Appendix A — Answer Key, Chapter 17: Mean as a Balance Point; Outliers

SOL 6.PS.2 · Covers textbook Chapter 17 and the companion workbook. Item numbers match the textbook; workbook items that repeat textbook problems share the same answers, and workbook-only items appear in the final section. Reasoning answers show an acceptable response, not the only wording.

Lesson 17.1 — Mean as a Balance Point

Guided practice

1. Mean $= \dfrac{1 + 1 + 2 + 4}{4} = \dfrac{8}{4} = 2$. Left distances: $1 + 1 = 2$; the $2$ sits on the balance point. Right distance: $2$. The sides match.
2. Mean $= \dfrac{7 + 8 + 9 + 12}{4} = \dfrac{36}{4} = 9$.
3. Mean $= \dfrac{3 + 3 + 4 + 6}{4} = \dfrac{16}{4} = 4$, so the balance point is $4$. Left: $1 + 1 = 2$; right: $2$.
4. Yes. Mean $= \dfrac{4 + 5 + 6 + 9}{4} = \dfrac{24}{4} = 6$. Left distances: $2 + 1 = 3$; right distance: $3$.
5. Because the total distance of the values below the mean equals the total distance of the values above it. If each dot were a weight on a board, a support at the mean would hold the board level.

Independent practice

6. Mean $= \dfrac{10 + 10 + 12 + 14 + 14}{5} = \dfrac{60}{5} = 12$.
7. Mean $= \dfrac{2 + 5 + 5 + 8}{4} = \dfrac{20}{4} = 5$. Left distance: $3$. Right distance: $3$. The two fives sit on the balance point and contribute $0$.
8. Sum $= 1 + 2 + 2 + 3 + 3 + 3 + 4 + 6 = 24$, so the mean is $\dfrac{24}{8} = 3$. Left distances: $2 + 1 + 1 = 4$. Right distances: $1 + 3 = 4$. Balanced.
9. The four values must total $4 \times 5 = 20$. Since $3 + 4 + 8 = 15$, the missing value is $20 - 15 = 5$.
10. The four values must total $4 \times 7 = 28$. Since $5 + 6 + 8 = 19$, the fourth value is $28 - 19 = 9$.
11. Mean $= \dfrac{68 + 70 + 72 + 74 + 76}{5} = \dfrac{360}{5} = 72$ degrees. Left distances: $4 + 2 = 6$. Right distances: $2 + 4 = 6$. The temperatures are spread evenly on both sides of the mean, so the balance point sits exactly in the middle of the data.
12. Sample answer: $2$ and $4$ have a mean of $3$, and $3$ is not in the data set. The balance point is a location on the number line, not one of the plotted values; it only has to make the left and right distances equal, and here both are $1$.

Exit ticket 17.1

1. Mean $= \dfrac{3 + 3 + 4 + 6}{4} = \dfrac{16}{4} = 4$.
2. Left total: $1 + 1 = 2$. Right total: $2$.
3. The three values must total $3 \times 5 = 15$. Since $2 + 6 = 8$, the missing value is $7$.
4. The balance point is the place on the number line where the dots would balance if each were an equal weight: the total distance of the dots on the left equals the total distance of the dots on the right. That place is the mean.

Lesson 17.2 — Measures of Center and Range

Guided practice

1. Mean $\dfrac{50}{5} = 10$; median $9$; mode $9$; range $14 - 8 = 6$.
2. Mean $\dfrac{120}{5} = 24$; median $24$; no mode; range $28 - 20 = 8$.
3. Mean $\dfrac{68}{8} = 8.5$; median $\dfrac{8 + 9}{2} = 8.5$; modes $7$ and $9$; range $12 - 6 = 6$.
4. Range $= 11 - 3 = 8$.
5. The median is the value in the middle position, so the values must be in order before you can tell which one is in the middle. In $18$, $12$, $20$, $15$, $15$ the middle number as written is $20$, but the true median is $15$.

