Cumulative frequency and quantiles

(a) 40

1. $4 + 10 + 26 = 40$.

• B1 Correct answer: 40.

(b) $(60, 40)$

1. Cumulative frequencies are plotted at the upper bound of each class.

• B1 Correct answer: $(60, 40)$.

(a) $30$ people

1. $$36+(25-20)/(30-20)(64-36)$$
2. $$80-50$$
3. Between 20 and 30 minutes, cumulative frequency increases by 64 $-$ 36 = 28.
4. At 25 minutes, halfway through that interval, cumulative frequency is estimated as 36 + 28/2 = 50.
5. Cumulative frequency counts times at or below the value, so the estimated number above 25 minutes is 80 $-$ 50 = 30.

• P1 Establishing $36+(25-20)/(30-20)(64-36)$ or an equivalent valid method.
• P1 Establishing $80-50$ or an equivalent valid method.
• A1 Correct answer: $30$ people

(a) 25

1. Read up from 70 to the curve: about 55 students scored 70 or less.
2. $$80 - 55 = 25$$

• M1 Reading the cumulative frequency at 70 (about 55).
• A1 Answer in the range 23 to 27.

(b) about 31%

1. $\frac{25}{80} \times 100 = 31.25\%$.

• B1 Answer in the range 28% to 34%.

(a) $17.0$ seconds

1. $$10+(30-18)/36\times 10$$
2. $$20+(90-54)/48\times 10$$
3. $$1.2\times (27.5-40/3)$$
4. The lower quartile is in the 10 to 20 interval: Q1 = 10 + ((30 $-$ 18)/(54 $-$ 18)) $\times$ 10 = 13⅓ seconds.
5. The upper quartile is in the 20 to 30 interval: Q3 = 20 + ((90 $-$ 54)/(102 $-$ 54)) $\times$ 10 = 27.5 seconds.
6. The original interquartile range is 27.5 $-$ 13⅓ = 14⅙ seconds.
7. Adding 3 to every time leaves the interquartile range unchanged; multiplying every time by 1.2 multiplies it by 1.2.
8. The new interquartile range is 1.2 $\times$ 14⅙ = 17.0 seconds.

• P1 Establishing $10+(30-18)/36\times 10$ or an equivalent valid method.
• P1 Establishing $20+(90-54)/48\times 10$ or an equivalent valid method.
• P1 Establishing $1.2\times (27.5-40/3)$ or an equivalent valid method.
• A1 Correct answer: $17.0$ seconds

(a) $12.5$ minutes

1. Interpolate the lower quartile in the 10 to 20 minute segment.
   $$10+(120/4-12)/(60-12)\times 10$$
2. Interpolate the upper quartile and subtract the lower quartile.
   $$20+(3\times 120/4-60)/(108-60)\times 10-13.75$$
3. Therefore $12.5$ minutes.

• M1 Interpolate the lower quartile in the 10 to 20 minute segment.
• M1 Interpolate the upper quartile and subtract the lower quartile.
• A1 Correct answer: $12.5$ minutes

(b) $36$

1. Interpolate cumulative frequency at 25, then take its complement.
   $$120-(60+108)/2$$
2. Therefore $36$.

• M1 Interpolate cumulative frequency at 25, then take its complement.
• A1 Correct answer: $36$

(a) $30$ minutes

1. Locate 90% of the total frequency on the vertical scale.
   $$0.9\times 140$$
2. Therefore $30$ minutes.

• M1 Locate 90% of the total frequency on the vertical scale.
• A1 Correct answer: $30$ minutes

(b) $\frac{2}{5}$

1. Interpolate the two cumulative counts and subtract.
   $$((70+126)/2-(14+70)/2)/140$$
2. Therefore $\frac{2}{5}$.

• M1 Interpolate the two cumulative counts and subtract.
• A1 Correct answer: $\frac{2}{5}$

(a) $23.75$ cm

1. Find the height with cumulative frequency 30, between 20 cm and 30 cm.
   $$20+(30-21)/24\times 10$$
2. Therefore $23.75$ cm.

• M1 Find the height with cumulative frequency 30, between 20 cm and 30 cm.
• A1 Correct answer: $23.75$ cm

(b) $9$

1. Estimate the cumulative frequency at 35 cm, then subtract from 60.
   $$60-(45+12\times 5/10)$$
2. Therefore $9$.

• M1 Estimate the cumulative frequency at 35 cm, then subtract from 60.
• A1 Correct answer: $9$

(a) School $B$ did better on average (higher median) and its marks were more consistent (smaller interquartile range: 12 against 28).

1. Average: $B$'s median (64) is higher than $A$'s (60), so $B$ generally scored higher.
2. Spread: IQR $A = 73 - 45 = 28$, IQR $B = 70 - 58 = 12$, so $B$'s marks were more consistent.

• M1 Both interquartile ranges, 28 and 12.
• C1 The medians compared in context.
• C1 The IQRs compared in context (B more consistent).