A regular tetrahedron dice has faces 1, 2, 3 & 4. The random variable X represents the number on the face. Find the mean and variance of X

if we think of a tetrahedron dice (4 sides from 1 - 4), we would say that the probability P(X=1)=P(X=2)=P(X=3)=P(X=4), so the probability for each side is 1/4

Does this mean that themean = (1+2+3+4)/4 = 2.50?

And therefore, is thevariance = 0.25x[(1-2.5)?^{2}+(2-2.5)^{2}+(3-2.5)^{2}+(4-2.5)^{2}] = 1.25

If This is correct, I hope to post what I have done for the rest of the question to ensure I'm on the right track. Thank you very much, even if you've just taken time to read my post,

Kind regards, Tom