Independent practice

6. Mean $\dfrac{20}{5} = 4$; median $3$; mode $3$; range $8 - 1 = 7$.
7. Mean $\dfrac{210}{6} = 35$; median $\dfrac{30 + 40}{2} = 35$; no mode; range $60 - 10 = 50$.
8. Mean $\dfrac{20}{4} = 5$; median $5$; mode $5$; range $5 - 5 = 0$.
9. Mean $\dfrac{30}{6} = 5$; median $\dfrac{5 + 5}{2} = 5$; mode $5$; range $9 - 2 = 7$.
10. Sum $= 0(3) + 1(7) + 2(5) + 3(3) + 4(1) + 5(1) = 35$ for 20 students, so the mean is $\dfrac{35}{20} = 1.75$. The 10th value is $1$ and the 11th is $2$, so the median is $1.5$. The tallest stack is above $1$, so the mode is $1$. Range $= 5 - 0 = 5$.
11. Sum $= 6 + 8 + 8 + 9 + 10 + 10 + 12 = 63$, so the mean is $\dfrac{63}{7} = 9$; median $9$; modes $8$ and $10$; range $12 - 6 = 6$. Sample answer: either the mean or the median works here, since both are $9$ and no score is far from the rest; reporting $9$ describes a typical score well.
12. Sample answer: $1$, $2$, $3$, $4$, $20$. The mean is $\dfrac{30}{5} = 6$ and the median is $3$. One value far above the rest pulls the balance point to the right while leaving the middle position unchanged, so the mean exceeds the median.

Exit ticket 17.2

1. Mean $\dfrac{28}{4} = 7$; median $\dfrac{7 + 7}{2} = 7$; mode $7$; range $10 - 4 = 6$.
2. Ordered: $2$, $4$, $6$, $7$, $9$. Median $= 6$.
3. Range $= 22 - 8 = 14$.
4. If no value repeats — or if every value appears the same number of times — there is no single value that occurs most often, so the set has no mode. For example, $4$, $6$, $8$, $10$ has no mode.

Lesson 17.3 — What Happens When Data Changes

Guided practice

1. Before: mean $\dfrac{21}{3} = 7$; median $7$. After removing $9$: the data are $5$, $7$, so mean $\dfrac{12}{2} = 6$ and median $\dfrac{5 + 7}{2} = 6$. Both dropped by $1$.
2. Before: mean $\dfrac{25}{5} = 5$; median $5$; no mode. After adding $5$: the data are $3$, $4$, $5$, $5$, $6$, $7$, so mean $\dfrac{30}{6} = 5$, median $\dfrac{5 + 5}{2} = 5$, and the mode is now $5$. The mean and median are unchanged because the added value equals both.
3. Before: mean $\dfrac{52}{4} = 13$; median $\dfrac{12 + 14}{2} = 13$. After changing $16$ to $24$: mean $\dfrac{60}{4} = 15$; median still $13$, since the middle two values are still $12$ and $14$.
4. Down. Adding a weight to the left of the balance point tips the board that way, so the mean decreases.
5. The mean uses every value, so any change to any value changes the sum and therefore the mean. The median uses only the middle position, so changing a value at one end may leave the middle value exactly where it was.

