AS & A Level · Practice questions · Variance and standard deviation, including coded data

Square, average, root.

Six original Cambridge-style questions on variance and standard deviation, including coded data and two data sets, 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

Get to \(n\), \(\sum x\) and \(\sum x^2\).

Mean: \(\bar{x} = \dfrac{\sum x}{n}\) (or \(\dfrac{\sum fx}{\sum f}\)).
Variance: \(\dfrac{\sum x^2}{n} - \bar{x}^2\); standard deviation is its square root.
Coded: mean \(= a + \dfrac{\sum (x - a)}{n}\); variance is the same as for the coded values.
Reverse: \(\sum x = n\bar{x}\) and \(\sum x^2 = n(\sigma^2 + \bar{x}^2)\).
Combine: add the totals, then use the formulas once.

Recall

[2 marks]

Find the mean and standard deviation of 3, 5, 6, 9, 12.

\(\bar{x} = 35 \div 5 = 7\); \(\sum x^2 = 9 + 25 + 36 + 81 + 144 = 295\). ✓
Variance \(= 59 - 49 = 10\), standard deviation \(= \sqrt{10} = 3.16\) (3 s.f.). ✓

Calculation

[2 marks]

For 15 values, \(\sum x = 270\) and \(\sum x^2 = 5100\). Find the mean and standard deviation.

\(\bar{x} = 270 \div 15 = 18\). ✓
Variance \(= 5100 \div 15 - 18^2 = 340 - 324 = 16\), standard deviation 4. ✓

Calculation

[3 marks]

The scores \(x\) of 20 players are: \(x = 1\): 2, 2: 5, 3: 8, 4: 4, 5: 1. Calculate the mean and standard deviation.

\(\sum fx = 57\), mean \(= 2.85\). ✓
\(\sum fx^2 = 183\), variance \(= 9.15 - 8.1225 = 1.0275\). ✓
Standard deviation \(= 1.01\) (3 s.f.). ✓

Calculation

[5 marks]

The masses \(x\) grams of 30 packets satisfy \(\sum (x - 100) = -60\) and \(\sum (x - 100)^2 = 1470\).

(a) Find the mean and standard deviation of the masses. (b) Find \(\sum x\) and \(\sum x^2\).

(a) Mean \(= 100 + (-60 \div 30) = 98\) g. ✓
Variance \(= \dfrac{1470}{30} - (-2)^2 = 49 - 4 = 45\); standard deviation \(= \sqrt{45} = 6.71\) g (3 s.f.). ✓
(b) \(\sum x = -60 + 3000 = 2940\). ✓
\(\sum x^2 = 30(45 + 98^2) = 30 \times 9649 = 289\,470\). ✓

Analysis

[5 marks]

20 girls have mean mark 52 and standard deviation 6. 30 boys have mean mark 47 and standard deviation 8. Find the mean and standard deviation of all 50 marks.

Girls: \(\sum x = 1040\), \(\sum x^2 = 20(36 + 2704) = 54\,800\). ✓
Boys: \(\sum x = 1410\), \(\sum x^2 = 30(64 + 2209) = 68\,190\). ✓
All: \(\sum x = 2450\), mean \(= 49\); \(\sum x^2 = 122\,990\). ✓
Variance \(= 2459.8 - 49^2 = 58.8\), standard deviation \(= 7.67\) (3 s.f.). ✓

Analysis

[6 marks]

A factory measures the lengths \(x\) cm of rods from two machines. For 10 rods from machine A, \(\sum x = 120\) and \(\sum x^2 = 1530\). When 15 rods from machine B are included, the mean of all 25 rods is 14 cm and the standard deviation is 3 cm.

Find the mean and standard deviation of the lengths of the 15 rods from machine B.

All 25 rods: \(\sum x = 25 \times 14 = 350\), \(\sum x^2 = 25(3^2 + 14^2) = 5125\). ✓
Machine B: \(\sum x = 350 - 120 = 230\), \(\sum x^2 = 5125 - 1530 = 3595\). ✓
Mean \(= 230 \div 15 = 15.3\) cm (3 s.f.). ✓
Variance \(= \dfrac{3595}{15} - \left(\dfrac{230}{15}\right)^2 = 239.67 - 235.11 = 4.56\), standard deviation \(= 2.13\) cm (3 s.f.). ✓

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.