Question

The continuous random variable Y has a probability density function given by: f(y)=k(5-y) for 0 ≤ y ≤ 5,0 otherwise, for some value of k>0. What is the value of k? Number

Answer 1

The answer is 2/25.

To find the value of k, we need to ensure that the probability density function (pdf) integrates to 1 over its entire range.

The given pdf is f(y) = k(5-y) for 0 ≤ y ≤ 5, and 0 otherwise.

Integrating the pdf from 0 to 5 should yield 1:
∫[0 to 5] k(5 - y) dy = 1

Integrating the expression yields:
k ∫[0 to 5] (5 - y) dy = 1
k [5y - (y^2/2)] evaluated from 0 to 5 = 1
k [5(5) - (5^2/2) - (0 - (0^2/2))] = 1
k [25 - (25/2)] = 1
k [25/2] = 1
k = 2/25

Therefore, the value of k is 2/25.