(126 \div 7 = 18\) rentals. 4. Tom's calculation incorrectly gives the weekday and weekend means equal weight.

Tom's calculation is incorrect. The correct daily average is \(18\) boat rentals.

- Distinguish a typical group size from the total amount needed. - Identify which measure, together with the number of groups, determines the sum. - A large group is still important when planning the total.

1. The total number of people is \(12 + 12 + 13 + 13 + 14 + 14 + 41 = 119\). 2. The mean group size is \(119 \div 7 = 17\) people. 3. The total can be reconstructed from the mean and number of groups: \(17 \times 7 = 119\). 4. The median of \(13\) describes a typical group but does not determine the total.

The mean is more useful for the total planning. With \(7\) groups, it gives \(17 \times 7 = 119\) portions.

- Compare the newspaper's value with most individual purchases. - Look for one purchase that strongly affects the mean. - Distinguish a correct calculation from an appropriate interpretation.

1. The mean is \((4 + 5 + 5 + 6 + 6 + 7 + 44) \div 7 = 77 \div 7 = \$11\). 2. The median is \(\$6\). 3. Six purchases are between \(\$4\) and \(\$7\), while one purchase is \(\$44\). 4. The reported mean is mathematically correct but misleading as a description of a typical purchase.

The report is mathematically correct but misleading. The median of \(\$6\) better describes a typical purchase.

- Compare each statement with the box, median, mean, and outliers. - The mean does not always show where most values lie. - Choose the statement that accounts for the entire displayed distribution.

1. The median is \(10\) minutes, and the box extends from \(8\) to \(12\) minutes. 2. The high outliers at \(30\) and \(40\) minutes pull the mean to \(14\) minutes. 3. Statement A mistakes the shifted mean for the location of most values. 4. Statement C is false because the outliers increase the mean. 5. Statement B agrees with the box and the median.

Statement B is best supported. The median of \(10\) minutes is at the center of the box, while high outliers raise the mean.

- Use the target mean and new sample size to find the required total. - Subtract the original total to recover the added value. - Insert the new value into the ordered plot before finding its center and tallest stack.

1. The original total is \(149\). 2. After adding one observation, there will be \(11\) values. A mean of \(15\) requires a total of \(15\times11=165\). 3. The new observation is \(165-149=16\). 4. The value \(15\) still has the greatest frequency, \(3\), so the mode is \(15\). 5. In the ordered list of \(11\) values, the \(6\)th value is \(15\), so the median is \(15\).

New observation: \(16\) Mode: \(15\) Median: \(15\)

- Use the weekly mean to find the total for seven days. - Add the six known temperatures and subtract from the required total. - For part b), consider only the net change in the sum.

1. A mean of \(58^\circ\text{F}\) for \(7\) days requires a total of \(58\times7=406\). 2. The Monday-through-Saturday total is \(54+58+53+60+56+62=343\). 3. Sunday’s temperature is \(406-343=63^\circ\text{F}\). 4. Correcting Monday subtracts \(4\), while correcting Tuesday adds \(4\). The total change is \(-4+4=0\), so the total and the mean remain unchanged.

a) \(63^\circ\text{F}\) b) The mean does not change because the two corrections have a net effect of \(0\) on the total.

- Find one mean with all seven values and another without the suspected error. - Compare both means with the cluster of realistic temperatures. - Consider how one unusually large value affects a mean.

1. The sum of all readings is \(54+56+52+58+54+56+90=420\), so the mean is \(420\div7=60^\circ\text{F}\). 2. Without the suspected error, the sum is \(420-90=330\), so the mean is \(330\div6=55^\circ\text{F}\). 3. The six realistic readings range from \(52^\circ\text{F}\) to \(58^\circ\text{F}\). The mean of \(55^\circ\text{F}\) lies within this cluster, while the \(90^\circ\text{F}\) outlier pulls the overall mean upward. Therefore, \(55^\circ\text{F}\) better represents the typical weather.

a) \(60^\circ\text{F}\) b) \(55^\circ\text{F}\) c) \(55^\circ\text{F}\) better describes the typical weather because the suspected outlier raises the mean of all seven readings.

- What total must five values have if their mean is \(20\)? - Which position contains the median when five values are ordered? - Set the median first, then choose the other values so the total is correct. - Order the values and check that none are repeated.

1. For five ordered values \(x_1 < x_2 < x_3 < x_4 < x_5\), the median is the middle value, so \(x_3 = 10\). 2. A mean of \(20\) requires a total of \(5 \times 20 = 100\). 3. Choose two different positive integers less than \(10\), such as \(1\) and \(2\). Their total with the median is \(1 + 2 + 10 = 13\). 4. The last two values must total \(100 - 13 = 87\). Choosing \(11\) gives the final value \(87 - 11 = 76\). 5. The data set \(1, 2, 10, 11, 76\) has median \(10\), mean \((1 + 2 + 10 + 11 + 76) \div 5 = 20\), and five different positive-integer values.

