a.20,3% b.18% c .15,6% d 21%
2006-08-20 08:14:11 · 3 answers · asked by guitarkhongday 1 in Science & Mathematics ➔ Mathematics
Yes, there's sufficient information given the reasonable assumption that the event of a ship docking during the day is independent of any other ship docking. In that case the number of ships on any day is a Poisson distribution. Such a probability distribution is determined completely by its mean value (5):
P(k) = (5^k/k!)exp(-5)
--> P(5) = .18
--> (b)
2006-08-20 10:45:59 · answer #1 · answered by shimrod 4 · 0⤊ 0⤋
I don't think you've given enough information to answer the question, or maybe I've just misunderstand. I think what you're saying is that you're given on average 5 ships moor in a day. You want to know the probability that exactly 5 ships moor. I think you need to know more about the distribution to answer the question.
2006-08-20 15:23:32 · answer #2 · answered by NordicGuru 3 · 0⤊ 0⤋
can you formulate a proper question?
2006-08-20 15:22:35 · answer #3 · answered by Blues Man 2 · 0⤊ 0⤋