Question

Learning Theory In a group project in learning theory, a mathematical model for the proportion \(P\) of correct responses after \(n\) trials was found to be $$ P=\frac{0.83}{1+e^{-0.2 n}} $$ (a) Find the limiting proportion of correct responses as \(n\) approaches infinity. (b) Find the rates at which \(P\) is changing after \(n=3\) trials and \(n=10\) trials.

Short answer

The limiting proportion as \(n\) approaches infinity is \(0.83\). The rates at which \(P\) is changing for \(n=3\) and \(n=10\) trials can be calculated by substituting the respective \(n\) values into the derivative of \(P\), obtained as \(\frac{0.166 * e^{-0.2n}}{(1+e^{-0.2n})^2}\).

Step-by-step solution

Find the Limiting Proportion of Correct Responses

To find the limiting proportion \(P\) as \(n\) approaches infinity, we simply take the limit of the function as \(n\) approaches infinity. In mathematical terms, we need to compute: \[ \lim_{n \to \infty} \frac{0.83}{1+e^{-0.2 n}} \] As numeral values increase, \(e^{-0.2n}\) approaches zero, because any number raised to a negative infinity power is zero. Therefore, our expression simplifies to: \[ \lim_{n \to \infty} \frac{0.83}{1+0} = \frac{0.83}{1} \]

Differentiate The Function

Firstly, calculate derivative of the given function \(P = 0.83 / (1 + e^{-0.2n})\) We will begin by noting the function as \(P = 0.83 * (1+e^{-0.2n})^{-1}\). Now using chain rule and the definition of the derivative of an exponential function, we get: \[ \frac{dP}{dn} = 0.83 * -1 * (1+e^{-0.2n})^{-2} * e^{-0.2n} * -0.2 \] Simplifying, \[ \frac{dP}{dn} = \frac{0.166 * e^{-0.2n}}{(1+e^{-0.2n})^2} \]

Evaluate The Derivative At Specific Points

Next, we substitute \(n = 3\) and \(n = 10\) into \(\frac{dP}{dn}\), respectively. When \(n = 3\), \[ \frac{dP}{dn} = \frac{0.166 * e^{-0.2*3}}{(1+e^{-0.2*3})^2} \] Similarly, when \(n = 10\), \[ \frac{dP}{dn} = \frac{0.166 * e^{-0.2*10}}{(1+e^{-0.2*10})^2} \] Compute these values for the rate of change of proportion at the specific trials.