Question

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

Answer 1

Answer:

The maximum variance is 250.

Step-by-step explanation:

Consider the provided function.

$V(x)=1000x(1-x)$

$V(x)=1000x-1000x^2$

Differentiate the above function as shown:

$V'(x)=1000-2000x$

The double derivative of the provided function is:

$V''(x)=-2000$

To find maximum variance set first derivative equal to 0.

$1000-2000x=0$

$x=\frac{1}{2}$

The double derivative of the function at $x=\frac{1}{2}$ is less than 0.

Therefore, $x=\frac{1}{2}$ is a point of maximum.

Thus the maximum variance is:

$V(x)=1000(\frac{1}{2})-1000{\frac{1}{2}}^2$

$V(x)=250$

Hence, the maximum variance is 250.