A population of rabbits is at its lowest in January $(t=0)$ and oscillates 21 above and below an average of 127 during the year. What is the population of rabbits in October?

Step 1 ：The problem describes a sinusoidal function, where the population of rabbits oscillates above and below an average. The amplitude of the oscillation is 21 (the amount it goes above and below the average), the average or midline is 127, and the period is 12 months (since it's described as being over the course of a year).

Step 2 ：We can model the population with the function \(P(t) = A\sin(B(t-h)) + k\), where \(A\) is the amplitude, \(B\) is the frequency, \(h\) is the horizontal shift, and \(k\) is the vertical shift or midline.

Step 3 ：In this case, \(A = 21\), \(B = \frac{2\pi}{12}\) (since the period is 12 months), \(h = 0\) (since the population is at its lowest in January, which we're treating as \(t=0\)), and \(k = 127\).

Step 4 ：We want to find the population in October, which is \(t=9\) (since January is \(t=0\)).

Step 5 ：So we need to calculate \(P(9) = 21\sin\left(\frac{2\pi}{12}\times 9\right) + 127\).

Step 6 ：Final Answer: The population of rabbits in October is \(\boxed{106}\).