The Midge Problem

The Midge Problem, A Mathematical Modeling Example
In 1981, two new varieties of a tiny biting insect called a midge were discovered in the jungles of Brazil by biologists W. L. Grogan and W. W. Wirth. They dubbed one kind of midge an Apf midge and the other an Af midge. The biologists found that the Apf midge is a carrier of a debilitating disease that causes swelling of the brain when a human is bitten by an infected midge. Although the disease is rarely fatal, the disability caused by the swelling may be permanent. The other form of the midge, the Af, is quite harmless and a valuable pollinator. In an effort to distinguish the two varieties, the biologists took measurements on the midges they caught. The two measurements taken were wing length and antenna length, both measured in centimeters. The data are provided below.
Af midges

Wing length (cm)	1.72	1.64	1.74	1.7	1.82	1.82	1.9	1.82	2.08
Antenna length (cm)	1.24	1.38	1.36	1.4	1.38	1.48	1.38	1.54	1.56

Apf midges

Wing length (cm)	1.78	1.86	1.96	2.0	2.0	1.96
Antenna length (cm)	1.14	1.2	1.3	1.26	1.28	1.18

Is it possible to distinguish an Af midge from an Apf midge on the basis of wing and antenna lengths? Source: “The Midge Problem,” Everybody’s Problems, Consortium, Number 55, Fall, 1995, COMAP, Inc., Lexington, MA.

Assign students to groups of 2 or 3 students so that all are involved. In this problem, students are asked to establish criteria to distinguish between two different, but related, data sets. You may want to handle this activity in one of three ways. Students can work with their group in class for 30 minutes and then the groups can share their ideas with the class. Alternately, students can work with their group in class and then write up their solutions over the next couple of days and submit solutions as an out-of-class assignment. If you choose to work the problem together with the class, it is still useful to give students some time to work together with a group so that a variety of ideas are generated. It seems that once you start down a solution path with the entire class, it is sometimes difficult to have students suggest alternate solutions.
Have students enter the data into lists on their calculators. For our purposes we will make the following list assignments:

L1= Af wing length

L2= Af antenna length

L3= Apf wing length

L4= Apf antenna length
Sample Solutions:

1. Fit a least squares line to the Af ordered pairs (wing length, antenna length) and a least squares line to the Apf ordered pairs (wing length, antenna length) and then take the average of these two slopes and intercepts to form a “border line.”

Af data:

Apf data:

Average line:

The point (1.72, 1.24) from the Af is slightly below boundary line so we would consider it an Apf midge and therefore mistakenly kill it. However, your students might argue that it is better to err on the side of caution. The criteria would be to keep any midge whose ordered pair (wing, antenna) is above the line and destroy those with ordered pair below the line. Mathematically, this can be written: if , then don’t kill the midge.
2. Another solution is found by looking at the ratios of antenna lengths and wing lengths for each type of midge and seeing if there is a relation. In L5 enter the data L2/L1 and in L6 enter the data L4/L3 using the formula at the top of these lists. Looking at L5 we see the values are about 0.7 or 0.8, whereas in L6 the values are about 0.6 or 0.65.
Let’s look at these two lists (L5 and L6) using back-to-back box and whisker plots. We notice that there is a significant difference in the plots of the two lists of ratios. We notice that the largest ratio for Apf midges is 0.663 and the smallest ratio for Af midges is 0.721. A student might decide that the average of these two numbers is a point of demarcation and say if the ratio is more than 0.692 then the midge is Af.
2.5 Keeping with this ratio theme, we can use the one variable statistics option on the calculator to find that the average ratio in L5 is 0.785 and the average ratio in L6 is 0.637. That is, for Af midges on average we have and for Apf midges we have. We can rewrite these ratios as equations of lines.

For an Af midge:

For an Apf midge:.

S

uperimpose these lines over the data to see if you believe they are reasonable models for each set of data. Now find the average of these lines to form the border: . With this idea we would say any ordered pair above the line is an Af midge or said another way, if then the midge is an Af midge.

3. A third solution is to look at the two “lowest” points for the Af data and the two “highest” points for the Apf data, write the equations of the lines through these points and use these two lines as boundaries.

Af midges: (1.72, 1.24) and (1.9, 1.38) give the line .

Apf midges: (1.96, 1.3) and (1.78, 1.14) give the line .

Therefore, if then the midge is Af and if then the midge is Apf.

4. If a midge border is desired, the method above can be modified by averaging the two lines. We would say if then the midge is Af.

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