October 31, 2000 Volume 29, No.8
|Title:||Assessing the Trend of the Florida Manatee via Aerial Surveys: 1982-Present|
|Speaker:||W. Bruce Craig, Purdue University|
|Time:||Thursday, November 2, 4:00pm. Cookies served at 3:45pm.|
Next Tuesday, November 7 is election day. Many readers will be able to vote in their first presidential election--an election that has not been this close since Kennedy/Nixon in 1960. Don't believe what the statisticians say, your vote does count!
This Week's Colloquium
In many animal population studies, the construction of a probability (stochastic) model provides an effective way to capture underlying
biological and sampling characteristics which contribute to the overall variation in the data. In this talk I will discuss a stochastic model to assess the population trend of the Florida manatee, along the east (Atlantic) coast of that state using aerial survey data collected at winter aggregation sites between 1982 and 1998. This model accounts for the method by which the manatees were counted, their possible movement between surveys, and the increase/decrease of the total population over time. Unlike the previous version (Craig et al., Biometrics, 1997), this model incorporates a Markov chain process to describe the population, thereby reducing the trend's dependency on the selected trend model. The resulting data show an overall increase in the population from 1982-89 and a leveling or even a decrease in this population in the 1990's.
This Week's Speaker
Bruce Craig was born and raised in Madison, WI, went to Washington University in St Louis as an undergrad (majors Math and Econ), came back to Madison for graduate school (MS and Ph.D. in Statistics) and arrived at Purdue in 1996. Besides the usual teaching and research duties, he is also the assistant director of the Purdue Statistical consulting service. He was a competitive swimmer through college and enjoys sports generally. Now, he is married and has a 4-yr old daughter and a 1 1/2-yr old son.
David Molnar is a new member in the math department this year. He was last seen at the University of Connecticut before arriving in Minnesota to teach at St. Olaf. This decision was partially made due to the high quality of brats we offer. His main area of focus is on problem solving and how general theories of problem solving trace back to the Greeks such as Pappus and G. Polya. David also has a strong interest in games. He believes that games give good illustrations of problem solving skills and strategies, which then can be applied to mathematical questions. Besides teaching math, he enjoys playing various games with his 8-yr old daughter Ava and is an avid Red Sox fan.
The annual Carlson Calculus Contest will be held on Nov 14 and 15 this year. The contest is open to any student who has not enrolled in Math 220 or a higher-level math course. Students will pick up an exam sometime between 4 and 7 PM and work in teams of three for two hours. The exam will consist of 8 or 10 problems which calculus students may find amusing. So, get a team together for the contest to be eligible for fantastic prizes. You may preregister by e-mailing Cliff Corzatt at firstname.lastname@example.org.
Last Week's Solutions
Problem: Determine a strategy for the youngest member of a clan of thieves for divvying up a pile of loot based on the following facts. 1) There are 40 thieves of different ages. 2) The youngest does the dividing. 3) If a majority decide they are not satisfied with the division then the youngest gets killed.
Solution provided by Paul Tlucek: First, let's assign numbers to each of the robbers, with the youngest getting #1, and the oldest getting #40. Then, consider the simplest scenario, that of just two robbers remaining, #39 and #40. There is no way for a majority to vote against #39 keeping all the money, so that is exactly what happens. Now, if #'s 38, 39 and 40 are still alive, 38 can keep all of the money except one coin to give to 40. 40 will then vote in favor of 38, because if he does not, he will get nothing. If #'s 37-40 are still alive, then 37 can keep all the money except for a coin given to 39. 39 will vote in favor of that, because if he does not, he will get nothing. If 36-40 are all alive, then 36 will keep all of the money, except for two coins, one for 38 and the other for 40. Those two will vote in favor of that, because if they do not, they will get nothing. If we follow this logic all the way down to the case of the youngest robber (#1), he will keep all the money, except for a gold coin given to each odd individual (3,5,7,9,11,etc.). All the odd numbered robbers will vote in favor of this because if they do not, they will get nothing when #2 divides the coins, and so the voting will be 20 in favor, 20 against, and neither side has the majority, so the division of the treasure is accepted.
*Also solved by Jonathon Kuipers and Robert Hillard.
Problem of the Week
Let a(1), a(2), ..., a(44) be 44 integers with 0 < a(1) < a(2) < a(3) < ...< a(44) < 126. Prove that at least one of the 43 differences d(i)=a(i+1)-a(i) occurs at least 10 times.
**Please submit all solutions to Cliff Corzatt (email@example.com) by noon on Friday.
Editor-in-Chief: Jill Dietz
Associate Editor: Jennifer Beilfuss
Problems Editor: Cliff Corzatt
MM Czar: Donna Brakke