Answer Key

7.6 Statistics

Pack A — Answers

#	Question	Answer
1	Find the mode of: 3, 7, 3, 9, 5, 3, 7. Justify by stating how many times it appears compared to all other values.	Mode = 3; it appears 3 times, which is more than any other value (7 appears twice, 9 and 5 appear once)
2	Find the median of: 5, 2, 8, 1, 6. Explain what you had to do before finding the middle value and why.	Median = 5; must sort first (1, 2, 5, 6, 8) so that the middle value represents the centre of the data
3	Three numbers have a mean of 6. Two of the numbers are 4 and 10. Find the third number and justify using the definition of mean.	4; because mean = total ÷ count, so total = 6 × 3 = 18; third number = 18 − 4 − 10 = 4
4	Find the range of: 12, 5, 18, 9, 3. Explain what the range tells us about the spread of the data.	Range = 15; this means the data spans 15 units — the values are fairly spread out
5	A bar chart shows books read: Fiction 14, Non-fiction 9, Comics 5. A student says "Fiction is more than double Non-fiction." Is this correct? Justify.	No — double 9 is 18, and 14 < 18, so Fiction is not more than double Non-fiction
6	A frequency table shows favourite colours: Red = 8, Blue = 12, Green = 5, Yellow = 3. (a) How many students were surveyed? (b) Is "favourite colour" categorical or numerical data? Justify.	(a) 28 students; (b) categorical — colours are labels, not numbers
7	A shoe-size frequency table shows: Size 4: 3; Size 5: 8; Size 6: 5; Size 7: 2 students. State the mode and explain why it is called the modal size.	Modal size = 5; it has frequency 8, higher than any other size — mode means most common
8	Find the median of: 3, 11, 7, 1, 9, 5. Explain why you average the two middle values instead of picking just one.	Median = 6; with an even count there is no single middle value, so we average the two middle values to find the centre
9	A pie chart shows how 30 students travel to school: Walking = 10, Bus = 15, Car = 5. What fraction walk? Justify that the fractions for all groups must add to 1.	$\dfrac{1}{3}$ walk; fractions: $\frac{10}{30} + \frac{15}{30} + \frac{5}{30} = \frac{30}{30} = 1$ ✓
10	A bar chart is drawn for: Mon = 4, Tue = 6, Wed = 2, Thu = 8. A student wants the vertical axis to go up in steps of 3. Explain why this is a poor choice and suggest a better scale step. Justify.	Steps of 3 are poor because the highest value (8) does not fall on a gridline; steps of 2 are better (gridlines at 2, 4, 6, 8 — all bars land exactly on gridlines)
11	A frequency table shows scores out of 10: Score 6: 4 students; Score 8: 3 students; Score 10: 3 students. Find the mean score.	7.6
12	Find the median of this data set: 14, 3, 21, 9, 17, 5, 11, 8, 25, 13, 7.	11
13	Two classes sit a spelling test. Class A scores: 6, 8, 7, 9, 5. Class B scores: 4, 10, 6, 9, 1. Compare the means and ranges of the two classes.	Class A: mean = 7, range = 4. Class B: mean = 6, range = 9. Class A has a higher mean and a smaller range, so it performed better and more consistently.
14	A survey of 30 people found 10 prefer comedy films. What angle (in degrees) should the comedy sector be in a pie chart?	120°
15	A two-way table shows student choices. Year 7: Swimming 14, Tennis 11, Total 25. Year 8: Swimming 9, Tennis 16, Total 25. How many students play tennis altogether?	27
16	The mean of four numbers is 8. Three of them are 5, 9, 6. Find the fourth number.	12
17	A pie chart for 36 students has a sector of 120° for "Football". How many students chose Football?	12
18	For the data set 4, 7, 4, 9, 4, 7, 3, 9, find (a) the mode and (b) the median.	(a) Mode = 4; (b) Median = 5.5
19	Five plants are measured (cm): 12.4, 9.7, 11.3, 10.6, 13.5. Find the mean height, rounding to 1 decimal place.	11.5 cm
20	A bar chart shows daily visitors to a café over 5 days: Mon 40, Tue 55, Wed 50, Thu 45, Fri 60. Calculate the mean number of daily visitors.	50
21	The mean of 5 test scores is 78. 4 of the scores are 65, 80, 70, 90. Find the missing score.	85
22	A pie chart has a sector of 90°. (a) What fraction of the data does it represent? (b) What percentage?	(a) $\dfrac{1}{4}$ (b) 25%
23	Team A sprint times (s): 11.2, 12.4, 11.8, 13.0, 11.6. Team B sprint times (s): 10.5, 14.1, 11.9, 12.5, 11.0. Compare the two distributions using the mean and range.	Team A: mean = 12.0 s, range = 1.8 s. Team B: mean = 12.0 s, range = 3.6 s. Both teams have the same mean speed but Team A is more consistent (smaller range).
