Question

Learning Thoery In a learning theory project, the proportion P of correct responses after n trials can be modeled by p = 0.83/(1+e-02n).

(a) Find the proportion of correct responses after 3 trials.
(b) Find the proportion of correct responses after 7 trials.
(c) use a graphing utility to graph the model.Find the number of trials required for the porportion of correct responses to be 0.75.
(d) Does the porportion of correct responses have a limit as n increase without bound?Explain your reasoning.

Answer 1

a)
`P(n=3) = (0.83)/(1+e^(-0.2(3)))= (0.83)/(1+ e^(-0.6)) = 0.536`

b)
`P(n=7) = (0.83)/(1+e^(-0.2(7)))= (0.83)/(1+ e^(-1.4)) = 0.666`

c)
`0.75 =(0.83)/(1+e^(-0.2n))`

`1+ e^(-0.2n) = (0.83)/(0.75)= (83)/(75)`

`e^(-0.2n) = (83)/(75)-1= (8)/(75)`

`ln e^(-0.2n) = ln ((8)/(75))`

`-0.2 n = ln((8)/(75))`

And then if we solve for t we got:

`n = (ln((8)/(75)))/(-0.2) = 11.19 trials`

d) If we find the limit when n tend to infinity for the function we have this:

`lim_(n \to \infty) (0.83)/(1+e^(-0.2t)) = 0.83`

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

Explanation:

For this case we have the following expression for the proportion of correct responses after n trials:

`P(n) = (0.83)/(1+e^(-0.2t))`

Part a

For this case we just need to replace the value of n=3 in order to see what we got:

`P(n=3) = (0.83)/(1+e^(-0.2(3)))= (0.83)/(1+ e^(-0.6)) = 0.536`

So the number of correct reponses after 3 trials is approximately 0.536.

Part b

For this case we just need to replace the value of n=7 in order to see what we got:

`P(n=7) = (0.83)/(1+e^(-0.2(7)))= (0.83)/(1+ e^(-1.4)) = 0.666`

So the number of correct responses after 7 weeks is approximately 0.666.

Part c

For this case we want to solve the following equation:

`0.75 =(0.83)/(1+e^(-0.2n))`

And we can rewrite this expression like this:

`1+ e^(-0.2n) = (0.83)/(0.75)= (83)/(75)`

`e^(-0.2n) = (83)/(75)-1= (8)/(75)`

Now we can apply natural log on both sides and we got:

`ln e^(-0.2n) = ln ((8)/(75))`

`-0.2 n = ln((8)/(75))`

And then if we solve for t we got:

`n = (ln((8)/(75)))/(-0.2) = 11.19 trials`

And we can see this on the plot attached.

Part d

If we find the limit when n tend to infinity for the function we have this:

`lim_(n \to \infty) (0.83)/(1+e^(-0.2t)) = 0.83`

So then the number of correct responses have a limit and is 0.83 as n increases without bound.