All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let X denote the number of people joining the queue in 4 minutes.

$X \sim \mathrm{Po}(4.8)$

$\mathrm{P}(X \ge 8) = 1 - \mathrm{P}(X \le 7) \approx 0.113$

(ii)
Let Y denotes number of people joining the queue in t seconds.

$Y \sim \mathrm{Po}(0,02t)$

$\mathrm{P}(Y \le 1) = 0.7$

$\mathrm{P}(Y = 0) + \mathrm{P}(Y = 1) = 0.7$

$\frac{e^{-0.02t}(0.02t)^0}{0!} + \frac{e^{-0.02t}(0.02t)^1}{1!}=0.7$

$e^{-0.02t}(1+0.02t) = 0.7$

Using the Graphing Calculator, t = 55

(iii)
Let W denote the number of people joining the queue in 15 minutes.

$W \sim \mathrm{Po}(18)$

Since $\lambda = 18 > 10, ~W \sim \mathrm{N}(18, 18)$ approximately.

Let V denote the number of people leaving the queue in 15 minutes.

$V \sim \mathrm{Po}(27)$

Since $\lambda = 27 > 10, ~V \sim \mathrm{N}(27, 27)$ approximately

$\Rightarrow W - V \sim \mathrm{N}(-9,45)$ approximately

$\mathrm{P}(W -V \ge -11)$

$= \mathrm{P}(W-V > -11.5)$ by continuity correction

$\approx 0.645$

(iv)
The average number of people joining the queue per minute may not remain constant throughout a period of long hours as they have different traveling time. There might be no flight departing too. Thus, it would not appropriate.