Question

F(x)=k(9x−x 0≤x≤9 and f(x)=0 if x>9. For what value of k is f a probability density function?

Answer 1

To make f(x) a probability density function, we need to find a value of k that satisfies the conditions of a pdf. However, in this case, there is no value of k that satisfies these conditions.

Step-by-step explanation:

A probability density function (pdf) must satisfy two conditions:

1. The function must be non-negative, meaning that f(x) ≥ 0 for all x.
2. The total area under the function must be equal to 1.

In this case, we have f(x) = k(9x−x), where 0 ≤ x ≤ 9, and f(x) = 0 if x > 9. We want to find the value of k that makes f(x) a pdf. To do this, we need to calculate the total area under the function and set it equal to 1:

∫[0,9] k(9x−x) dx = 1

Integrating the function, we get:

k[-18x^2/2 + 9x] from 0 to 9 = 1

-81k + 81k = 1

Simplifying, we find:

0 = 1

Since this equation is not true, there is no value of k that makes f(x) a probability density function.