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Montano1993 [528]
3 years ago
15

Changes in airport procedures require considerable planning. Arrival rates of aircrft are important factors that muct be taken i

nto account. Suppose small aircraft arrive at a certain airport, according to a Poisson process, at the rate of 5.5 per hour(a) What is the probability that exactly 4 small aircraft arrive during a 1-hour period?(b) What is the probability that at least 4 arrive duringa 1-hour period?(c) If we define a working day as 12 hours, what isthe probability that at least 75 small aircraft arrive during a working day?
Mathematics
1 answer:
Irina18 [472]3 years ago
7 0

Answer:

a) 0.1558

b) 0.7983

c) 0.1478

Step-by-step explanation:

If we suppose that small aircraft arrive at the airport according to a <em>Poisson process</em> <em>at the rate of 5.5 per hour</em>  and if X is the random variable that measures the number of arrivals in one hour, then the probability of k arrivals in one hour is given by:

\bf P(X=k)=\displaystyle\frac{(5.5)^ke^{-5.5}}{k!}

(a) What is the probability that exactly 4 small aircraft arrive during a 1-hour period?

\bf P(X=4)=\displaystyle\frac{(5.5)^4e^{-5.5}}{4!}=0.1558

(b) What is the probability that at least 4 arrive during a 1-hour period?

\bf P(X\geq4)=1-P(X

(c) If we define a working day as 12 hours, what is the probability that at least 75 small aircraft arrive during a working day?

If we redefine the time interval as 12 hours instead of one hour, then the rate changes from 5.5 per hour to 12*5.5 = 66 per working day, and the pdf is now

\bf P(X=k)=\displaystyle\frac{(66)^ke^{-66}}{k!}

and we want <em>P(X ≥ 75) = 1-P(X<75)</em>. But

\bf P(X

hence

P(X ≥ 75) = 1-0.852 = 0.1478

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Answer:

1000 times

Step-by-step explanation:

Given:

The Sun is roughly 10^2 times as wide as the Earth.

The Star KW Sagittarii is roughly 10^5 times as wide as the Earth.

Question asked:

About how many times as wide as the Sun is KW Sagittarii?

Solution:

Let the width of the earth =  x

As the Sun is roughly 10^2 times as wide as the Earth, hence the width of the sun = x\times10^{2}

And as the Star KW Sagittarii is roughly 10^5 times as wide as the Earth, hence the width of the Star = x\times10^{5}

Now, to find that how many times width of the Star  KW Sagittarii is as respect to the width of the Sun, we will simply divide:

Width of the Star  KW Sagittarii =  x\times10^{5}

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\frac{x\times10^{5} }{x\times10^{2} }

x canceled by x

\frac{10^{5} }{10^{2} } =10^{5-2} =10^{3} =10\times\ 10\times10=1000

Therefore, Star  KW Sagittarii is 1000 times wider than Sun.

<em>First of all we calculated width of Sun in terms of width of earth and then calculated the width of the Star in terms of earth and for comparison we did simple division that showed that the Star KW Sagittarii is 1000 times wider than the Sun.</em>

<em />

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