24	6 friends share a total of 48 sweets equally. Then 12 more sweets are added and shared equally among the same group. How many sweets does each person have now?	10
25	Eight students score: 5, 6, 6, 7, 7, 7, 8, 40. (a) Find the mean, median, and mode. (b) Which average best represents the data? Explain.	(a) Mean = 10.75, Median = 7, Mode = 7. (b) Median or mode — the score of 40 is an outlier that distorts the mean.
26	A survey of 60 people found: Category A = 20, Category B = 25, Category C = 15. Calculate the angle for each sector to draw a pie chart.	A = 120°, B = 150°, C = 90°
27	The mean of seven numbers is 12. Six of the numbers have a sum of 74. Write and solve an equation to find the seventh number.	10
28	A pie chart shows that 35% of 60 students chose Science as their favourite subject. How many students is that?	21
29	A company has 10 employees. The 2 managers each earn £100000; the remaining 8 staff each earn £20000. (a) Calculate the actual mean salary. (b) Explain why the mean might be misleading as a "typical" salary.	(a) Mean = £36 000. (b) The mean is pulled up by the two high manager salaries; 8 out of 10 employees earn only £20 000. The median (£20 000) is more representative.
30	A data set has 5 whole numbers. The mode is 6, the range is 8, and the mean is 7. Write one possible data set.	One possible set: 3, 6, 6, 9, 11 — but verify against constraints. A correct set: 3, 6, 6, 8, 12 — nope. Valid example: 3, 6, 6, 10, 10 gives mean=(3+6+6+10+10)÷5=35÷5=7, mode=6 (appears twice, but so does 10 — bimodal). Better: 3, 6, 6, 8, 12: range=9 (no). Try: 3, 6, 6, 7, 13: range=10 (no). Try: 4, 6, 6, 8, 11: sum=35, mean=7 ✓, range=7 (no). Try: 3, 6, 6, 9, 11: sum=35 ✓, mean=7 ✓, range=8 ✓, mode=6 ✓. Answer: 3, 6, 6, 9, 11.
31	Five positive integers have mean 11, median 10, and range 6. List one possible set of numbers.	One valid set: 8, 9, 10, 13, 15 — check: sum=55, mean=11 ✓; median=10 ✓; range=15−8=7 (no). Try: 8, 9, 10, 12, 16: sum=55 ✓, range=8 (no). Try: 9, 10, 10, 12, 14: sum=55 ✓, range=14−9=5 (no). Try: 8, 10, 10, 13, 14: sum=55 ✓, range=6 ✓, median=10 ✓. Answer: 8, 10, 10, 13, 14.
32	Daily temperatures (°C) for one week: 3, −2, 0, 4, −1, 5, −3. Find the mean and range.	Mean = $\dfrac{6}{7}$ ≈ 0.9 °C (or exact: $\dfrac{6}{7}$ °C). Range = 8 °C.
33	A market stall records daily profit (+) or loss (−) in £ over 5 days: 35, −12, 48, −20, 9. Find the mean daily profit/loss.	£12 profit
34	A student has sat 4 assessments with a current mean of 65. She wants her overall mean to be 70 after a total of 6 assessments. What total must she score across the remaining 2 assessments?	She must score a total of 160 in the remaining 2 assessments (a mean of 80 per assessment).
35	Two archers each shoot 6 arrows. Archer A distances from bullseye (cm): 3, 5, 4, 6, 3, 3. Archer B: 2, 8, 1, 9, 2, 8. Compare using mean and range, and state which archer you would pick for a competition.	Archer A: mean = 4 cm, range = 3 cm. Archer B: mean = 5 cm, range = 8 cm. Choose Archer A — lower mean (more accurate on average) and lower range (more consistent).
36	A class of 40 students chose a sport. Results: Football = 16, Tennis = 8, Swimming = 10, Other = 6. Calculate the angle and percentage for each sector.	Football: 144°, 40%. Tennis: 72°, 20%. Swimming: 90°, 25%. Other: 54°, 15%.
37	The sum of 12 data values is 156. Without dividing, use prime factorisation to check whether the mean will be a whole number. Then calculate the mean.	Mean = 13 (a whole number)
38	A pie chart shows how a school spends its budget of £200000. Teaching staff = 60%, Resources = 20%, Premises = 15%, Other = 5%. (a) Calculate the amount spent on each category. (b) What angle represents Premises in the pie chart?	(a) Teaching: £120 000, Resources: £40 000, Premises: £30 000, Other: £10 000. (b) Premises = 54°.
39	Lengths of 30 leaves are grouped: 1–3 cm: 6 leaves; 3–5 cm: 14 leaves; 5–7 cm: 10 leaves. Use midpoints to estimate the mean length.	4.3 cm
40	A data set is: $x$, 8, 14, $2x$, 6. The mean of the data set is 11. Form an equation and solve it to find $x$.	$x = 9$

