MATH SOLVE

4 months ago

Q:
# A "planet transit" is a rare celestial event in which a planet appears to cross in front of its star as seen from Earth. The planet transit causes a noticeable dip in the star's brightness, allowing scientists to detect a new planet even though it is not directly visible. The National Aeronautics and Space Administration (NASA) recently launched its Kepler mission, designed to discover new planets in the Milky Way by detecting extrasolar planet transits. After one year of the mission in which 3,000 stars were monitored, NASA announced that 5 planet transits were detected. (NASA, American Astronomical Society, Jan. 4, 2010.) Assume that the number of planet transits discovered for every 3,000 stars follows a Poisson distribution with λ=5. What is the probability that, in the next 3,000 stars monitored by the Kepler mission, more than 4 planet transits will be seen?

Accepted Solution

A:

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 whichx 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.