(a) [2 marks]

EITHER recognising that half the total frequency is 10 (may be seen in an ordered list or indicated on the frequency table) (M1).
OR \( 5 + 1 + 4 = 3 + x \).
OR \( \sum f = 20 \).
THEN \( x = 7 \) (A1).

(b) [3 marks]

METHOD 1
\( \sigma^2 = \frac{5 \times (2 – 4.3)^2 + 1 \times (3 – 4.3)^2 + 4 \times (4 – 4.3)^2 + 3 \times (5 – 4.3)^2 + 7 \times (6 – 4.3)^2}{20} = 2.51 \) (M1A1).
\( \sigma = \sqrt{2.51} = 1.58429… \approx 1.58 \) (A1).
METHOD 2
\( \sigma^2 = \frac{5 \times 2^2 + 1 \times 3^2 + 4 \times 4^2 + 3 \times 5^2 + 7 \times 6^2}{20} – 4.3^2 = 2.51 \) (M1A1).
\( \sigma = \sqrt{2.51} = 1.58429… \approx 1.58 \) (A1).