The Student Room Group

v.easy probability question

maybe i'm just tired, but this 'should be' easy question is troubling me!

let f(x)=kx2f(x)=kx^2, where k is a constant and 1<=x<=1-1<= x <=1

what is the value of k for f(x) to be a probability density function?


i'm thinking it could either be -1 or 1? but the next question goes on to ask about 0.5 etc and so i think i'm wrong!

p.s. <= means "less than or equal to".

thanks :smile:
Reply 1
A probability density function has 2 main properties: (a) It's always greater than or equal to 0 (b) The area under it must be equal to 1.

If k was -1, it would abuse at least property (a) because f(x) = -x^2 and f(1) = -1, which is less than 0.

Latest