Independent practice

6. Before: mean $\dfrac{30}{5} = 6$; median $6$; range $10 - 2 = 8$. After adding $12$: mean $\dfrac{42}{6} = 7$; median $\dfrac{6 + 8}{2} = 7$; range $12 - 2 = 10$.
7. Before: mean $\dfrac{110}{4} = 27.5$; median $\dfrac{25 + 30}{2} = 27.5$; range $35 - 20 = 15$. After removing $20$: mean $\dfrac{90}{3} = 30$; median $30$; range $35 - 25 = 10$.
8. Before: mean $\dfrac{50}{5} = 10$; median $9$; mode $8$; range $15 - 8 = 7$. After changing $15$ to $10$: the data are $8$, $8$, $9$, $10$, $10$, so mean $\dfrac{45}{5} = 9$; median $9$; modes $8$ and $10$; range $10 - 8 = 2$.
9. Add $6$. A value equal to the mean sits exactly at the balance point, so it does not tip the board: the new sum is $24$ over $4$ values, and $\dfrac{24}{4} = 6$.
10. The four values must total $4 \times 11 = 44$. The current sum is $30$, so add $14$.
11. Before: mean $\dfrac{500}{5} = 100$; median $100$; range $110 - 90 = 20$. After the $130$: mean $\dfrac{630}{6} = 105$; median $\dfrac{100 + 105}{2} = 102.5$; range $130 - 90 = 40$. One unusually high game raised her mean by $5$ and doubled the range, while the median moved only $2.5$, so the sixth game says more about her best night than about her typical night.
12. The mean. It is computed from every value, and the balance point must shift toward a weight placed far from it. The median only moves to the next position in the ordered list, and the mode does not change at all unless the new value creates a repeat.

Exit ticket 17.3

1. New sum $= 3 + 5 + 7 + 9 = 24$, so the new mean is $\dfrac{24}{4} = 6$.
2. Median of $4$, $6$, $8$, $10$ is $\dfrac{6 + 8}{2} = 7$. After removing $10$: the data are $4$, $6$, $8$, so the median is $6$.
3. New data: $6$, $8$, $10$, $20$. Mean $= \dfrac{44}{4} = 11$.
4. The mean decreases. The new value sits to the left of the balance point, so it pulls the balance toward the lower end.

Lesson 17.4 — Outliers and Their Effect

Guided practice

1. The outlier is $40$. With it: mean $\dfrac{72}{6} = 12$; median $\dfrac{6 + 7}{2} = 6.5$; range $40 - 5 = 35$. Without it: mean $\dfrac{32}{5} = 6.4$; median $6$; range $8 - 5 = 3$.
2. The mean changed most among the centers, from $12$ to $6.4$; the range changed most overall, from $35$ to $3$. The median moved only from $6.5$ to $6$.
3. The outlier is $30$. Without it the data are $70$, $72$, $73$, $74$, $75$, $76$, $78$: mean $\dfrac{518}{7} = 74$; median $74$; range $78 - 70 = 8$. (With the outlier: mean $\dfrac{548}{8} = 68.5$; median $\dfrac{73 + 74}{2} = 73.5$; range $48$.)
4. No. A high outlier raises the mean, but a low outlier lowers it. In $30$, $86$, $88$, $90$, $92$, $94$, the outlier $30$ pulls the mean down from $90$ to $80$.
5. The dots for most of the data form a cluster. A wide empty space followed by a single dot shows that one value lies far from all the others, which is exactly what an outlier is.

Independent practice

6. The outlier is $30$. With it: mean $\dfrac{50}{5} = 10$; median $5$; mode $5$; range $30 - 4 = 26$. Without it: mean $\dfrac{20}{4} = 5$; median $5$; mode $5$; range $6 - 4 = 2$.
7. The outlier is $1$. With it: mean $\dfrac{190}{5} = 38$; median $47$; range $49 - 1 = 48$. Without it: mean $\dfrac{189}{4} = 47.25$; median $\dfrac{47 + 48}{2} = 47.5$; range $49 - 45 = 4$.
8. The median and the mode. The median depends only on the middle position, which a single far-off value shifts by at most one step, and the mode counts repeats, which a lone value does not create.
9. The outlier is $45$. The mean will be greater than the median, because the high value pulls the balance point up while leaving the middle position near the cluster.
10. The mode is the most frequently occurring value. A single outlier appears once, so it almost never becomes the most frequent value, and it does not change how often the other values occur.
11. With the outlier: mean $\dfrac{60}{5} = \$12$; median $\$6$. Without the $36: mean $\dfrac{24}{4} = \$6$; median $\$5.50$. The median, about $\$6$, better describes a typical allowance, because the mean of $\$12$ is higher than four of the five actual allowances.
12. They should find out where the value came from. If it is a recording or equipment error — a stopwatch left running, a mistyped number — they should correct it or remove it and say so in their report. If it is a genuine measurement, they should keep it and report both the mean and the median, so readers can see the typical value and the effect of the unusual one. Either way they must state what they did and why.

