Posted by **student** on Friday, July 13, 2007 at 7:46pm.

Find expected value for the random variable.

its suppost to be a table 6 X 2 i used the .... to represent separation

z.....3......6.....9.....12.......15

p(z)..0.14..0.29..0.36..0.11...0.10

So what i did is i said

e(x) = 3 (0.14)+ 6(0.29)+ etc to all the rest. and my result was 8.22 am i correct that would be the expected value for the reandom variable.