Mr Lee is planning to go fishing this weekend. Assuming that the number of fish caught per hour follows a Poisson distribution with mean 0.6, find the number of complete hours that Mr Lee needs to fish so that the probability of catching more than two fish exceeds 80 %.

Answer 1

Well, lets find the probability of catching 0, 1 , or 2 fish in an hour:
p(0) = e^-.6 (.6)^0/0! = .549*1/1 = .549
p(1) = e^-.6(.6)^1/1! = .549*.6 = .329
p(2) =e^-.6(.6)^2/2! =.549*.36/2 = .099
so total p = .977
so chances of catching more than 2 fish in an hour = 1-.977 = .023

Answer 2

I want my chances of 0, 1 or 2 fish to be less than 0.20 so I have at least a 0.80 probability of catching more than 2.
Try 6 hours
L = .6*6 = 3.6 fish expected in 6 hours
p(0,1,2) = (1+3.6+3.6^2/2)e^-3.6
= 11.08(.015) = .304
that is more than 0.20 so I better fish another hour
Try 7 hours
L = .6*7 = 4.2 fish expected in 7 hours
(1+4.2 +8.82)e^-4.2
14.02 * .015
.21 Wow is that close but just an hour longer
Try 8 hours
L = .6*8 = 4.8
(1+4.8+4.8^2 /2) e^-4.8
17.32 * .00823
.142
so 8 hours does it easily. We only have a 14% probability of catching less than 3 fish in eight hours

Answer 3

Yall wrong the answer is 7.132