Histograms with unequal widths

(a) 3

1. Frequency density $=$ frequency $\div$ class width $= 15 \div 5 = 3$.

• M1 Frequency ÷ class width: $15 \div 5$.
• A1 Correct answer: 3.

(b) 0.4

1. Width 20: $8 \div 20 = 0.4$.

• B1 Correct answer: 0.4.

(a) $1.8$

1. $$18/(25-15)$$
2. Frequency density = frequency $\div$ class width.
3. The class width is 25 $-$ 15 = 10 minutes.
4. Frequency density = 18 $\div$ 10 = 1.8.

• P1 Establishing $18/(25-15)$ or an equivalent valid method.
• A1 Correct answer: $1.8$

(a) 17

1. Frequency $=$ frequency density $\times$ width.
2. Half of the 20 to 30 class.
   $$1.8 \times 5 = 9$$
3. Half of the 30 to 50 class.
   $$0.8 \times 10 = 8$$
4. $$9 + 8 = 17$$

• P1 A frequency from area: $1.8 \times 5$ or $0.8 \times 10$.
• P1 Both parts, 9 and 8.
• A1 Correct answer: 17.

(a) No: frequency is shown by area, not height. The 30 to 50 bar is twice as wide, so it holds 16 patients against 18.

1. In a histogram, frequency $=$ density $\times$ width (the area of the bar).
2. 20 to 30: $1.8 \times 10 = 18$. 30 to 50: $0.8 \times 20 = 16$.
3. His conclusion happens to be true (18 > 16), but the reason is wrong: a bar more than twice as tall does not mean more than twice as many.

• M1 Both frequencies from areas, 18 and 16.
• C1 The reasoning is wrong because frequency is area, not height (with the areas as evidence).

(a) $1$

1. Divide frequency by class width, not by the upper boundary.
   $$10/(15-5)$$
2. Therefore $1$.

• P1 Divide frequency by class width, not by the upper boundary.
• A1 Correct answer: $1$

(b) The middle bar: densities are 4/5, 1 and 2/5, respectively.

1. Calculate the outer class densities to compare with the middle one.
   $$4/5$$
2. Calculate the final class density.
   $$6/15$$
3. The middle bar: densities are 4/5, 1 and 2/5, respectively.

• P1 Calculate the outer class densities to compare with the middle one.
• P1 Calculate the final class density.
• C1 Correct conclusion with supporting reasoning: The middle bar: densities are 4/5, 1 and 2/5, respectively.

(a) $180$

1. Calculate each rectangle’s drawn area, which is proportional to frequency.
   $$10\times 2+15\times 4+5\times 6$$
2. Allocate the total using the middle area’s share.
   $$330\times 60/110$$
3. Therefore $180$.

• P1 Calculate each rectangle’s drawn area, which is proportional to frequency.
• P1 Allocate the total using the middle area’s share.
• A1 Correct answer: $180$

(b) The widths differ, so frequency is proportional to bar area. Height alone represents frequency density up to the vertical scale.

1. The widths differ, so frequency is proportional to bar area. Height alone represents frequency density up to the vertical scale.

• C1 Correct conclusion with supporting reasoning: The widths differ, so frequency is proportional to bar area. Height alone represents frequency density up to the vertical scale.

(a) $36$

1. Calibrate one centimetre of bar height from the known density.
   $$(32/20)/8$$
2. Multiply the new height by the scale and width.
   $$12\times 0.2\times 15$$
3. Therefore $36$.

• P1 Calibrate one centimetre of bar height from the known density.
• P1 Multiply the new height by the scale and width.
• A1 Correct answer: $36$

(b) $68$

1. Add the two frequencies.
   $$32+36$$
2. Therefore $68$.

• P1 Add the two frequencies.
• A1 Correct answer: $68$

(a) $8 + 14 + 18 + 16 + 9 = 65$

1. $10 < t \le 20$: $1.4 \times 10 = 14$.
2. $30 < t \le 50$: $0.8 \times 20 = 16$.
3. $8 + 14 + 18 + 16 + 9 = 65$.

• P1 Both missing frequencies from density times width.
• A1 The total shown as 65.

(b) 25.8 minutes

1. The median is the $65 \div 2 = 32.5$th value.
2. Cumulative: 8, 22, 40, ... so it lies in $20 < t \le 30$.
3. It is $32.5 - 22 = 10.5$ values into a class of 18.
   $$20 + \tfrac{10.5}{18} \times 10 = 25.83\ldots$$

• P1 Locating the median class by cumulative frequency.
• P1 Interpolating within the class.
• A1 Correct answer: 25.8 minutes.