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 %.
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
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
Yall wrong the answer is 7.132
To find the number of complete hours that Mr. Lee needs to fish, so that the probability of catching more than two fish exceeds 80%, we need to use the Poisson probability formula.
The Poisson distribution is defined by a parameter λ, which represents the average number of events occurring in a fixed interval. In this case, the mean number of fish caught per hour is given as 0.6, so λ = 0.6.
Let's denote X as the random variable representing the number of fish caught per hour. According to the problem, we need to find the number of hours, denoted as k, such that P(X > 2) > 0.8.
To solve this problem, we can use the cumulative probability function of the Poisson distribution. The cumulative probability function calculates the probability of getting less than or equal to a certain number of events. To find the probability of getting more than two fish, we can subtract the cumulative probability of getting two or fewer fish from 1.
The formula for the cumulative probability of the Poisson distribution is:
P(X ≤ k) = e^(-λ) * (λ^0/0! + λ^1/1! + ... + λ^k/k!)
We can rewrite P(X > 2) as 1 - P(X ≤ 2) to calculate the probability of catching more than two fish.
To solve for k, we can start with k = 1 (assuming Mr. Lee fishes for 1 hour), then increment k until P(X > 2) exceeds 0.8.
Here's the step-by-step process:
1. Start with k = 1 and calculate the cumulative probability of getting two or fewer fish using the formula above: P(X ≤ 2) = e^(-0.6) * (0.6^0/0! + 0.6^1/1! + 0.6^2/2!).
2. Subtract the cumulative probability from 1 to find P(X > 2): P(X > 2) = 1 - P(X ≤ 2).
3. If P(X > 2) > 0.8, then we have found the number of hours Mr. Lee needs to fish. If not, increment k by 1 and repeat steps 1 and 2.
Repeat steps 1 to 3 until P(X > 2) exceeds 0.8.
Note that the calculation can also be done using a computer program or a statistical calculator that supports the Poisson distribution.