In: Statistics and Probability
Audrey and Diana go fishing at the Lyndon Fishing Pond. Upon arrival the owner informs them that the pond is stocked with an infinite number of independent fish, and that a typical fisher catches fish at a Poisson rate of 2 fish per hour. There are 8 other people fishing there that day. Diana has the same skill level as a typical fisher but Audrey catches on average twice as many fish as a typical fisher.
(a) (2) Find the mean and variance of the total number of fish caught over 6 hours.
For the rest of the question, assume that 100 fish were caught
that day.
(b) (3) Show that we have the following probabilities. Use those
rounded probabilities in parts b), c) and d):
i. The probability a fish was caught by Audrey is 0.182
ii. The probability a fish was caught by Diana is 0.091
iii. The probability a fish was caught by someone else is 0.727
(c) (2) Find the probability that Audrey catches 15 fish and Diana catches 15 fish
(d) (2) Find the probability that Audrey and Diana catch 30 fish together
(e) (2) Given that Audrey catches 15 fish, find the probability that Diana catches 15 fish
(f) (2) Explain logically the difference between the probabilities in (c), (d), and (e)
(there are more than 4 parts, as per policy i am answering first 4 parts, i.e. , a, b (i,ii,iii) )
a.
λ = 6*(total rate of fish caught per hour) = 5*( rateaudrey + ratediana + 8* rateaverage)
= 6*(2*rateaverage + rateaverage + 8*rateaverage)
= 66*rateaverage
= 66*2 fish per hour
= 132 fish per hour
Mean = λ = 132 fish/hour
variance = λ = 132 fish/hour
b.
P(fish caught by that person) = rate of that person / total rate
= rate of that person / ( rateaudrey + ratediana + 8* rateaverage)
= rate of person / (2*rateaverage + rateaverage + 8*rateaverage)
= rate of person / 11*2
= (rate of person)*0.04545
i.
P(audrey) = rate of audrey * 0.04545
= (2*2)*0.04545
= 0.182
ii.
P(diana) = rate of diana * 0.04545
= (2)*0.04545
= 0.091
iii.
P(someone else) = rate of other 8 * 0.04545
= (8*2)*0.04545
= 0.727
(please UPVOTE)