Question:

Stepwise Solution:

$15 + f_1 + 21 + f_2 + 17 = 100 \Rightarrow 53 + f_1 + f_2 = 100 \Rightarrow f_1 + f_2 = 47$ -- (eq 1). [1.0 Mark]

Class marks $x_i$: $10, 30, 50, 70, 90$. [0.5 Mark]

$\sum f_i x_i = (15 \times 10) + (f_1 \times 30) + (21 \times 50) + (f_2 \times 70) + (17 \times 90)$
$= 150 + 30 f_1 + 1050 + 70 f_2 + 1530 = 2730 + 30 f_1 + 70 f_2$. [1.5 Marks]

Mean $= \dfrac{2730 + 30 f_1 + 70 f_2}{100} = 53 \Rightarrow 2730 + 30 f_1 + 70 f_2 = 5300 \Rightarrow 30 f_1 + 70 f_2 = 2570 \Rightarrow 3 f_1 + 7 f_2 = 257$ -- (eq 2). [1.0 Mark]

Multiply eq 1 by 3: $3 f_1 + 3 f_2 = 141$.
Subtract from eq 2: $4 f_2 = 116 \Rightarrow f_2 = 29$.
Then $f_1 = 47 - 29 = 18$. Missing frequencies: $f_1 = 18, f_2 = 29$. [1.0 Mark]

Marking Scheme:

• Finding eq 1: $f_1 + f_2 = 47$: 1.0 Mark
• Class marks and $\sum f_i x_i = 2730 + 30 f_1 + 70 f_2$: 1.5 Marks
• Finding eq 2: $3 f_1 + 7 f_2 = 257$: 1.0 Mark
• Solving linear system for $f_1 = 18, f_2 = 29$: 1.5 Marks