AS & A Level · Practice questions · Stem-and-leaf, box plots, histograms and cumulative frequency

Draw it, then read it.

Six original Cambridge-style questions on histograms, stem-and-leaf diagrams, box plots and cumulative frequency graphs, at Paper 5 standard.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Boundaries first, then density or position.

Boundaries: for rounded data, extend each class by half a unit each way before finding widths.
Histograms: height = frequency ÷ width; frequency = width × height.
Stem-and-leaf: order the data, give a key for both sides, read left leaves backwards.
Cumulative frequency: plot at upper boundaries; read at \(\tfrac{n}{4}\), \(\tfrac{n}{2}\), \(\tfrac{3n}{4}\) or \(\tfrac{k}{100}n\).
Compare: one comment on average, one on spread, both in context.

Recall

[2 marks]

Lengths of leaves are measured to the nearest centimetre and grouped as 5-9, 10-14, 15-24. State the class boundaries and class width of the class 15-24.

Boundaries 14.5 and 24.5. ✓
Width \(= 24.5 - 14.5 = 10\). ✓

Calculation

[3 marks]

The times \(t\) seconds taken by 60 people are: \(0 \le t < 5\): 10, \(5 \le t < 10\): 16, \(10 \le t < 20\): 24, \(20 \le t < 40\): 10.

On a histogram the bar for \(5 \le t < 10\) is 8 cm high. Find the heights of the other three bars.

Frequency densities: \(10 \div 5 = 2\), \(16 \div 5 = 3.2\), \(24 \div 10 = 2.4\), \(10 \div 20 = 0.5\). ✓
3.2 is drawn as 8 cm, so 1 unit of density is 2.5 cm. ✓
Heights: \(2 \times 2.5 = 5\) cm, \(2.4 \times 2.5 = 6\) cm, \(0.5 \times 2.5 = 1.25\) cm. ✓

Analysis

[4 marks]

The reaction times, in hundredths of a second, of 11 boys and 11 girls are:

Boys: 21, 24, 25, 28, 30, 31, 33, 35, 36, 40, 46
Girls: 19, 22, 23, 23, 26, 27, 27, 29, 32, 34, 38

(a) Draw a back-to-back stem-and-leaf diagram with a key. (b) Find the median and interquartile range for each group and compare the two groups.

(a) Stems 1 to 4. Boys (left): \(\mid 2 \mid\) leaves 8 5 4 1; \(\mid 3 \mid\) leaves 6 5 3 1 0; \(\mid 4 \mid\) leaves 6 0. Girls (right): 1 | 9; 2 | 2 3 3 6 7 7 9; 3 | 2 4 8. Key: 1 | 2 | 6 means 21 for boys and 26 for girls. ✓
(b) Boys: \(Q_1 = 25\) (3rd), median 31 (6th), \(Q_3 = 36\) (9th), IQR 11. ✓
Girls: \(Q_1 = 23\), median 27, \(Q_3 = 32\), IQR 9. ✓
The girls reacted faster on average (median 27 against 31) and their times were slightly less spread out (IQR 9 against 11). ✓

Calculation

[4 marks]

The speeds \(v\) km/h of 120 cars give cumulative frequencies \(v \le 40\): 8, \(v \le 50\): 30, \(v \le 60\): 78, \(v \le 70\): 108, \(v \le 80\): 120. No car travelled slower than 30 km/h.

Estimate (a) the median speed, (b) the number of cars travelling faster than 65 km/h, (c) the 80th percentile.

(a) 60th car: \(50 + \dfrac{60 - 30}{48} \times 10 = 56.25\), about 56.3 km/h. ✓
(b) At 65 km/h: \(78 + 0.5 \times 30 = 93\), so \(120 - 93 = 27\) cars. ✓
(c) 80% of 120 is 96: \(60 + \dfrac{96 - 78}{30} \times 10 = 66\) km/h. ✓

Analysis

[4 marks]

A histogram of the masses \(m\) grams of some apples has bars \(0 \le m < 20\) height 0.6, \(20 \le m < 30\) height 2.8, \(30 \le m < 35\) height 4 and \(35 \le m < 50\) height 1.2 (heights are frequency densities).

(a) Find the total number of apples. (b) Estimate the number of apples with masses between 25 g and 40 g.

(a) Frequencies: \(20 \times 0.6 = 12\), \(10 \times 2.8 = 28\), \(5 \times 4 = 20\), \(15 \times 1.2 = 18\). Total 78. ✓
(b) \(5 \times 2.8 + 20 + 5 \times 1.2 = 14 + 20 + 6 = 40\) apples. ✓

Analysis

[4 marks]

The lengths, in cm, of 60 fish are summarised: minimum 12, lower quartile 18, median 21, upper quartile 27, maximum 40.

(a) Describe the skewness of the distribution, giving a reason. (b) A second sample of 60 fish is to be compared with this one. Give one advantage and one disadvantage of using box-and-whisker plots rather than histograms for the comparison.

(a) \(Q_3 - Q_2 = 6\) is greater than \(Q_2 - Q_1 = 3\), and the upper whisker (13) is longer than the lower (6), so the data are positively skewed. ✓
(b) Advantage: two box plots on one scale show the medians, quartiles and spreads side by side, which makes comparison easy. ✓
Disadvantage: the detailed shape of each distribution (for example, how many fish are in each length band) is lost. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.