SOLUTION: Two tourists are enjoying the view from San Francisco’s Golden Gate Bridge. They have decided to spend some time waiting for ships to pass under the bridge. Suppose that ship arr

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Question 1197449: Two tourists are enjoying the view from San Francisco’s Golden Gate Bridge. They have decided to spend some time waiting for ships to pass under the bridge. Suppose that ship arrivals at the Golden Gate (leaving or entering San Francisco Bay) are a Poisson distribution at a mean rate of Lambda symbol = 0.4 per hour.
a. What is the probability that the tourists will see 2 to 4 ships passing beneath them if they spend 1 hour waiting for the ships?
b. What is the probability that the tourists will see at least 2 ships passing beneath them if they spend 30 minutes waiting for the ships?
c. What is the probability that the tourists will see at most 3 ships passing beneath them if they spend 2 hours waiting for the ships?

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi  
Poisson distribution at a mean rate of Lambda symbol = 0.4 'per hour'.

spend '1 hour' waiting for the ships
P%28x%2C+lambda%29+=%28+lambda%5Ex%29%2A%28e%5E%28-.4%2A1%29+%29%2Fx%21+
Using TI or similarly an inexpensive calculator like an Casio fx-115 ES plus
P(2 ≤ x ≤ 4) = P(x=2) + P(x=3) + P(x=4) = .0536 + .007 + .0007


spend '30 minutes' waiting for the ships 
P%28x%2C+lambda%29+=%28+lambda%5Ex%29%2A%28e%5E%28-.4%2F2%29+%29%2Fx%21+
P(x ≥ 2) = 1 -  P(x≤ 1) = 1 - (P(x=0) + P(x=1) )
P(x ≥ 2) = 1 -  %28%28.4%5E0%2F0%21%29%2B%28.4%5E1%2F1%21%29+%2Ae%5E%28-.4%2F2%29%29 

spend '2 hours' waiting for the ships 
P%28x%2C+lambda%29+=%28+lambda%5Ex%29%2A%28e%5E%28-lambda%2A2%29+%29%2Fx%21+
P( x ≤ 3 ) = P(x=0) + P(x=1) + P(x=2)
p( x ≤ 3) = %28%28.4%5E0%2F0%21%29%2B%28.4%5E1%2F1%21%29%2B%28.4%5E2%2F2%21%29%29+%2Ae%5E%28-.4%2A2%29
Wish You the Best in your Studies.