Pack B — Answers

#	Question	Answer
1	Find the mode of: 4, 6, 4, 8, 6, 4, 2. Justify by stating how many times it appears compared to all other values.	Mode = 4; it appears 3 times, which is more than any other value (6 appears twice, 8 and 2 appear once)
2	Find the median of: 9, 3, 7, 1, 5. Explain what you had to do before finding the middle value and why.	Median = 5; must sort first (1, 3, 5, 7, 9) so the middle value is meaningful
3	Three numbers have a mean of 6. Two of the numbers are 3 and 7. Find the third number and justify using the definition of mean.	8; because total = 6 × 3 = 18; third number = 18 − 3 − 7 = 8
4	Find the range of: 20, 7, 15, 2, 11. Explain what the range tells us about the spread of the data.	Range = 18; the data spans 18 units between the lowest (2) and highest (20) values
5	A bar chart shows books read: Fiction 20, Non-fiction 9, Comics 7. A student says "Fiction is more than double Non-fiction." Is this correct? Justify.	No — double 9 is 18, and 20 > 18 but barely; Fiction is more than double since 20 > 18 ✓ — actually yes for Pack B
6	A frequency table shows favourite colours: Red = 10, Blue = 7, Green = 9, Yellow = 4. (a) How many students were surveyed? (b) Is "favourite colour" categorical or numerical data? Justify.	(a) 30 students; (b) categorical — colours are labels/categories, not numerical values
7	A shoe-size frequency table shows: Size 5: 6; Size 6: 11; Size 7: 4; Size 8: 1 students. State the mode and explain why it is called the modal size.	Modal size = 6; it has frequency 11, the highest of all sizes
8	Find the median of: 8, 2, 14, 6, 10, 4. Explain why you average the two middle values instead of picking just one.	Median = 7; with 6 values there is no single middle, so we average the 3rd and 4th values
9	A pie chart shows how 40 students travel to school: Walking = 20, Bus = 12, Car = 8. What fraction walk? Justify that the fractions for all groups must add to 1.	$\dfrac{1}{2}$ walk; fractions: $\frac{20}{40} + \frac{12}{40} + \frac{8}{40} = \frac{40}{40} = 1$ ✓
10	A bar chart is drawn for: Mon = 5, Tue = 10, Wed = 15, Thu = 5. A student wants the vertical axis to go up in steps of 3. Explain why this is a poor choice and suggest a better scale step. Justify.	Steps of 3 are poor because 15 is not a multiple of 3 neatly fitting the chart; steps of 5 are better (gridlines at 5, 10, 15 — all bars land on gridlines)
11	A frequency table shows scores out of 10: Score 5: 5 students; Score 7: 4 students; Score 9: 1 students. Find the mean score.	6.2
12	Find the median of this data set: 22, 6, 18, 4, 30, 12, 8, 26, 14, 20, 10.	14
13	Two classes sit a spelling test. Class A scores: 5, 7, 6, 8, 4. Class B scores: 2, 9, 8, 7, 4. Compare the means and ranges of the two classes.	Class A: mean = 6, range = 4. Class B: mean = 6, range = 7. The means are equal but Class A has a smaller range, so it is more consistent.
14	A survey of 40 people found 15 prefer comedy films. What angle (in degrees) should the comedy sector be in a pie chart?	135°
15	A two-way table shows student choices. Year 7: Swimming 12, Tennis 18, Total 30. Year 8: Swimming 15, Tennis 5, Total 20. How many students play tennis altogether?	23
16	The mean of four numbers is 10. Three of them are 7, 12, 9. Find the fourth number.	12
17	A pie chart for 48 students has a sector of 90° for "Football". How many students chose Football?	12
18	For the data set 5, 2, 5, 8, 2, 5, 6, 8, find (a) the mode and (b) the median.	(a) Mode = 5; (b) Median = 5.5
19	Five plants are measured (cm): 8.2, 11.5, 9.8, 12.1, 10.4. Find the mean height, rounding to 1 decimal place.	10.4 cm
20	A bar chart shows daily visitors to a café over 5 days: Mon 30, Tue 70, Wed 60, Thu 50, Fri 40. Calculate the mean number of daily visitors.	50
21	The mean of 6 test scores is 72. 5 of the scores are 60, 75, 80, 68, 73. Find the missing score.	76
22	A pie chart has a sector of 72°. (a) What fraction of the data does it represent? (b) What percentage?	(a) $\dfrac{1}{5}$ (b) 20%
