# Thread: Expected values and variance

1. ## Expected values and variance

So, the problem is:
X is a random variable with E(x) = 100 and V(x) = 15. Find:
a) E(x^2)
b) E(3x+10)
c) E(-x)
d) V(-x)
e) std dev(-x)

For part a), i used the equation that V(x) = E(x^2) - (E(x))^2, so 15 = E(x^2) - 10000, therefore E(x^2) = 10015

I'm not sure at all where to start with b, any pointers?

for c, E(-x) would equal -100, right? I think this is assuming that the pmf still exists for x having negative values, and that the probabilities are the same...is this correct?

using the same equation as part a, i found V(-x) to equal 15 as well, since squaring a negative number still results in a positive one, and so the std. dev of (-x) would be sqrt(15)?

Thanks!

2. Originally Posted by mistykz
So, the problem is:
X is a random variable with E(x) = 100 and V(x) = 15. Find:
a) E(x^2)
b) E(3x+10)
c) E(-x)
d) V(-x)
e) std dev(-x)

For part a), i used the equation that V(x) = E(x^2) - (E(x))^2, so 15 = E(x^2) - 10000, therefore E(x^2) = 10015

I'm not sure at all where to start with b, any pointers?
Since that is linear, E(3x+10)= 3E(x)+ 10.

for c, E(-x) would equal -100, right? I think this is assuming that the pmf still exists for x having negative values, and that the probabilities are the same...is this correct?
Yes, E(-x) is -100. No, it is not necessary for x to have negative values for this to make sense. E(f(x))= $\displaystyle \sum f(x)*p(x)$ where the sum (or integral for continuous distributions) is taken over all possible values of x For example, suppose x has value 1 with prob 1/2 and value 2 with prob 1/2. Then E(x)= p(1)(-1)+ p(2)(-2)= (1/2)(-1)+ (1/2)(-2)= -3/2.

using the same equation as part a, i found V(-x) to equal 15 as well, since squaring a negative number still results in a positive one, and so the std. dev of (-x) would be sqrt(15)?

Thanks!