suppose that p is the probability that a randomly selected person is left handed. The value (1-p) is the probability that the person is not left-handed. In the sample of 100 people, the function V(p)= 1000 p(1-p) represents the variance of the number of left-handed people in a group of 100.
a) What value of p maximizes the variance?
b) What is the maximum variance?

The start of the question doesn't matter entirely, although is interesting to read. What we are trying to do is find the value for [tex]p[/tex] such that [tex]1000p(1-p)[/tex] is maximized. Once we have that [tex]p[/tex], we can easily find the answer to part b.

Finding the value that maximizes [tex]1000p(1-p)[/tex] is the same as finding the value that maximizes [tex]p(1-p)[/tex], just on a smaller scale. So, we really want to maximize [tex]p(1-p)[/tex]. To do this, we will do a trick called completing the square.

[tex]p(1-p)=p-p^2=-p^2+p=-(p^2-p)=-(p^2-p+1/4)-(-1/4)=-(p-1/2)^2+1/4[/tex].

Because there is a negative sign in front of the big squared term, combined with the fact that a square is always positive, means we need to find the value of [tex]p[/tex] such that the inner part of the square term is equal to [tex]0[/tex].

[tex]p-1/2=0\\p=1/2[/tex].

So, the answer to part a is [tex]\boxed{1/2}[/tex].

We can then plug [tex]1/2[/tex] into the equation for p to find the answer to part b.

[tex]1000(1/2)(1-1/2)=1000(1/2)(1/2)=1000*1/4=250[/tex].

So, the answer to part b is [tex]\boxed{250}[/tex].

And we're done!