Question

a) Cari kobarangkalian terdapat 2 atau kurang kapal terbang yang mendarat Find the probabiity that there are 2 or fower flights arriving or departing from the (2 markah/marks atau berlepas daripada landasan tersebut dalam tempoh satu jam runway in an hour (b) Apakah jumlah kapal terba ng yang djangka merndarat dan berlepas daliam tempoh 4 jam? What is the oxpected numbor of flights aniving and doparting in 4 hours? (1 markah/mark) (c) Cari kebarangkalian bahawa junlah yang djangka tersebut (yang diperoleh sebelum ini) memang mendarat atau berlepas dalam tempoh 4 jam. Find the probability that this expected number of flights (obtained previously) actually amrive or depart in a given 4 hour period (2 markah/marks) 5. Bandingkan di antara "reka bentuk eksperimen secara rawak sepenuhnya" dan "reka bentuk dalam kawalan yang ketat apabila digunakan untuk mengumpul subsampel dari populasi. Kemudian, berikan contoh kaedah pensampelan untuk setiap reka bentuk tersebut Differentiate between completely randomized experimental design and rigorously controlled design when is used to collect subsamples from population. Then, provide an example method of sampling for each design (5 markahlmarks) 316

Answer 1

The number of flights arrive $X$ can be modelld by Poisson distribution.

a) The Poisson PMF is $P\left ( X=x \right )=e^{-\lambda }\frac{\lambda ^x}{x!};x=0,1,2,3,......$ . Here $\lambda =2$.

The required probability

$\sum_{x=0}^{2}P\left ( X=x \right )=\sum_{x=0}^{2}e^{-\lambda }\frac{\lambda ^x}{x!};\\ {\color{Blue} \sum_{x=0}^{2}P\left ( X=x \right )=0.9856123}$

b) Consider the Poisson process in $t$ hours. The Piosson parameter in $t$ hours is $\lambda t$.

The Poisson PMF is $P\left ( X=x \right )=e^{-\lambda t}\frac{\left (\lambda t \right ) ^x}{x!};x=0,1,2,3,......$

Expected value of the above Poisson process is $E\left ( X \right )=\lambda t=2\times 4={\color{Blue} 8}$

c) The Poisson PMF is $P\left ( X=x \right )=e^{-\lambda t}\frac{\left (\lambda t \right ) ^x}{x!};x=0,1,2,3,......$ . Here $\lambda =2,t=4$.

The required probability $P\left ( X=8 \right )=e^{-\lambda t }\frac{\left (\lambda t \right )^x}{x!}$ .

$P\left ( X=8 \right )=e^{-8}\frac{\left (2\times 4 \right )^8}{8!}\\ {\color{Blue} P\left ( X=8 \right )=0.1395865}$