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.

KS Comments:

The question is pretty standard Poisson. So long as students are careful with continuity correction, I don’t see much problems with this questions. They should answer (iv) contextually.

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