Question

Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α= 8 per hour, so that the number of arrivals during a time period of t hours is a Poisson rv with parameter λ = 8t.

(a) What is the probability that exactly 6 small aircraft arrive during a 1-hour period? At least 6? At least 10?

(b) What are the expected value and standard deviation of the number of small aircraft that arrive during a 90-min period? 

(c) What is the probability that at least 20 small aircraft arrive during a 2.5-hour period? That at most 10 arrive during this period?

 

 

Answer 1

Solution

Given : \( \lambda=\mu=8t \)

Formula poisson probability :  \( P(X=k)=\frac{\lambda ^ke^{-\lambda }}{k!} \)

(a) What is the probability that exactly 6 small aircraft arrive during a 1-hour period

For \( t=1\implies \lambda=\mu=8(1)=8 \)

Evaluate the formula at  \( k=0,1,2,..,10 \)

For  \( k=0\implies P(X=0)=\frac{8^0e^{-8}}{0!}\approx 3.3546\times 10^{-4} \)

\( k=1\implies P(X=1)=\frac{8e^{-8}}{1!}\approx 0.0027 \)

\( .................................... \)

\( k=5\implies P(X=5)=\frac{8^5e^{-8}}{5!}\approx 0.0916 \)

\( k=6\implies P(X=6)=\frac{8^6e^{-8}}{6!}\approx 0.1221 \)

\( .................................... \)

\( k=9\implies P(X=9)=\frac{8^9e^{-8}}{9!}\approx 0.1241 \)

Addition rule for disjoint or multually exclusive events :  

\( P(A\hspace{2mm}or\hspace{2mm} B)=P(A)+P(B) \)

Add the corresponding probabilities : 

\( \implies P(X\geq 6)=1-P(X<6)=1-P(X=0)-P(X=1)-...-P(X=5) \)

                             \( =1-3.3546\times 10^{-4}+0.0027+...+0.0916=0.8088 \)

\( \implies P(X\leq 10)=1-P(X<10)=1-P(X=0)-P(X=1)-...-P(X=9) \)

                               \( =1-3.3546\times 10^{-4}-0.0027-...-0.1241=0.2834 \)

Therefore.  

\( P(X=6)=0.1221\hspace{3mm},P(X\geq 6)=0.8088\hspace{3mm}, P(X\leq 10)=0.2824 \)

(b) What are the expected value and standard deviation of the number of small aircraft that arrive during a 90-min period

we have  \( t=90min=1.5hours \)

The mean (expected value) is the product of the rate per hour and the number of hours 

\( \implies \lambda=\mu=8t=8(1.5)=12 \)

The standard deviation of a Poisson distribution is the square roof of the mean

\( \implies \sigma=\sqrt{\mu}=\sqrt{\lambda}=\sqrt{12}=2\sqrt{3}\approx 3.4641 \)

Therefore.  Expected value is 12 and standard deviation is  

\( \sqrt{12}=2\sqrt{3}\approx 3.4641 \)

(c) What is the probability that at least 20 small aircraft arrive during a 2.5-hour period? That at most 10 arrive during this period

For  \( t=2.5\implies \lambda=\mu=8t=8(2.5)=20 \)

Evaluate the formula at  \( k=0,1,2,...,19 \)

For \( k=0\implies P(X=0)=\frac{20^0e^{-20}}{0!}\approx 2.0612\times 10^{-9} \)

\( k=1\implies P(X=1)=\frac{20^1e^{-20}}{1!}\approx 4.1223\times 10^{-8} \)

\( .................................... \)

\( k=10\implies P(X=1)=\frac{20^{10}e^{-20}}{10!}\approx 0.0058 \)

\( .................................... \)

\( k=19\implies P(X=1)=\frac{20^{19}e^{-20}}{19!}\approx 0.0888 \)

Addition rule for disjoint or multually exclusive events :  

\( P(A\hspace{2mm}or\hspace{2mm})=P(A)+P(B) \)

Add the corresponding probabilities : 

\( \implies P(X\leq 19)=P(X=0)+P(X=1)+...+P(X=19) \)

\( =2.0612\times 10^{-9}+4.1223\times 10^{-8}+...+0.0888=0.4703 \)

\( \implies P(X\leq 10)=P(X=0)+P(X=1)+...+P(X=10) \)

                        \( =2.0612\times 10^{-9}+4.1223\times 10^{-8}+...+0.0058=0.0108 \)

Complement rule ; \( P(\bar{A})=1-P(A) \)

\( \implies P(X\geq 20)=1-P(X<20)=1-P(X\leq 19)=1-0.4703=0.5297 \)

Therefore.  \( P(X\geq 20)=0.5297\hspace{3mm},P(X\leq 10)=0.0108 \)

Therefor.

a).\( P(X=6)=0.1221\hspace{3mm},P(X\geq 6)=0.8088\hspace{3mm}, P(X\leq 10)=0.2824 \)

b).Expected value is 12 and standard deviation is \( \sqrt{12}=2\sqrt{3}\approx 3.4641 \)

c). \( P(X\geq 20)=0.5297\hspace{3mm},P(X\leq 10)=0.0108 \)