Answer:
a)[tex] P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536[/tex]
b) [tex] P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666[/tex]
c) [tex] 0.75 =\frac{0.83}{1+e^{-0.2n}}[/tex]
[tex] 1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}[/tex]
[tex]e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}[/tex]
[tex] ln e^{-0.2n} = ln (\frac{8}{75})[/tex]
[tex] -0.2 n = ln(\frac{8}{75})[/tex]
And then if we solve for t we got:
[tex] n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials[/tex]
d) If we find the limit when n tend to infinity for the function we have this:
[tex] lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83[/tex]
So then the number of correct responses have a limit and is 0.83 as n increases without bound.
Step-by-step explanation:
For this case we have the following expression for the proportion of correct responses after n trials:
[tex] P(n) = \frac{0.83}{1+e^{-0.2t}}[/tex]
Part a
For this case we just need to replace the value of n=3 in order to see what we got:
[tex] P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536[/tex]
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:
[tex] P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666[/tex]
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:
[tex] 0.75 =\frac{0.83}{1+e^{-0.2n}}[/tex]
And we can rewrite this expression like this:
[tex] 1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}[/tex]
[tex]e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}[/tex]
Now we can apply natural log on both sides and we got:
[tex] ln e^{-0.2n} = ln (\frac{8}{75})[/tex]
[tex] -0.2 n = ln(\frac{8}{75})[/tex]
And then if we solve for t we got:
[tex] n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials[/tex]
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:
[tex] lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83[/tex]
So then the number of correct responses have a limit and is 0.83 as n increases without bound.
