Question

Suppose that the number of airplanes arriving at an airport per minute is a Poisson process. The mean number of airplanes arriving per minute is 3. The probability that exactly 6 planes arrive in the next minute is 0.05041.True / False.

Answer 1

Answer:

$P(X=6)$

And using the probability mass function we got:

$P(X=6) =3^6 \frac{e^{-3}}{6!}=0.05041$

So then we can conclude that the statement is True.

Step-by-step explanation:

Definitions and concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

$P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}$

And the parameter $\lambda= 3$ represent the average ocurrence rate per unit of time.

Solution to the problem

For this case we want to find this probability:

$P(X=6)$

And using the probability mass function we got:

$P(X=6) =3^6 \frac{e^{-3}}{6!}=0.05041$

So then we can conclude that the statement is True.

Answer 2

Answer:

The given statement is True.

Step-by-step explanation:

We are given that the number of airplanes arriving at an airport per minute is a Poisson process with the mean number of airplanes arriving per minute is 3.

Let X = Distribution of number of airplanes arriving at an airport per minute

So, X ~ Poisson($\lambda$)

The mean of Poisson distribution is given by, E(X) = $\lambda$ = 3

which means, X ~ Poisson(3)

The probability distribution function of a Poisson random variable is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!}; for x=0,1,2,3...$

Probability that exactly 6 planes arrive in the next minute = P(X = 6)

    P(X = 6) = $\frac{e^{-3}*3^{6}}{6!}$ = $\frac{e^{-3}*729}{720}$ = 0.05041

Therefore, the given statement is true.