23	Team A sprint times (s): 9.8, 10.4, 10.2, 9.6, 10.0. Team B sprint times (s): 8.9, 11.7, 10.3, 9.1, 10.0. Compare the two distributions using the mean and range.	Team A: mean = 10.0 s, range = 0.8 s. Team B: mean = 10.0 s, range = 2.8 s. Equal means; Team A is more consistent (smaller range).
24	8 friends share a total of 64 sweets equally. Then 24 more sweets are added and shared equally among the same group. How many sweets does each person have now?	11
25	Eight students score: 2, 3, 3, 4, 4, 4, 5, 30. (a) Find the mean, median, and mode. (b) Which average best represents the data? Explain.	(a) Mean = 6.875, Median = 4, Mode = 4. (b) Median or mode — the score of 30 is an outlier that distorts the mean.
26	A survey of 80 people found: Category A = 30, Category B = 20, Category C = 30. Calculate the angle for each sector to draw a pie chart.	A = 135°, B = 90°, C = 135°
27	The mean of seven numbers is 15. Six of the numbers have a sum of 89. Write and solve an equation to find the seventh number.	16
28	A pie chart shows that 45% of 80 students chose Science as their favourite subject. How many students is that?	36
29	A company has 8 employees. The 2 managers each earn £90000; the remaining 6 staff each earn £15000. (a) Calculate the actual mean salary. (b) Explain why the mean might be misleading as a "typical" salary.	(a) Mean = £33 750. (b) The two managers earning £90 000 each inflate the mean well above what most staff earn (£15 000). The median (£15 000) better represents a typical salary.
30	A data set has 5 whole numbers. The mode is 4, the range is 10, and the mean is 6. Write one possible data set.	One valid set: 1, 4, 4, 7, 14: sum=30, mean=6 ✓, range=13 (no). Try: 1, 4, 4, 9, 12: sum=30 ✓, range=11 (no). Try: 2, 4, 4, 8, 12: sum=30 ✓, range=10 ✓, mode=4 ✓. Answer: 2, 4, 4, 8, 12.
31	Five positive integers have mean 9, median 8, and range 10. List one possible set of numbers.	One valid set: 4, 7, 8, 10, 16 — sum=45, mean=9 ✓; median=8 ✓; range=12 (no). Try: 5, 7, 8, 10, 15: sum=45 ✓, range=10 ✓, median=8 ✓. Answer: 5, 7, 8, 10, 15.
32	Daily temperatures (°C) for one week: −4, 1, −2, 6, −5, 3, −1. Find the mean and range.	Mean = $−\dfrac{2}{7}$ ≈ −0.3 °C. Range = 11 °C.
33	A market stall records daily profit (+) or loss (−) in £ over 5 days: 60, −25, 40, −15, −10. Find the mean daily profit/loss.	£10 profit
34	A student has sat 5 assessments with a current mean of 60. She wants her overall mean to be 65 after a total of 8 assessments. What total must she score across the remaining 3 assessments?	She must score a total of 220 in the remaining 3 assessments (a mean of approximately 73.3 per assessment).
35	Two archers each shoot 6 arrows. Archer A distances from bullseye (cm): 4, 6, 5, 7, 4, 4. Archer B: 1, 10, 2, 9, 1, 7. Compare using mean and range, and state which archer you would pick for a competition.	Archer A: mean = 5 cm, range = 3 cm. Archer B: mean = 5 cm, range = 9 cm. Same mean; choose Archer A for consistency (smaller range).
36	A class of 50 students chose a sport. Results: Football = 20, Tennis = 10, Swimming = 15, Other = 5. Calculate the angle and percentage for each sector.	Football: 144°, 40%. Tennis: 72°, 20%. Swimming: 108°, 30%. Other: 36°, 10%.
37	The sum of 15 data values is 225. Without dividing, use prime factorisation to check whether the mean will be a whole number. Then calculate the mean.	Mean = 15 (a whole number)
38	A pie chart shows how a school spends its budget of £150000. Teaching staff = 50%, Resources = 25%, Premises = 15%, Other = 10%. (a) Calculate the amount spent on each category. (b) What angle represents Premises in the pie chart?	(a) Teaching: £75 000, Resources: £37 500, Premises: £22 500, Other: £15 000. (b) Premises = 54°.
39	Lengths of 40 leaves are grouped: 0–4 cm: 8 leaves; 4–8 cm: 22 leaves; 8–12 cm: 10 leaves. Use midpoints to estimate the mean length.	6.2 cm
40	A data set is: $x$, 5, 11, $2x$, 9. The mean of the data set is 8. Form an equation and solve it to find $x$.	$x = 5$

