Question

A sample plot of grassland in surveyed for a particular species of insect. Over a 12 week period in the summer, the following data are collected Week 2 10 12 176 230 504 Population 403 255 478 398 (a.) What is the population average given from the table above? (b.) A reasonable fit to the data is given by the cubic polynomial f(t) = ao +ait+a2t2+a3t3. Use Wolfram Alpha to determine the best fit model. Again, the command in Wolfram Alpha is сubic fit {{ti,pi}, {tz, P2}, ..., ftr,pт}} where t, pi} are the time and population given by the table above. Use the polynomial approximation to predict the maximum and minimum populations (c.) To find the average population using the polynomial curve you obtained in part (b), to compute the following definite integral we want 12 1 f(t) dt Paverage 12 Jo Evaluate this integral and comment on the difference between the two averages you obtain in parts (a) and (c). Which value do you think is better? Why?

Answer 1

a) average population using the table can be computed in the below fashion.

$Average \ Population = \frac{1}{7}(403 + 255 + 176 + 230 + 478 + 504 + 398) = 349.14$

Near est Integer Answer349

b) Using the wolfram alpha, we can compute the cubic fit for the polynomial by using the function.

cubic fit {{0,403},{1,255},{2,176},{5,230},{9,478},{10,504},{12,398}}

f (t) 2.02541t839.4071t- 181.472t 400.731

The maxima occurs nears t=10 and the minima occurs near t=3

f(10) 2.02541 (10)3 39.4071 (10)2- 181.472 (10)400.731 501.3 _

c) calculating the average using the integration

$P_{average} = \frac{1}{12} \int_0^{12} f(t)dt = \frac{1}{12} \int_0^{12} (-2.02541t^3 + 39.4071t^2 - 181.472t + 400.731)dt = \frac{1}{12} [-0.5063525t^4 + 13.1357t^3 - 90.736t^2 + 400.731t] _0^{12}$

$P_{average} = \frac{1}{12} [-0.5063525(12)^4 + 13.1357(12)^3 - 90.736(12)^2 + 400.731(12)] = 328.46$

The better approximation will be coming from the P(average) value since the data is missing for few of the weeks, which can be better modelled by the function as compared to taking just the sum and averaging it out.

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