Histograms

Frequency density = frequency ÷ class width

$50$–$70$: $20 \div 20 = 1$

$70$–$80$: $50 \div 10 = 5$

$80$–$90$: $60 \div 10 = 6$

$90$–$110$: $30 \div 20 = 1.5$

Draw bars of these heights with the correct widths.

(a) Frequency = class width × frequency density:

$10 \times 0.4 = 4$, $10 \times 0.6 = 6$, $5 \times 1 = 5$, $20 \times 0.2 = 4$

(b) Total $= 4 + 4 + 6 + 5 + 4 = 23$, so the lower quartile is the $\frac{23+1}{4} = 6$th value.

The first class holds 4 values, so we need 2 more from the $5 < t \le 15$ class, which holds 4 values over 10 minutes.

Lower quartile $= 5 + \frac{2}{4} \times 10 = 10$ minutes

The £300k–£400k bar holds 20 houses over a width of 100, so its height stands for 0.2 per £1000.

That sets the scale: the bar is 2 large squares tall, so each large square up is 0.1.

0–100: $0.14 \times 100 = 14$

100–150: $0.6 \times 50 = 30$

150–200: $0.8 \times 50 = 40$

Total $= 14 + 30 + 40 = 84$

(a) Frequency density = frequency ÷ class width:

$20 \div 10 = 2$, $70 \div 20 = 3.5$, $22 \div 5 = 4.4$, $30 \div 15 = 2$, $8 \div 10 = 0.8$

Draw bars of these heights with no gaps; label the vertical axis 'frequency density'.

(b) 20 to 30: half of 70 = 35; 30 to 35: 22; 35 to 40: $\frac{5}{15}$ of 30 = 10

Total $= 35 + 22 + 10 = 67$, so the fraction is $\frac{67}{150}$

Known frequencies: 0–5: 11; 10–20: $10 \times 1.5 = 15$; 30–50: $20 \times 0.3 = 6$

Left: $60 - 11 - 15 - 6 = 28$, shared equally: 14 each.

5–10: $14 \div 5 = 2.8$; 20–30: $14 \div 10 = 1.4$

(a) The 30–40 bar holds 28 students, so its fd is $28 \div 10 = 2.8$: this sets the vertical scale.

For 40–60: fd $= 24 \div 20 = 1.2$, so draw a bar of height 1.2 from 40 to 60.

(b) 0–20: fd 1.6, so $1.6 \times 20 = 32$. 20–30: fd 3.6, so $3.6 \times 10 = 36$.

Total $= 32 + 36 + 28 + 24 = 120$

(a) The 12–16 bar has frequency 10 and width 4, so its frequency density is 2.5. The 10–12 bar is 1.5 high, so its frequency is $1.5 \times 2 = 3$.

(b) $9 \div 2 = 4.5$ and $7 \div 5 = 1.4$, so draw bars of those heights.

(c) Frequencies: $4, 8, 16, 13, 2$ (total 43). The median is the 22nd value. After 16 s there are 12 students, so we need 10 more of the 16 in the 16–20 class: $16 + \frac{10}{16} \times 4 = 18.5$ seconds.

(a) Frequency density = frequency ÷ class width:

$90\div1 = 90$, $48\div0.5 = 96$, $22\div0.5 = 44$, $8\div1 = 8$, $12\div2 = 6$

(b) On the grid, the 1–2 bar is 5 times as tall as the 3–5 bar, but the 3–5 bar is twice as wide.

Area ratio $= 5\times1 : 1\times2 = 5 : 2$

So the number is $\frac{2}{5}n = 0.4n$

Frequencies = width × density:

$0.8 \times 10 = 8$, $0.4 \times 50 = 20$, $0.4 \times 100 = 40$, $0.2 \times 60 = 12$ (total 80)

Mean $= \frac{6.8 \times 8 + 7.4 \times 20 + 7.8 \times 40 + 8.1 \times 12}{80} = \frac{611.6}{80} = 7.645$ m

Frequencies from the bars (width × height): $2.5 \times 1.2 = 3$, $2.5 \times 2 = 5$, $5 \times 2.8 = 14$, $12.5 \times 0.8 = 10$

Known total $= 32$, which is 80% of the trees, so there are $32 \div 0.8 = 40$ trees.

Missing class: 20% of 40 $= 8$ trees.

Frequency density $= 8 \div 2.5 = 3.2$, so draw a bar of height 3.2 from 10 to 12.5.

Frequency = frequency density × width:

0–60: $2 \times 60 = 120$; 60–90: $5 \times 30 = 150$; 90–120: $9 \times 30 = 270$; 120–135: $6 \times 15 = 90$; 135–180: $2 \times 45 = 90$

Total $= 720$. Factory (60–120 g): $150 + 270 = 420$

$x = \frac{420}{720} \times 360 = 210$

(a) Frequency density = frequency ÷ class width:

$120 \div 20 = 6$, $90 \div 30 = 3$, $120 \div 30 = 4$, $140 \div 70 = 2$, $100 \div 50 = 2$

(b) The median is the $285$th value. $120 + 90 = 210$, so it is $75$ into the 50–80 class.

$50 + \frac{75}{120} \times 30 = 68.75 \approx 69$ miles

Frequency = frequency density × class width:

$0.8 \times 5 = 4$, $1.6 \times 10 = 16$, $2.2 \times 10 = 22$, $1.2 \times 15 = 18$. Total $= 60$.

20% of 60 $= 12$ people.

In the top class, 12 people cover $12 \div 1.2 = 10$ marks, so the score is $40 - 10 = 30$.