Answer:
a) p = x/n
b) p = 2/5
c) see attached
Explanation:
a) You apparently want to maximize L(p) with respect to p, assuming values of x and n are fixed. The derivative of L(p) can be found using the product rule. Then the value of p that maximizes L(p) can be found by setting the derivative to zero.
[tex]\dfrac{dL}{dt}=\dfrac{d}{dt}\left(p^{x}(1-p)^{(n-x)}\right)=xp^{(x-1)}(1-p)^{(n-x)}-(n-x)p^{x}(1-p)^{(n-x-1)}\\\\=L(p)\dfrac{x-np}{p(1-p)}[/tex]
Setting this to zero, we have ...
[tex]x-np=0\\\\x=np\\\\p=\dfrac{x}{n}[/tex]
b) For x=2, n=5, the likelihood function L(p) is maximized for ...
... p = x/n = 2/5
c) See the attached plot. The peak (maximum likelihood) is at p=0.4 = 2/5.
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Comment on the problem
This problem involves derivatives, something that you don't expect in a middle school math curriculum.
