AS & A Level · Practice questions · Measures of central tendency and variation

Sum it up in two numbers.

Six original Cambridge-style questions on the mean, median, mode, range and interquartile range, 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

Totals for means, positions for medians.

Mean: \(\bar{x} = \dfrac{\sum fx}{\sum f}\); for grouped data use midpoints.
Missing or combined values: work with totals, \(\sum x = n\bar{x}\).
Median and quartiles: find the position, then count along the cumulative frequencies.
Choose: outliers or skew means median and IQR; otherwise mean and standard deviation.
Compare: one average and one spread statement, in context.

Recall

[2 marks]

Find the mean, median and mode of 5, 8, 8, 9, 11, 12, 15.

Mean \(= \dfrac{68}{7} = 9.71\) (3 s.f.). ✓
Median 9 (4th value), mode 8. ✓

Calculation

[3 marks]

The numbers of siblings \(x\) of 40 students are: \(x = 0\): 8, 1: 15, 2: 12, 3: 4, 4: 1. Find the mean, median and mode.

\(\sum fx = 0 + 15 + 24 + 12 + 4 = 55\), mean \(= 55 \div 40 = 1.375 = 1.38\) (3 s.f.). ✓
Median: the 20th and 21st values. Cumulative frequencies 8, 23, ... so both are 1. Median 1. ✓
Mode 1. ✓

Calculation

[4 marks]

The masses \(m\) grams of 40 eggs are: \(50 \le m < 60\): 5, \(60 \le m < 65\): 12, \(65 \le m < 70\): 15, \(70 \le m < 80\): 8.

Estimate (a) the mean mass, (b) the median mass.

(a) Midpoints 55, 62.5, 67.5, 75. \(\sum fm = 275 + 750 + 1012.5 + 600 = 2637.5\). ✓
Mean \(\approx 2637.5 \div 40 = 65.9\) g (3 s.f.). ✓
(b) Median: 20th egg. Cumulative frequencies 5, 17, 32, 40, so it is the 3rd of 15 in \(65 \le m < 70\). ✓
\(65 + \dfrac{3}{15} \times 5 = 66\) g. ✓

Calculation

[3 marks]

Five numbers have mean 14. Four of them are 10, 12, 15 and 19.

(a) Find the fifth number. (b) A sixth number \(x\) is added and the mean becomes 15. Find \(x\).

(a) Total \(= 5 \times 14 = 70\); \(70 - 56 = 14\). ✓
(b) New total \(= 6 \times 15 = 90\), so \(x = 90 - 70 = 20\). ✓

Analysis

[3 marks]

A class has 25 girls and 15 boys. The mean score of the girls is 58 and the mean score of the whole class is 55.3. Find the mean score of the boys.

Class total \(= 40 \times 55.3 = 2212\); girls' total \(= 25 \times 58 = 1450\). ✓
Boys' total \(= 2212 - 1450 = 762\), so the boys' mean is \(762 \div 15 = 50.8\). ✓

Analysis

[6 marks]

The times, in seconds, of two sprinters in 9 races are:

Ali: 12.1, 12.3, 12.4, 12.4, 12.6, 12.7, 12.8, 13.0, 15.9
Bo: 12.0, 12.2, 12.5, 12.6, 12.7, 12.9, 13.0, 13.1, 13.2

(a) Find the mean, median and interquartile range of each sprinter's times. (b) Ali fell in one race. Which average should be used to choose the faster sprinter? Justify your choice and compare the two sprinters.

(a) Ali: mean \(= 116.2 \div 9 = 12.9\) s; median 12.6 s; \(Q_1 = \tfrac{1}{2}(12.3 + 12.4) = 12.35\), \(Q_3 = \tfrac{1}{2}(12.8 + 13.0) = 12.9\), IQR 0.55 s. ✓
Bo: mean \(= 114.2 \div 9 = 12.7\) s (3 s.f.); median 12.7 s; \(Q_1 = 12.35\), \(Q_3 = 13.05\), IQR 0.7 s. ✓
(b) 15.9 is an outlier caused by the fall, so it distorts Ali's mean. Use the median (and IQR). ✓
Ali's median (12.6 s) is lower, so he is generally faster, and his smaller IQR shows he is more consistent. Choose Ali. ✓

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.