Data tables, bar/pie/time-series charts and frequency polygons

(a) $90$°

1. $$15/60\times 360$$
2. A full pie chart represents $360^\circ.$
3. Climbing accounts for 15/60 = 1/4 of the members.
4. The sector angle is 360 $\div$ 4 = $90^\circ.$

• P1 Establishing $15/60\times 360$ or an equivalent valid method.
• A1 Correct answer: $90$°

(a) A line graph with day on the horizontal axis

1. The measurements form a time series.
2. A line graph with day on the horizontal axis shows how the level changes over time.

• B1 Correct answer: A line graph with day on the horizontal axis

(a) $120^\circ$

1. Each person is $360 \div 60 = 6^\circ$.
2. $$20 \times 6 = 120$$

• M1 $\frac{20}{60} \times 360$ or $20 \times 6$.
• A1 Correct answer: $120^\circ$.

(b) $90^\circ$

1. $15 \times 6 = 90$. Check all four: $90 + 120 + 60 + 90 = 360$.

• B1 Correct answer: $90^\circ$.

(a) 25

1. $72^\circ$ is 18 students, so each student is $72 \div 18 = 4^\circ$.
2. $$100 \div 4 = 25$$

• P1 Degrees per student, 4 (or students per degree, 0.25).
• P1 Using it on the $100^\circ$ sector.
• A1 Correct answer: 25.

(b) 90

1. $360 \div 4 = 90$.

• B1 Correct answer: 90.

(a) $50$%

1. Compare the change with the first week’s sales.
   $$(54-36)/36\times 100$$
2. Therefore $50$%.

• M1 Compare the change with the first week’s sales.
• A1 Correct answer: $50$%

(b) A time-series line graph keeps weeks in order and shows the rise and fall over time. A pie chart shows shares of a whole.

1. A time-series line graph keeps weeks in order and shows the rise and fall over time. A pie chart shows shares of a whole.

• C1 Correct conclusion with supporting reasoning: A time-series line graph keeps weeks in order and shows the rise and fall over time. A pie chart shows shares of a whole.

(a) $(15,9)$

1. Use the midpoint of the second class for the horizontal coordinate.
   $$\frac{10+20}{2}$$
2. Therefore $(15,9)$.

• M1 Use the midpoint of the second class for the horizontal coordinate.
• A1 Correct answer: $(15,9)$

(b) $20$

1. Add all class frequencies.
   $$6+9+5$$
2. Therefore $20$.

• M1 Add all class frequencies.
• A1 Correct answer: $20$

(a) $37.5$%

1. Add both bus entries and divide by the total.
   $$(18+12)/80\times 100$$
2. Therefore $37.5$%.

• M1 Add both bus entries and divide by the total.
• A1 Correct answer: $37.5$%

(b) $180$°

1. Divide by the 40 girls and multiply by 360°.
   $$20/40\times 360$$
2. Therefore $180$°.

• M1 Divide by the 40 girls and multiply by 360°.
• A1 Correct answer: $180$°

(c) Yes: 20 of 40 girls (50%) walk, compared with 14 of 40 boys (35%).

1. Find the proportion of boys who walk.
   $$14/40\times 100$$
2. Yes: 20 of 40 girls (50%) walk, compared with 14 of 40 boys (35%).

• P1 Find the proportion of boys who walk.
• C1 Correct conclusion with supporting reasoning: Yes: 20 of 40 girls (50%) walk, compared with 14 of 40 boys (35%).

(a) No: 10 students in class $A$ and 12 in class $B$.

1. Class $A$.
   $$\tfrac{120}{360} \times 30 = 10$$
2. Class $B$.
   $$\tfrac{96}{360} \times 45 = 12$$
3. 12 is more than 10, so Jo is wrong: a pie chart shows proportions, not numbers.

• P1 Football in class $A$: 10.
• P1 Football in class $B$: 12.
• C1 "No", with both numbers, 10 and 12, compared.