Thanks!

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]]>Refer to this comment last year for some heads up on continuity correction.

In short, for the CC in Q7(iv), P(`R` < 25) becomes P(`R` < 24.5) âˆµ 25 is not included in the interval.

For Q12(iii), P("at most 11 people left the queue overall by 0945") as described in the suggested solution vs P("at least 24 people in the queue by 0945") yields the same outcome.

I suppose your workings is as follows, in which case is fine:

X ~ Po(27) → X ~ N(27, 27) approximately

Y ~ Po(18) → X ~ N(18, 18) approximately

No. of people in queue at 0945,

W ~ N(35 + 18 − 27, 18+27)

W ~ N(26, 45)

P(W ≥ 24) ≈ 0.617 (3sf)

As for Q11(i), this is actually a question on Permutations & Combinations where P(`R` = 4) is the probability of getting a *combination* of 4 women in the committee of 10 (^{18}C_{4} × ^{12}C_{6}), out of every possible combination of choosing a committee of 10 from 30 people (^{30}C_{10}).

If you do B(10, 0.6), P(`R` = 4) would then refer to the probability of picking a woman 4 times out of 10 independent tries. This scenario is different from the first and will give you a different answer which will result in you having difficulty in proving part (ii).

Yes, a binomial model requires the calls to be independent of each other i.e. who I call now will have no bearing on who I call next and vice versa.

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