Exit ticket 17.4

1. The outlier is $52$.
2. With: mean $\dfrac{90}{5} = 18$. Without: mean $\dfrac{38}{4} = 9.5$.
3. With: median $10$. Without: median $\dfrac{9 + 10}{2} = 9.5$. It changed by only $0.5$.
4. Put the data in order or plot it on a number line, then look for a value separated from the cluster by a noticeable gap. The greatest value is not automatically an outlier; it is an outlier only when it stands far apart from the rest.

Chapter 17 Review

Part A — Mean as a balance point (6.PS.2a)

1. Mean $= \dfrac{3 + 5 + 7 + 9}{4} = \dfrac{24}{4} = 6$. Left distances: $3 + 1 = 4$. Right distances: $1 + 3 = 4$.
2. Mean $= \dfrac{2 + 3 + 3 + 8}{4} = \dfrac{16}{4} = 4$, so the balance point is $4$.
3. The three values must total $3 \times 8 = 24$. Since $6 + 9 = 15$, the missing value is $9$.
4. If the support were placed below the least value, every dot would be on one side and the board would tip. The same is true above the greatest value. Balance is only possible somewhere between the two extremes.

Part B — Effect of adding, removing, or changing a value (6.PS.2b)

5. Before: mean $\dfrac{40}{5} = 8$; median $8$; range $8$. After adding $20$: mean $\dfrac{60}{6} = 10$; median $\dfrac{8 + 10}{2} = 9$; range $20 - 4 = 16$.
6. Before: mean $\dfrac{50}{5} = 10$; median $10$. After removing $14$: mean $\dfrac{36}{4} = 9$; median $\dfrac{8 + 10}{2} = 9$.
7. Before: mean $\dfrac{80}{4} = 20$; median $\dfrac{18 + 21}{2} = 19.5$. After changing $26$ to $30$: mean $\dfrac{84}{4} = 21$; median still $19.5$.
8. Add $5$, the value of the mean. The new sum is $25$ over $5$ values, and $\dfrac{25}{5} = 5$, so the balance point does not move; the added weight sits directly on it.
9. The median depends only on which value holds the middle position. Changing a value at one end leaves the ordered positions of the middle values untouched, so the median stays where it was even though the sum, and therefore the mean, changes.

Part C — Outliers (6.PS.2c)

10. The outlier is $60$. With it: mean $\dfrac{132}{6} = 22$; median $\dfrac{15 + 15}{2} = 15$; mode $15$; range $60 - 12 = 48$. Without it: mean $\dfrac{72}{5} = 14.4$; median $15$; mode $15$; range $16 - 12 = 4$.
11. The outlier is $30$. With it: mean $\dfrac{480}{6} = 80$; median $\dfrac{88 + 90}{2} = 89$; range $94 - 30 = 64$. Without it: mean $\dfrac{450}{5} = 90$; median $90$; range $94 - 86 = 8$.
12. No outlier. The values rise steadily with no gap, so none stands far from the rest. Mean $= \dfrac{147}{6} = 24.5$; median $= \dfrac{24 + 25}{2} = 24.5$.
13. Most affected: the mean and the range. The mean uses every value, so a weight far from the balance point tips it, and the range is computed from the extremes, one of which is the outlier. Least affected: the median and the mode, because the median depends only on the middle position and the mode depends on repeats.

