# 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.

## Yes, you are correct. To find the expected value for a random variable, you multiply each possible value of the random variable by its corresponding probability, and then sum up the results.

In this case, the random variable has the values 3, 6, 9, 12, and 15, with the corresponding probabilities 0.14, 0.29, 0.36, 0.11, and 0.10 respectively.

To find the expected value, you can calculate:

E(X) = 3 * 0.14 + 6 * 0.29 + 9 * 0.36 + 12 * 0.11 + 15 * 0.10

E(X) = 0.42 + 1.74 + 3.24 + 1.32 + 1.50

E(X) = 8.22

So your result of 8.22 is indeed the expected value of the random variable in this case. Well done!