Adding 2 to each value of the random variable [tex]X[/tex] makes a new random variable [tex]X+2[/tex]. Its mean would be
[tex]E[X+2]=E[X]+E[2]=E[X]+2[/tex]
since expectation is linear, and the expected value of a constant is that constant. [tex]E[X][/tex] is the mean of [tex]X[/tex], so the new mean would be
[tex]E[X+2]=10+2=12[/tex]
The variance of a random variable [tex]X[/tex] is
[tex]V[X]=E[X^2]-E[X]^2[/tex]
so the variance of [tex]X+2[/tex] would be
[tex]V[X+2]=E[(X+2)^2]-E[X+2]^2[/tex]
We already know [tex]E[X+2]=12[/tex], so simplifying above, we get
[tex]V[X+2]=E[X^2+4X+4]-12^2[/tex]
[tex]V[X+2]=E[X^2]+4E[X]+4-12^2[/tex]
[tex]V[X+2]=(V[X]+E[X]^2)+4E[X]-140[/tex]
Standard deviation is the square root of variance, so [tex]V[X]=3^2=9[/tex].
[tex]\implies V[X+2]=(9+10^2)+4(10)-140=9[/tex]
so the standard deviation remains unchanged at 3.
NB: More generally, the variance of [tex]aX+b[/tex] for [tex]a,b\in\mathbb R[/tex] is
[tex]V[aX+b]=a^2V[X]+b^2V[1][/tex]
but the variance of a constant is 0. In this case, [tex]a=1[/tex], so we're left with [tex]V[X+2]=V[X][/tex], as expected.
