Elevator passenger arrivals. A study of the arrival process of people using elevators at a multilevel office building was conducted and the results reported in Building Services Engineering Research and Technology (October 2012). Suppose that at one particular time of day, elevator passengers arrive in batches of size 1 or 2 (i.e., either 1 or 2 people arrive at the same time to use the elevator). The researchers assumed that the number of batches, n, arriving over a specific time period follows a Poisson process with mean $\mathbf{\lambda }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{1}$. Now let xn represent the number of passengers (either 1 or 2) in batch n and assume the batch size has probabilities $\mathbf{p}\mathbf{=}\mathbf{P}\left(x_n=1\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{and}\mathbf{}\mathbf{q}\mathbf{=}\mathbf{P}\left(x_n=2\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6}$. Then, the total number of passengers arriving over a specific time period is $\mathbf{y}\mathbf{=}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{x}}_{\mathbf{n}}$. The researchers showed that if ${\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{x}}_{\mathbf{n}}$are independent and identically distributed random variables and also independent of n, then y follows a compound Poisson distribution.

a. Find $\mathbf{P}\left(y=0\right)$, i.e., the probability of no arrivals during the time period. [Hint: y = 0 only when n = 0.]

b. Find $\mathbf{P}\left(y=1\right)$, i.e., the probability of only 1 arrival during the time period. [Hint: y = 1 only when n = 1 and ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{1}$.]

Step by step solution

01

Given information

The number of batches, n arriving over a specific time period follows a Poisson distribution with a mean,$\lambda =1.1$

The random variable y is the total number of passengers arriving in a specific period.

i.e., $\mathrm{y}={\mathrm{x}}_1+{\mathrm{x}}_2+...+{\mathrm{x}}_{\mathrm{n}}$

02

Finding the probability

a.

The random variable y is the total number of passengers arriving in a specific period.

i.e.,$\mathrm{y}={\mathrm{x}}_1+{\mathrm{x}}_2+...+{\mathrm{x}}_{\mathrm{n}}$

Here,$\lambda =1.1$

The random variable y takes the value 0 when

Then the probability is,=0

role="math" localid="1660276312895" $$\begin{gathered} P\left(y=0\right)=\frac{e^{-\lambda }{\lambda }^y}{y!} \\ =\frac{e^{-1.1}{\left(1.1\right)}^0}{0!} \\ =\frac{0.332871x1}{1} \\ =0.332871 \\ \approx 0.3329 \\ P\left(y=0\right)=0.3329 \end{gathered}$$

Therefore, the probability of no arrivals during the time is, 0.3329

03

Finding the probability

b.

The random variable y be the total number of passengers arriving in a specific period.

i.e.,$\mathrm{y}={\mathrm{x}}_1+{\mathrm{x}}_2+...+{\mathrm{x}}_{\mathrm{n}}$

Here,$\lambda =1.1$

The random variable y takes the value 1 when $\mathrm{n}=1\mathrm{and}{\mathrm{x}}_1=1$

Then the probability is,

$$\begin{gathered} P\left(y=0\right)=\frac{e^{-\lambda }{\lambda }^y}{y!} \\ =\frac{e^{-1.1}{\left(1.1\right)}^0}{0!} \\ =\frac{0.332871x1.1}{1} \\ \frac{=0.3661581}{1} \\ =0.3661581 \\ \approx 0.3662 \\ P\left(y=0\right)=0.3662 \end{gathered}$$

Therefore, the probability of only one arrival during the time is, 0.3662

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