Adding 2 to each value of the random variable
makes a new random variable
. Its mean would be
![E[X+2]=E[X]+E[2]=E[X]+2](https://tex.z-dn.net/?f=E%5BX%2B2%5D%3DE%5BX%5D%2BE%5B2%5D%3DE%5BX%5D%2B2)
since expectation is linear, and the expected value of a constant is that constant.
is the mean of
, so the new mean would be
![E[X+2]=10+2=12](https://tex.z-dn.net/?f=E%5BX%2B2%5D%3D10%2B2%3D12)
The variance of a random variable
is
![V[X]=E[X^2]-E[X]^2](https://tex.z-dn.net/?f=V%5BX%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2)
so the variance of
would be
![V[X+2]=E[(X+2)^2]-E[X+2]^2](https://tex.z-dn.net/?f=V%5BX%2B2%5D%3DE%5B%28X%2B2%29%5E2%5D-E%5BX%2B2%5D%5E2)
We already know
, so simplifying above, we get
![V[X+2]=E[X^2+4X+4]-12^2](https://tex.z-dn.net/?f=V%5BX%2B2%5D%3DE%5BX%5E2%2B4X%2B4%5D-12%5E2)
![V[X+2]=E[X^2]+4E[X]+4-12^2](https://tex.z-dn.net/?f=V%5BX%2B2%5D%3DE%5BX%5E2%5D%2B4E%5BX%5D%2B4-12%5E2)
![V[X+2]=(V[X]+E[X]^2)+4E[X]-140](https://tex.z-dn.net/?f=V%5BX%2B2%5D%3D%28V%5BX%5D%2BE%5BX%5D%5E2%29%2B4E%5BX%5D-140)
Standard deviation is the square root of variance, so
.
![\implies V[X+2]=(9+10^2)+4(10)-140=9](https://tex.z-dn.net/?f=%5Cimplies%20V%5BX%2B2%5D%3D%289%2B10%5E2%29%2B4%2810%29-140%3D9)
so the standard deviation remains unchanged at 3.
NB: More generally, the variance of
for
is
![V[aX+b]=a^2V[X]+b^2V[1]](https://tex.z-dn.net/?f=V%5BaX%2Bb%5D%3Da%5E2V%5BX%5D%2Bb%5E2V%5B1%5D)
but the variance of a constant is 0. In this case,
, so we're left with
, as expected.