Problems — Worked Solutions

**Constructing a data set from constraints.** A teacher gives six students a test. You know: - The mean score is 14. - The median score is 13. - The range is 12. - The mode is 13. - All scores are whole numbers between 1 and 20 inclusive. (a) Show that the total of all six scores is 84. (b) The scores in order are: $a,\ b,\ 13,\ 13,\ d,\ e$. Find all possible sets of scores that satisfy every condition. Explain your reasoning step by step.

Answer

Two valid sets: {8, 11, 13, 13, 19, 20} and {8, 12, 13, 13, 18, 20}. See working for full derivation.

(a) Mean = 14, six scores: total = $14 \times 6 = \mathbf{84}$. (b) Ordered scores: $a < b < 13 = 13 < d < e$ (with 13 appearing at least twice for the mode, and median = mean of 3rd and 4th = $\frac{13+13}{2} = 13$ ✓). Range = $e - a = 12$. So $a + b + 13 + 13 + d + e = 84 \Rightarrow a + b + d + e = 58$. Also $e = a + 12$. Substituting: $a + b + d + (a+12) = 58 \Rightarrow 2a + b + d = 46$. Constraints: $a \geq 1$, $b > a$, $b < 13$, $d > 13$, $e = a+12 \leq 20$ so $a \leq 8$. Try $a = 8$: $e = 20$. $b + d = 30 - 2(8) = 14$... wait: $2(8) + b + d = 46 \Rightarrow b + d = 30$. With $b \in \{9,10,11,12\}$ and $d > 13$: $b = 12, d = 18$: valid (9, 12, 13, 13, 18, 20: $b=12 < 13$ ✓, $d=18>13$ ✓, $e=20 \leq 20$ ✓). $b = 11, d = 19$: (9,11,13,13,19,20... wait $a=8$ here). Let me redo: $a=8,e=20$: need $b+d=30$, $8 < b < 13$, $d > 13$. $b=12,d=18$: set $\{8,12,13,13,18,20\}$; sum$=84$ ✓, range$=12$ ✓. $b=11,d=19$: $\{8,11,13,13,19,20\}$; sum$=84$ ✓. $b=10, d=20$: $d=20=e$ — would require two 20s but $d < e$: invalid. Try $a=7$: $e=19$. $b+d=32$. $b<13, d>13$: $b=12,d=20$: $\{7,12,13,13,20,19\}$ — order problem ($d < e$: $20 > 19$): invalid. $b=13$: $b$ must be $< 13$: invalid. So $b \leq 12, d \leq 19$ (since $e=19$): $b=12,d=20$ invalid; $b=12,d=19$ but $d < e = 19$: invalid. Hmm, so $d < e=19$, meaning $d \leq 18$: $b + d \leq 12 + 18 = 30 < 32$: no solution for $a=7$. So the valid sets are those with $a=8$: $\{8,12,13,13,18,20\}$ and $\{8,11,13,13,19,20\}$. Both have sum 84 ✓, median 13 ✓, mode 13 ✓, range 12 ✓.

**The shifting median.** A class of 15 students takes a maths test. Their scores are listed in order, and the median is 72. (a) What position is the median in a list of 15 values? (b) Five new students join the class. All five score 85. The teacher lists all 20 scores in order and finds the new median. Will the new median definitely be higher than 72, or could it stay at 72, or could it go either way? Explain with a specific example. (c) Now suppose instead those five new students all scored 65. What happens to the median? Again, could it stay at 72? (d) Find the **minimum** number of new students (all scoring 85) you would need to add to a class of 15 whose median is 72 to **guarantee** the new median is above 72. Justify your answer.

Answer

(a) 8th position. (b) New median is definitely higher than 72 — it sits at position 10.5 (mean of 10th and 11th), both of which are ≥ 72, and with 5 students above the old median the 10th and 11th values are both above 72. (c) New median could stay at 72 or drop below it. (d) Adding 8 students scoring 85 gives 23 values; median is the 12th — guaranteed above original 8th. The answer is 8.

(a) In an ordered list of 15 values, the median is the $\frac{15+1}{2} = \mathbf{8}$th value. (b) After adding 5 students scoring 85, there are 20 values. The new median = mean of the 10th and 11th values. The original 8th value was 72. Adding five 85s pushes all original values 8th–15th down to positions (roughly) 8th–15th, but the five 85s appear after all original values ≤ 85. Key insight: originally, at least 8 values are ≤ 72 (the 8th value and all before it). After adding five 85s, those ≤72 values occupy positions 1–8 (at most). So the 10th and 11th values are both ≥ 72, and since the five new students scored strictly above 72, both positions are at least 72. If the 9th and 10th original values were 72, the new median would be higher. In fact the 10th and 11th positions in the combined list are among the original values from position 8 onwards or the new 85s — all ≥ 72, and with 5 extras at 85 the 10th and 11th are guaranteed to be above the 8th original (72). The new median is **definitely ≥ 72**, and in most realistic cases **strictly above** 72. **Simpler argument**: originally 7 values are below the median. After adding five 85s, there are still just 7 values below 72. The new median (10th value of 20) is at least the 10th smallest overall — but 7 values are below 72 and 5 are above, so the 10th value is one of the original values from position 3 onwards, which are ≥ the original 8th (72). The new median is ≥ 72 and likely higher if any of positions 9–11 in the original list exceeded 72. (c) Adding five students scoring 65: now there are 20 values, and 5 new values below the old median (72). Originally 7 values were below 72; now up to 12 values could be below 72 (7 original + 5 new). The 10th and 11th values in the new ordered list might both be below 72, so the median could **drop below** 72 or stay at 72 depending on the original data. Example where it drops: original = 60,62,64,66,68,70,72,72,74,76,78,80,82,84,86. Median = 8th = 72. Add five 65s: new ordered list includes 60,62,64,65,65,65,65,65,66,68,70,72,72,72,… The 10th value is 68 < 72. New median = (68+70)÷2 = 69 < 72. (d) We need the median position of $(15+k)$ values to fall above position 8 in the original list. The new median position is $\frac{15+k+1}{2}$ for odd totals (or mean of two central values for even totals). For the median to be **guaranteed** above 72, position $\lceil\frac{15+k}{2}\rceil > 8$, i.e. $15+k > 16$, so $k > 1$. But we also need the actual value there to exceed 72. With $k$ new values of 85: they all sit above the original values ≤ 72. The new 8th position is the same original value only if $k \le 7$ (since original positions 1–8 are still in the first 8 spots of the combined list). When $k = 8$: total = 23 values; median = 12th value. The original list has 8 values at or below 72 in positions 1–8. Adding eight 85s shifts these: positions 1–8 remain the 8 original values ≤ 72, position 9 onward includes original values ≥ 72 and the eight 85s. The 12th value > original 8th (72). So $k = \mathbf{8}$ is needed to guarantee the new median is strictly above 72.

