posted by Anonymous .
Suppose that the populations of the three species are distributed as shown in the following table. You capture a frog, note its gender and species, and then release it. This process is repeated until you have captured and recorded 50 frogs.
Species %total pop. Males Females
bullfrog 30% 60% 40%
spring peeper 50% 50% 50%
mink frog 20% 50% 50%
a) Determine the probability that there will be at least five female bullfrogs in sample.
b)Determine the probability that there will not be any mink frogs in sample.
c) Suppose that there were 30 spring peepers in the sample. Determine whether this is unusual enough to cause you to reconsider your original estimate of their proportion of the frog population.
Excek spreadsheets are very helpful for these kinds of problem.
a) the probability of selecting a female is .3*.4 + .5*.5 + .2*.5 = .47. Ergo, the probability of picking a male is .53.
The probability of picking all males is .53^50 = something really small.
for n males it (50-choose-n)*.53^(50-n)*.47^n. So, for n=1 its 50*.53^(49) * .47 = 1.725E(-13). Repeat for n=2,n=3, and n=4. the sum of these 5 really small values is the probability of getting 46 or more male. I get 2.33E(-9). 1-minus this is the probability of getting at least 5 females.
b) The probability of NOT picking milk frog in a single pick is .8 The probability in 50 picks is .8^50. I get 1.427E(-5).
c) Largely repeat the steps in a)