A random variable X has PX x x15 for x 1 2 3 4 5 and 0 ot

A random variable X has P(X = x) = x/15 for x = 1, 2, 3, 4, 5 and 0 otherwise. Find its mean and variance

x= P(x) x*P(x) (x-mean) (x-mean)^2 P(x)*(x-mean)^2 1 0.066667 0.066667 -2.66667 7.11111111 0.474074074 2 0.133333 0.266667 -1.66667 2.77777778 0.37037037 3 0.2 0.6 -0.66667 0.44444444 0.088888889 4 0.266667 1.066667 0.333333 0.11111111 0.02962963 5 0.333333 1.666667 1.333333 1.77777778 0.592592593 Mean= 3.666667 Variance= 1.555555556