**Pie chart sense-check.** A class of 36 students was asked to name their favourite season. The pie chart shows: Spring = 60°, Summer = 150°, Autumn = 90°, Winter = 60°. (a) What fraction of the class chose Summer? (b) How many students chose Autumn? (c) A student says "More than half the class chose Summer." Is she correct? Explain.

Answer

(a) 5/12 (b) 9 students (c) No — Summer is 150° out of 360°, which is less than half (180°).

(a) $\dfrac{150}{360} = \dfrac{5}{12}$ of the class chose Summer. (b) Autumn sector = $\dfrac{90}{360} = \dfrac{1}{4}$. Number of students = $\dfrac{1}{4} \times 36 = \mathbf{9}$. (c) Half the circle = 180°. Summer = 150° < 180°, so Summer represents less than half. The student is **incorrect**.

**Comparing two classes.** Class A quiz scores: 55, 62, 68, 71, 74, 74, 80, 82. Class B quiz scores: 40, 55, 70, 72, 74, 76, 88, 93. For each class calculate: (a) the mean (b) the median (c) the range. Then write two sentences comparing the two classes.

Answer

Class A: mean = 70.75, median = 72.5, range = 27. Class B: mean = 71, median = 73, range = 53. Class B has a slightly higher mean and median but a much larger range, so Class A is more consistent.

**Class A** (8 values, already ordered): sum = 55+62+68+71+74+74+80+82 = 566; mean = 566÷8 = **70.75**. Median = mean of 4th and 5th = (71+74)÷2 = **72.5**. Range = 82−55 = **27**. **Class B**: sum = 40+55+70+72+74+76+88+93 = 568; mean = 568÷8 = **71**. Median = (72+74)÷2 = **73**. Range = 93−40 = **53**. Class B's mean and median are slightly higher, showing marginally better scores on average. However, Class A's range (27) is much smaller than Class B's (53), so Class A performed more consistently.

**Find the data set (reasoning).** Five whole numbers satisfy all of these conditions: - Mean = 8 - Median = 7 - Mode = 7 - Range = 9 Find a possible set of five numbers. Show how you verified each condition.

Answer

One valid set: 4, 7, 7, 9, 13.

Total = 8 × 5 = 40. The middle (3rd) value must be 7 (median). At least two values must equal 7 (mode). Choose two 7s as 2nd and 3rd values, with minimum value $m$ and maximum $m + 9$ (range = 9). Values so far (ordered): $m$, 7, 7, ?, $m+9$. Remaining 5th value contributes to the total: $m + 7 + 7 + ? + (m+9) = 40 \Rightarrow 2m + ? + 23 = 40 \Rightarrow ? = 17 − 2m$. For the ordering to hold: $7 \le ? \le m+9$. Try $m = 4$: $? = 17 − 8 = 9$; check $7 \le 9 \le 13$ ✓. **Set: 4, 7, 7, 9, 13.** Verify: sum = 40 ✓, mean = 8 ✓, median = 7 ✓, mode = 7 ✓, range = 13−4 = 9 ✓.

**Pie chart construction.** A survey asked 120 people how they prefer to watch films: | Method | Number of people | |--------|-----------------| | Cinema | 30 | | Streaming | 54 | | DVD | 18 | | TV broadcast | 18 | (a) Calculate the pie chart angle for each method. (b) What percentage prefer Streaming? (c) If the pie chart is drawn, what does the Cinema sector look like compared to the Streaming sector?

Answer

(a) Cinema 90°, Streaming 162°, DVD 54°, TV 54°. (b) 45%. (c) Cinema (90°) is a right-angle sector; Streaming (162°) is nearly half the pie — considerably larger.

Multiplier = 360 ÷ 120 = 3°. (a) Cinema: 30 × 3 = **90°**. Streaming: 54 × 3 = **162°**. DVD: 18 × 3 = **54°**. TV: 18 × 3 = **54°**. Check: 90+162+54+54 = 360° ✓. (b) Streaming % = $\dfrac{54}{120} \times 100 = \mathbf{45\%}$. (c) The Cinema sector is a quarter of the circle (90°), while Streaming is almost half (162°), so Streaming appears nearly twice as large as Cinema.

**Misleading mean.** An estate agent says: "The mean house price on Maple Street is £240 000." There are 5 houses on the street. Four sell for £180 000 each, and one sells for £480 000. (a) Verify the estate agent's claim by calculating the mean. (b) Explain why the mean might be misleading. (c) Calculate the median house price. Which average better represents the street?

Answer

(a) Mean = £240 000 ✓ (b) One very high price pulls the mean up. (c) Median = £180 000; the median better represents the typical price.

