~(b) (i) If 5% of the candidates passed the examination, then (100 - 5)% = 95% passed with a mark \(\leq\) the lowest mark for distinction.
\(\text{95% of 200} = \frac{95}{100} \times 200 \)
= 190 candidates. From the ogive, 190 corresponds to 79.5 + 8 = 87.5 marks (to one decimal place)
(ii) From the ogive, the number of candidates who scored at most 45% is 69. Hence, the probability of selecting a candidate who scored at most 45%
= \(\frac{69}{200} = 0.345 \approxeq 0.3\) (1 d.p)
