Question

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?

Answer 1

Answer:

a) 1/2

b) 250

Step-by-step explanation:

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 $p$ such that $1000p(1-p)$ is maximized. Once we have that $p$, we can easily find the answer to part b.

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

$p(1-p)=p-p^2=-p^2+p=-(p^2-p)=-(p^2-p+1/4)-(-1/4)=-(p-1/2)^2+1/4$.

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 $p$ such that the inner part of the square term is equal to $0$.

$p-1/2=0\\p=1/2$.

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

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

$1000(1/2)(1-1/2)=1000(1/2)(1/2)=1000*1/4=250$.

So, the answer to part b is $\boxed{250}$.

And we're done!