(a) Sum = 4 × £180 000 + £480 000 = £720 000 + £480 000 = £1 200 000. Mean = £1 200 000 ÷ 5 = **£240 000** ✓. (b) Four out of five houses cost only £180 000. The one house at £480 000 is an outlier that raises the mean above what most buyers would actually pay. The mean is misleading as a "typical" price. (c) Ordered prices: £180 000, £180 000, **£180 000**, £180 000, £480 000. Median = 3rd value = **£180 000**. The median better represents the typical house price on this street.

**The heights puzzle.** Five friends measure their heights. You are told: - The mean height is 160 cm. - The range of heights is 20 cm. - Exactly three of the friends are 160 cm tall. (a) Show that the sum of all five heights is 800 cm. (b) Let the shortest friend have height $s$ cm and the tallest have height $t$ cm. Write two equations connecting $s$ and $t$. (c) Solve to find $s$ and $t$. Are there other possible solutions? (d) **Extension:** If instead two of the five friends are 160 cm tall (not three), and the mean and range are the same, find all possible values of $s$ and $t$. How many solutions are there now?

Answer

(a) 5 × 160 = 800. (b) s + t = 320 and t − s = 20. (c) s = 150 cm, t = 170 cm — the solution is unique. (d) With two at 160 cm: s + t = 480 − 2×160... wait: two at 160 means the remaining two (s and t) must sum to 800 − 3×160 = 320 with t − s = 20, giving the same unique answer s = 150, t = 170. With only two at 160: s + (remaining) + 160 + 160 + t = 800, so s + remaining + t = 480; many solutions exist depending on the 3rd value.

(a) Mean = 160, five friends: total = $160 \times 5 = \mathbf{800}$ cm. (b) Three friends are each 160 cm, contributing $3 \times 160 = 480$ cm. The remaining two friends (shortest $s$ and tallest $t$) must satisfy: $$s + t = 800 - 480 = 320$$ The range is $t - s = 20$. (c) Adding the two equations: $2t = 340$, so $t = \mathbf{170}$ cm. Subtracting: $2s = 300$, so $s = \mathbf{150}$ cm. Check: $150 + 160 + 160 + 160 + 170 = 800$ ✓. Range $= 170 - 150 = 20$ ✓. Mean $= 800 \div 5 = 160$ ✓. This solution is **unique** — the system of two equations with two unknowns has exactly one solution. There are no other possibilities. (d) If exactly **two** of the five friends are 160 cm tall, let the three others have heights $s$, $m$, $t$ where $s < m < t$. We need: $$s + m + t + 160 + 160 = 800 \Rightarrow s + m + t = 480$$ Range $= t - s = 20$, so $t = s + 20$. Substituting: $s + m + (s+20) = 480 \Rightarrow 2s + m = 460$. For valid whole-number solutions we need $s < 160 < t$ (otherwise one of the 160 cm people isn't actually the mode), $m \neq 160$, and $m > s$. Since $m = 460 - 2s$, and $s < 160$, $t = s+20 < 180$: if $s = 150, t = 170, m = 460-300 = 160$ — but $m = 160$ gives three people at 160 again (contradiction). So $m \neq 160$: $460 - 2s \neq 160 \Rightarrow s \neq 150$. Any other value of $s < 150$ or $150 < s < 160$ gives a valid $m$. There are **infinitely many solutions** — adding a third free variable removes the uniqueness.

**Adjusting data.** A teacher records the marks of 8 students: 56, 63, 70, 48, 72, 65, 59, 67. (a) Find the current mean. (b) The teacher adds 5 bonus marks to every student's score. Without listing all new scores, find the new mean. Explain your reasoning. (c) Find the range before and after the bonus marks are added.

Answer

(a) Mean = 62.5 (b) New mean = 67.5 (each score increases by 5, so the mean increases by 5) (c) Range stays the same at 24 both times.

(a) Sum = 56+63+70+48+72+65+59+67 = 500. Mean = 500 ÷ 8 = **62.5**. (b) Adding 5 to every score increases the total by 5 × 8 = 40. New mean = (500+40) ÷ 8 = 540 ÷ 8 = **67.5**. Alternatively: if every value increases by 5, the mean also increases by 5. (c) Before: Max = 72, Min = 48. Range = 72 − 48 = **24**. After adding 5: Max = 77, Min = 53. Range = 77 − 53 = **24**. The range is unchanged — adding the same constant to every value shifts all scores equally, so the spread remains the same.

**The moving mean mystery.** A student takes maths tests one at a time and tracks her running mean after each test. - After Test 1: mean = 80 - After Test 2: mean = 75 - After Test 3: mean = 80 - After Test 4: mean = 78 (a) Find the score she got on each of the four tests. Show your method. (b) She wants her mean to be exactly 80 after Test 5. What must she score? (c) **Investigate:** After $n$ tests the mean is $M_n$. After test $n+1$ the mean becomes $M_{n+1}$. Write a formula for the $(n+1)$th test score in terms of $n$, $M_n$, and $M_{n+1}$. (d) Use your formula to explain: if a student's mean goes **up** after a new test, what does that tell you about the score she just got?

