Conditioning on random variable
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Fish are caught at a rate of 3 every 2 hours. If a fisherman spends anywhere from 3 to 5.5 hours fishing on random day, find expectation and variance of number of fish he catches
probability
asked Dec 4 '18 at 15:33
Murph JonesMurph Jones
713
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$begingroup$
Fish are caught at a rate of 3 every 2 hours. If a fisherman spends anywhere from 3 to 5.5 hours fishing on random day, find expectation and variance of number of fish he catches
probability
asked Dec 4 '18 at 15:33
Murph JonesMurph Jones
713
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According to what distribution are the fish caught?
$endgroup$
– bob
Dec 4 '18 at 16:02
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$begingroup$
Fish are caught at a rate of 3 every 2 hours. If a fisherman spends anywhere from 3 to 5.5 hours fishing on random day, find expectation and variance of number of fish he catches
probability
asked Dec 4 '18 at 15:33
Murph JonesMurph Jones
713
$endgroup$
Fish are caught at a rate of 3 every 2 hours. If a fisherman spends anywhere from 3 to 5.5 hours fishing on random day, find expectation and variance of number of fish he catches
probability
probability
asked Dec 4 '18 at 15:33
Murph JonesMurph Jones
713
asked Dec 4 '18 at 15:33
Murph JonesMurph Jones
713
asked Dec 4 '18 at 15:33
Murph JonesMurph Jones
713
asked Dec 4 '18 at 15:33
Murph JonesMurph Jones
713
asked Dec 4 '18 at 15:33
Murph JonesMurph Jones
713
713
$begingroup$
According to what distribution are the fish caught?
$endgroup$
– bob
Dec 4 '18 at 16:02
add a comment |
$begingroup$
According to what distribution are the fish caught?
$endgroup$
– bob
Dec 4 '18 at 16:02
$begingroup$
According to what distribution are the fish caught?
$endgroup$
– bob
Dec 4 '18 at 16:02
$begingroup$
According to what distribution are the fish caught?
$endgroup$
– bob
Dec 4 '18 at 16:02
add a comment |
1 Answer
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Assuming that exactly 3 fish are caught every two hours, you would expect that if given from 3 to 5.5 hours, we would expect to catch anywhere between 4.5 and 8.25 fish. Assuming the fisherman decides his time fishing randomly between 3 and 5.5 hours, the number of caught fish could be represented by $X sim U[4.5, 8.25]$.
For uniform distribution: $$E[X] = dfrac{a+b}{2} = dfrac{4.5+8.25}{2} = 6.375$$ $$Var[X] = dfrac{(b-a)^2}{12} = dfrac{(8.25-4.5)^2}{12} approx 1.172$$
answered Dec 4 '18 at 16:07
bobbob
1089
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1 Answer
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$begingroup$
Assuming that exactly 3 fish are caught every two hours, you would expect that if given from 3 to 5.5 hours, we would expect to catch anywhere between 4.5 and 8.25 fish. Assuming the fisherman decides his time fishing randomly between 3 and 5.5 hours, the number of caught fish could be represented by $X sim U[4.5, 8.25]$.
For uniform distribution: $$E[X] = dfrac{a+b}{2} = dfrac{4.5+8.25}{2} = 6.375$$ $$Var[X] = dfrac{(b-a)^2}{12} = dfrac{(8.25-4.5)^2}{12} approx 1.172$$
answered Dec 4 '18 at 16:07
bobbob
1089
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add a comment |
$begingroup$
Assuming that exactly 3 fish are caught every two hours, you would expect that if given from 3 to 5.5 hours, we would expect to catch anywhere between 4.5 and 8.25 fish. Assuming the fisherman decides his time fishing randomly between 3 and 5.5 hours, the number of caught fish could be represented by $X sim U[4.5, 8.25]$.
For uniform distribution: $$E[X] = dfrac{a+b}{2} = dfrac{4.5+8.25}{2} = 6.375$$ $$Var[X] = dfrac{(b-a)^2}{12} = dfrac{(8.25-4.5)^2}{12} approx 1.172$$
answered Dec 4 '18 at 16:07
bobbob
1089
$endgroup$
add a comment |
$begingroup$
Assuming that exactly 3 fish are caught every two hours, you would expect that if given from 3 to 5.5 hours, we would expect to catch anywhere between 4.5 and 8.25 fish. Assuming the fisherman decides his time fishing randomly between 3 and 5.5 hours, the number of caught fish could be represented by $X sim U[4.5, 8.25]$.
For uniform distribution: $$E[X] = dfrac{a+b}{2} = dfrac{4.5+8.25}{2} = 6.375$$ $$Var[X] = dfrac{(b-a)^2}{12} = dfrac{(8.25-4.5)^2}{12} approx 1.172$$
answered Dec 4 '18 at 16:07
bobbob
1089
$endgroup$
Assuming that exactly 3 fish are caught every two hours, you would expect that if given from 3 to 5.5 hours, we would expect to catch anywhere between 4.5 and 8.25 fish. Assuming the fisherman decides his time fishing randomly between 3 and 5.5 hours, the number of caught fish could be represented by $X sim U[4.5, 8.25]$.
For uniform distribution: $$E[X] = dfrac{a+b}{2} = dfrac{4.5+8.25}{2} = 6.375$$ $$Var[X] = dfrac{(b-a)^2}{12} = dfrac{(8.25-4.5)^2}{12} approx 1.172$$
answered Dec 4 '18 at 16:07
bobbob
1089
answered Dec 4 '18 at 16:07
bobbob
1089
answered Dec 4 '18 at 16:07
bobbob
1089
answered Dec 4 '18 at 16:07
bobbob
1089
1089
add a comment |
add a comment |
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According to what distribution are the fish caught?
$endgroup$
– bob
Dec 4 '18 at 16:02