SOLUTION: To estimate the number of fish in a small lake, researchers caught 200 fish, tagged them, and returned them to the lake. The next day they caught 100 fish. Of those caught, 25 had

Question 1011694: To estimate the number of fish in a small lake, researchers caught 200 fish, tagged them, and returned them to the lake. The next day they caught 100 fish. Of those caught, 25 had a tag. If there were 800 fish in the lake, what is the probability that 25 of the 100 fish in the second catch would have been tagged?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
you tag 200 fish from the lake.
assuming that there are 800 fish in the lake, what is the probability that a sample of 100 fish taken from the lake would contain exactly 25 tagged fish.

if you tag 200 fish from the lake, and there are 800 fish in the lake, then the proportion of tagged fish to total fish is .25.

you subsequently catch 100 fish and want to know the probability that exactly 25 of those fish will be tagged.

this is a binomial distribution.

the formula for a binomial distribution is:

p(x) = c(n,x) * p^x * q^(n-x)

x = 25
n = 100
p = .25
q = 1 - .25 = .75

the formula becomes:

p(25) = c(100,25) * .25^25 * .75^75

if your calculator can handle it, you would get p(25) = .0917996918 which we'll round off to .0918.

you get p(25) = .0918.

c(n,x) is the combination formula of n! / (x! * (n-x)!)

it tells you how many possible sets you can get of x elements each out of a total of n elements when order is not important. sets that contain the same elements but in a different order are considered to be the same set and are only counted once. this is different from a permutation where sets that have the same elements but in a different order are counted as different sets.
for example:
abc and cba are counted as 1 set by the combination formula.
abc and cba are counted as 2 sets by the permutation formula.

if your calculator can't handle it, then you would use the normal approximation to the binomial.

in that case, you would do the following.

m = n*p = 100*.25 = 25
s = sqrt(n*p*q) = sqrt(100*.25*.75) = 4.330127019

m1 = m - .5 = 25 - .5 = 24.5
m2 = m + .5 = 25 + .5 = 25.5

z1 = (24.5 - 25) / 4.330127019 = -.154700538
z2 = (25.5 - 25) / 4.330127019 = .154700538

you would then use your z-score calculator to find the area under the distribution curve between a z-factor of -.154700538 and .154700538.

the result will be that the area under the distribution curve = .0919275415 which you would round off to .0919.

with the binomial distribution formula, you got .0918.
with the normal approximation to the binomial distribution formula, you got .0919.

they're very close to each other, so either one would have sufficed.

assuming the binomial distribution formula is the more accurate, the normal approximation of .0919 would be off .0001 from the binomial approximation of .0918 which is about a tenth of a percent off.
that's pretty good.

the first thing that you needed to do to solve this problem was to understand that it is a binomial distribution type of problem.

the second thing that you needed to know is that you could use the normal approximation to the binomial if you had to.

here's some references you might like to look at for further information.

http://www.stattrek.com/probability-distributions/binomial.aspx

http://www.mathsisfun.com/data/binomial-distribution.html

http://regentsprep.org/Regents/math/algtrig/ATS7/BLesson3.htm

http://onlinestatbook.com/2/normal_distribution/normal_approx.html

there are plenty more on the web.
just do a search for (binomial distribution) or (normal approximation to the binomial) and you'll get tons of them.

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