Answer

(a) Test 1 = 80, Test 2 = 70, Test 3 = 90, Test 4 = 72. (b) She must score 88. (c) Score = (n+1)·M_{n+1} − n·M_n. (d) If the mean goes up, the new score is above the previous mean.

(a) After each test, the total = (number of tests) × (mean at that point). - After Test 1: total = $1 \times 80 = 80$. Score 1 = **80**. - After Test 2: total = $2 \times 75 = 150$. Score 2 = $150 - 80 = \mathbf{70}$. - After Test 3: total = $3 \times 80 = 240$. Score 3 = $240 - 150 = \mathbf{90}$. - After Test 4: total = $4 \times 78 = 312$. Score 4 = $312 - 240 = \mathbf{72}$. Scores: 80, 70, 90, 72. (b) Target after Test 5: mean = 80, so total = $5 \times 80 = 400$. Current total after 4 tests = 312. Score 5 = $400 - 312 = \mathbf{88}$. (c) After $n$ tests: total = $n \cdot M_n$. After $n+1$ tests: total = $(n+1) \cdot M_{n+1}$. The new score is the difference: $$\text{Score}_{n+1} = (n+1) \cdot M_{n+1} - n \cdot M_n$$ Check with (a): Score 2 = $2 \times 75 - 1 \times 80 = 150 - 80 = 70$ ✓. Score 3 = $3 \times 80 - 2 \times 75 = 240 - 150 = 90$ ✓. (d) If $M_{n+1} > M_n$ (the mean went up), then: $$\text{Score}_{n+1} = (n+1)M_{n+1} - nM_n > (n+1)M_n - nM_n = M_n$$ So the new score is **strictly greater than the previous mean**. A rising mean tells you the latest score was above the old mean — which makes sense intuitively: dragging the average up requires a score above it.

**Dual statistics.** A scientist records temperatures at two weather stations over 5 days. Station A (°C): 4.2, −1.5, 3.8, −0.6, 2.1 Station B (°C): 6.5, −3.0, 5.2, −1.8, 3.1 (a) Find the mean temperature at each station. (b) Find the range at each station. (c) Station B has a higher mean. Does that mean Station B is warmer overall? Explain using the range.

Answer

(a) Station A mean = 1.6 °C; Station B mean = 2.0 °C. (b) Station A range = 5.7 °C; Station B range = 9.5 °C. (c) Station B has a higher mean but also a much larger range — it experiences bigger swings between warm and cold days, so it is less stable.

(a) Station A: sum = 4.2 + (−1.5) + 3.8 + (−0.6) + 2.1 = 8.0. Mean = 8.0 ÷ 5 = **1.6 °C**. Station B: sum = 6.5 + (−3.0) + 5.2 + (−1.8) + 3.1 = 10.0. Mean = 10.0 ÷ 5 = **2.0 °C**. (b) Station A: Max = 4.2, Min = −1.5. Range = 4.2 − (−1.5) = **5.7 °C**. Station B: Max = 6.5, Min = −3.0. Range = 6.5 − (−3.0) = **9.5 °C**. (c) Although Station B has a higher mean (warmer on average), its range is 9.5 °C compared to 5.7 °C. This means Station B experiences much greater temperature swings — it can be colder or hotter than Station A on any given day. The range shows Station A is more stable.

**Design your own data.** Create a data set of exactly **6 values** satisfying ALL of the following: - The mean is 12.5. - The median is 11.5. - The mode is 10. - The range is 15. - All values are positive whole numbers. Verify every condition and explain the reasoning you used to construct the set.

Answer

One valid set: 5, 10, 10, 13, 15, 20.

Total needed = 12.5 × 6 = 75. The median of 6 values is the mean of the 3rd and 4th values; we need (3rd + 4th) ÷ 2 = 11.5, so 3rd + 4th = 23. Mode = 10, so 10 appears at least twice — place it in positions 1 and 2 (or 2 and 3). Range = 15, so max − min = 15. Try: positions 1–2 = 10, 10 (mode satisfied). Then 3rd + 4th = 23. Choose 3rd = 10 would make mode 10 appear three times (still fine); 4th = 13. Or 3rd = 11, 4th = 12. Let's keep 3rd = 11, 4th = 12. Min so far = 10. For range = 15, max = min + 15. If min = 5 (position 1 is 5, not 10), we need 10 to appear twice elsewhere. Revised: positions ordered = 5, 10, 10, 13, ?, 20. Range = 20 − 5 = 15 ✓. Median = (10+13)÷2 = 11.5 ✓. Mode = 10 ✓. Sum = 5+10+10+13+?+20 = 58+?. We need total = 75, so ? = 17. But then we need 5th value ≤ 6th: 17 ≤ 20 ✓, and 4th ≤ 5th: 13 ≤ 17 ✓. **Final set: 5, 10, 10, 13, 17, 20.** Verify: Sum = 5+10+10+13+17+20 = 75 ✓. Mean = 75÷6 = 12.5 ✓. Median = (10+13)÷2 = 11.5 ✓. Mode = 10 (appears twice, most frequent) ✓. Range = 20−5 = 15 ✓.