One possible data set is \(1, 2, 10, 11, 76\). Its median is \(10\), its mean is \(20\), and all five values are different positive integers.

- Read all seven values from the graph and order them. - How do you find the median when there is an odd number of values? - With eight values, which two positions determine the median? - Test possible whole-number scores for the eighth game.

1. The graph shows \(2, 1, 4, 1, 3, 2,\) and \(5\) goals. 2. In order, the values are \(1, 1, 2, 2, 3, 4, 5\). The middle value is \(2\), so the median is \(2\) goals. 3. With eight values, the median is the mean of the fourth and fifth values in order. If the eighth score is \(0\), \(1\), or \(2\), those middle values are both \(2\), so the median is \(2\). 4. If the eighth score is \(3\) or more, the middle values are \(2\) and \(3\), giving a median of \(2.5\). Therefore, the possible eighth-game scores are \(0, 1,\) and \(2\).

a) \(2\) goals b) \(0\), \(1\), or \(2\) goals

- Find the total needed for seven values to have a mean of \(10\). - More than one pair can work. - Check every condition on the two missing ratings.

1. The required total is \(10 \times 7 = 70\) points. 2. The five known ratings total \(6 + 8 + 9 + 11 + 14 = 48\) points. 3. The two missing ratings must total \(70 - 48 = 22\). 4. One possible pair is \(10\) and \(12\). They are different whole numbers in the required range and have a sum of \(22\).

One possible pair is \(10\) and \(12\).

- Consider where the new value can appear in the ordered list. - Analyze ranges of values rather than testing only a few examples. - Check the boundary where the two middle positions change.

1. With six values, the median is the mean of the third and fourth values in order. 2. If \(x \ge 12\), the third and fourth values remain \(8\) and \(12\). 3. Their mean is \((8 + 12) \div 2 = 10\). 4. If \(x < 12\), the median is less than \(10\). Therefore, every nonnegative whole number \(x \ge 12\) works.

All nonnegative whole numbers \(x \ge 12\) are possible.

- After the removal, there is an even number of values. - First test removals to the left of the middle. - More than one value may work.

1. After one value is removed, eight values remain, so the median is the mean of the fourth and fifth values. 2. If \(2\), \(4\), \(5\), or \(6\) is removed, the middle values are \(7\) and \(8\). 3. Their mean is \((7 + 8) \div 2 = 7.5\). 4. Removing \(7\) or any greater value produces different middle values, so no other removal works.

\(2\), \(4\), \(5\), or \(6\) can be removed.

- Test values near the point where the middle position moves from the threes to the twos. - Track both the total number of values and the middle position. - Show why a smaller value does not work.

1. If \(x=4\), there are \(4 + 4 + 5 + 3 = 16\) values. The middle values, in positions \(8\) and \(9\), are \(2\) and \(3\), so the median is \((2 + 3) \div 2 = 2.5\). 2. If \(x=5\), there are \(17\) values. The median is the ninth value, and the block of twos occupies positions \(5\) through \(9\). 3. Therefore, the median is \(2\) when \(x=5\). Since every smaller value of \(x\) gives a median greater than \(2\), this is the least possible value.

The least possible value is \(x=5\).

- Translate the stacks of dots into an ordered list. - Determine how many values remain and which positions set the new median. - Test removals from different parts of the distribution and look for groups that have the same effect.

1. The plotted data in order are \(0, 1, 1, 2, 3, 3, 3, 4, 5\). 2. After one observation is removed, \(8\) values remain, so the median is the mean of the fourth and fifth values. 3. Removing \(3\), \(4\), or \(5\) leaves \(2\) and \(3\) as the two middle values, giving \((2+3)\div2=2.5\). 4. Removing \(0\), \(1\), or \(2\) leaves the two middle values equal to \(3\), so those removals do not work.

A value of \(3\), \(4\), or \(5\) could be removed.

- Identify the likely mode after removing one nonmodal dot. - Express the new mean using the original total minus the removed value. - Check that the resulting stack remains uniquely tallest.

1. The original total is \(15\). After removing one value \(r\), the new mean is \(\frac{15-r}{6}\). 2. Removing a \(2\) would leave values \(1\) and \(2\) tied for greatest frequency, so any valid removal must leave \(2\) as the unique mode. 3. Set the new mean equal to \(2\): \(\frac{15-r}{6}=2\). Then \(15-r=12\), so \(r=3\). 4. The new data are \(1,1,2,2,2,4\), with mean \(2\) and unique mode \(2\).

Remove the dot at \(3\). The new mean and unique mode are both \(2\).

- 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 \((5 + 5 + 5 + 10 + 10 + 10 + 15 + 15 + 15 + 65 + 65) \div 11 = 220 \div 11 = 20\).

- 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 \(Q_3 = 10\).