Look at the definition of mean, if p(x)=0 far all x<0, then the integrand of defining integral for the mean is never negative, so the integral is positive.
If the probability that a rv is negative is zero, the mean is of necessity positive (or zero).
given by >>>>
F(x)=
0 x<=0
x\8 0<=x<4 Mr F says: That 4 should be a 2.
x^2\16 2<=x<4
1 x>=4
find >>>>>>>>>
1-mean of (-2-3x)
2-p(1<x<3)
my solutions
1-mean = -41\4
2- p(1<x<3)= 1\4
The hard evidence supporting CaptainB (not that he needs supporting since his comments are bombproof) is that the mean (assuming my correction is correct) is 31/12.
Look at the definition of mean, if p(x)=0 far all x<0, then the integrand of defining integral for the mean is never negative, so the integral is positive.
If the probability that a rv is negative is zero, the mean is of necessity positive (or zero).
RonL
Ahh.. I see my confusion I was misreading the question you want the mean of which is of course negative since the mean of x is positive, and so is equal to
Ahh.. I see my confusion I was misreading the question you want the mean of which is of course negative since the mean of x is positive, and so is equal to