Respuesta :
Answer:
There is a 55.95% probability that, in the next 3,000 stars monitored by the Kepler mission, more than 4 planet transits will be seen.
Step-by-step explanation:
Assume that the number of planet transits discovered for every 3,000 stars follows a Poisson distribution with λ=5.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
[tex]e = 2.71828[/tex] is the Euler number
[tex]\lambda[/tex] is the mean in the given time interval.
For this problem, we have that [tex]\lambda = 5[/tex]
What is the probability that, in the next 3,000 stars monitored by the Kepler mission, more than 4 planet transits will be seen?
That is [tex]P(X > 4)[/tex]. We either see 4 or less planets, or we see more than 4. The sum of the probabilities is decimal 1. So
[tex]P(X \leq 4) + P(X > 4) = 1[/tex]
[tex]P(X > 4) = 1 - P(X \leq 4)[/tex]
In which
[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]
So
[tex]P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-5}*5^{0}}{(0)!} = 0.0067[/tex]
[tex]P(X = 1) = \frac{e^{-5}*5^{1}}{(1)!} = 0.0337[/tex]
[tex]P(X = 2) = \frac{e^{-5}*5^{2}}{(2)!} = 0.0842[/tex]
[tex]P(X = 3) = \frac{e^{-5}*5^{3}}{(3)!} = 0.1404[/tex]
[tex]P(X = 4) = \frac{e^{-5}*5^{4}}{(4)!} = 0.1755[/tex]
[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0067 + 0.0337 + 0.0842 + 0.1404 + 0.1755 = 0.4405[/tex]
Finally
[tex]P(X > 4) = 1 - P(X \leq 4) = 1 - 0.4405 = 0.5595[/tex]
There is a 55.95% probability that, in the next 3,000 stars monitored by the Kepler mission, more than 4 planet transits will be seen.