Part D — Mixed application and reasoning

14. Sum $= 20 + 25 + 25 + 30 + 35 + 45 + 65 = 245$, so the mean is $\dfrac{245}{7} = 35$; median $30$; mode $25$; range $65 - 20 = 45$. The outlier is $65$. Without it, the mean is $\dfrac{180}{6} = 30$, so the outlier raises the mean by $5$ minutes and lifts it above the median, making homework time look longer than it is for most of the group.
15. There is probably at least one low outlier, and possibly a few low scores. Most scores cluster near or above $88$, but one or more very low scores pull the balance point down to $80$ while the middle position stays high. Whenever the mean is well below the median, look for values far below the cluster.
16. Counterexample: in $30$, $86$, $88$, $90$, $92$, $94$, removing the outlier $30$ raises the mean from $80$ to $90$. The error is assuming outliers are always high. A low outlier drags the mean down, so removing it makes the mean larger.
17. First decide whether the value is real. A time of $600$ seconds next to a field of runners near $60$ seconds is almost certainly a recording error, probably a stopwatch left running, so the class should check the record and either correct it or remove it, and state clearly in the report what they did. If the value turned out to be real — a runner who walked the course — they should keep it and report the median along with the mean. The median is useful because it barely moves when one extreme value is present, so it still describes a typical runner.

Workbook-only items

Page 2, balance point. Sum $12$; number of values $3$; mean $4$. Distance table: $2$ is $2$ to the left; $4$ is $0$, on the point; $6$ is $2$ to the right. Left total $2$; right total $2$; yes, they match.

Page 3, balance table.

Page 3, missing values. $3$, $4$, 5, $8$ has a mean of $5$. The fourth value is 9.

Page 3, explain. The balance point is a position on the number line, not a data value. It only has to make the left and right distances equal, and that position may fall between two dots — as with $2$ and $4$, whose mean is $3$.

Page 5, measures table.

Page 5, careful. Order (put in order from least to greatest).

Page 6, line plot. Total students $20$; sum of values $35$; mean $1.75$; median $1.5$; mode $1$; range $5$.

Page 6, more practice.

Page 8, predict table. Add a value greater than the mean: up. Add a value less than the mean: down. Add a value equal to the mean: no change. Remove a value greater than the mean: down.

Page 8, table A. Before: mean $8$, median $8$, range $8$. After: mean $10$, median $9$, range $16$.

Page 8, table B. Before: mean $10$, median $10$, range $8$. After: mean $9$, median $9$, range $6$.

Page 8, table C. Before: mean $20$, median $19.5$, range $11$. After: mean $21$, median $19.5$, range $15$.

Page 9, change practice. 1. Mean $7$, median $7$, range $10$. 2. Mean $30$, median $30$, range $10$. 3. Mean $9$, median $9$, modes $8$ and $10$, range $2$. 4. Add $6$. 5. Add $14$. 6. Before: mean $100$, median $100$, range $20$; after: mean $105$, median $102.5$, range $40$.

Page 9, explain. The mean uses every value, so any change changes the sum. The median uses only the middle position, so a change at one end can leave it exactly where it was.

Page 11, circle the outlier. a) $40$ b) $30$ c) none d) $36$

Page 11, with and without table.

Page 11, which changed most. The mean and the range.

Page 12, outlier effects. 1. Outlier $30$. With: mean $10$, median $5$, mode $5$, range $26$. Without: mean $5$, median $5$, mode $5$, range $2$. 2. Outlier $1$. With: mean $38$, median $47$, range $48$. Without: mean $47.25$, median $47.5$, range $4$. 3. Outlier $36$. With: mean $\$12$, median $\$6$. Without: mean $\$6$, median $\$5.50$. The median describes a typical allowance better, because the mean of $\$12$ is higher than four of the five actual amounts.

Page 12, decide. The $600$ seconds is almost certainly a recording error, so the class should check the record and correct or remove the value. Either way they must report what they did and why, and reporting the median alongside the mean shows readers a typical time.

Pages 14–15, review. Same answers as the textbook Chapter 17 Review above, items 1–17. Page 15 item 14 also asks for the outlier: $65$, which raises the mean from $30$ to $35$ and lifts it above the median of $30$.