Respuesta :
Answer:
(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.
(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.
(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.
Step-by-step explanation:
Let the random variable X = number of aircraft arrive at a certain airport during 1-hour period.
The arrival rate is, λt = 8 per hour.
(a)
For t = 1 the average number of aircraft arrival is:
[tex]\lambda t=8\times 1=8[/tex]
The probability distribution of a Poisson distribution is:
[tex]P(X=x)=\frac{e^{-8}(8)^{x}}{x!}[/tex]
Compute the value of P (X = 6) as follows:
[tex]P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214[/tex]
Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.
Compute the value of P (X ≥ 6) as follows:
[tex]P(X\geq 6)=1-P(X<6)\\=1-\sum\limits^{5}_{x=0}(\frac{e^{-8}(8)^{x}}{x!})\\=1-0.19123\\=0.80877\\\approx0.8088[/tex]
Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.
Compute the value of P (X ≥ 10) as follows:
[tex]P(X\geq 10)=1-P(X<10)\\=1-\sum\limits^{9}_{x=0}(\frac{e^{-8}(8)^{x}}{x!})\\=1-0.71663\\=0.28337\approx0.2834[/tex]
Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.
(b)
For t = 90 minutes = 1.5 hour, the value of λ, the average number of aircraft arrival is:
[tex]\lambda t=8\times 1.5=12[/tex]
The expected value of the number of small aircraft that arrive during a 90-min period is 12.
The standard deviation is:
[tex]SD=\sqrt{\lambda t}=\sqrt{12}=3.464[/tex]
The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.
(c)
For t = 2.5 the value of λ, the average number of aircraft arrival is:
[tex]\lambda t=8\times 2.5=20[/tex]
Compute the value of P (X ≥ 20) as follows:
[tex]P(X\geq 20)=1-P(X<20)\\=1-\sum\limits^{19}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=1-0.47025\\=0.52975\\\approx0.5298[/tex]
Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.
Compute the value of P (X ≤ 10) as follows:
[tex]P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108[/tex]
Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.
The probability distributions are as follows;
a. P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.
b. The standard deviation of the number of small aircraft that arrive during a 90 minutes period is 3.464.
c. P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.
What is Poisson distribution?
It is the most important topic of probability. It is used to calculate the probability for an event with a mean rate value.
Suppose small aircraft arrive at a certain airport according to a Poisson process with a rate α = 8 per hour, so that the number of arrivals during a time period of t hours is a Poisson rv with parameter μ = 8t.
Let X be the random variable of aircraft arriving at a certain airport during one hour period.
a. The probability that exactly 6 small aircraft arrive during a 1-hour period, at least 6, and at least 10.
For t = 1, the mean number of aircraft arrival will be
αt = 8 × 1 = 8
The probability distribution of a Poisson distribution will be
[tex]P(X = x) = \dfrac{e^{-8}*8^x}{x!}\\[/tex]
For x = 6, we have
[tex]\rm P(X = 6) = \dfrac{e^{-8}*8^6}{6!}\\\\P(X = 6) = 0.12214[/tex]
For exact 6 small aircraft, we have
[tex]\rm P(X \geq 6) = 1- P(X < 6)\\\\P(X \geq 6) = 1- \Sigma _{x=0}^{5} \dfrac{e^{-8}*8^x}{x!}\\\\P(X \geq 6) = 0.8088[/tex]
For X ≥ 10, we have
[tex]\rm P(X \geq 10) = 1- P(X < 6)\\\\P(X \geq 10) = 1- \Sigma _{x=0}^{9} \dfrac{e^{-8}*8^x}{x!}\\\\P(X \geq 10) = 0.2834[/tex]
b.The expected value and standard deviation of the number of small aircraft that arrive during a 90-min period
For t = 90 minutes = 1.5 hours, the value of α, the average of aircraft arrival will be
αt = 8 × 1.5 = 12
Then the expected value will be
[tex]\rm Standard \ deviation = \sqrt{\alpha *t} = \sqrt{12} = 3.464[/tex]
The standard deviation of the number of small aircraft that arrive during a 90 minutes period is 3.464.
c. 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, the value of α, the mean number of aircraft arrival will be
αt = 8 × 2.5 = 20
Then P(X ≥ 20), we have
[tex]\rm P(X\geq 20) = 1- P(X < 20)\\\\P(X\geq 20) = 1- \Sigma _{x= 0}^{19} \dfrac{e^{-20}*20^x}{x!}\\\\P(X \geq 20) = 0.5298[/tex]
And P(X ≤ 10), we have
[tex]\rm P(X \leq 10) = \Sigma _{x= 0}^{10} \dfrac{e^{-20}*20^x}{x!}\\\\P(X \leq 10) = 0.0108[/tex]
More about the Poisson distribution link is given below.
https://brainly.com/question/5673802
