# Question

Let X be a discrete random variable with a finite number of possible values, say, x1, x2, . . . , xm. For convenience, set pk = P(X = xk ), for k = 1, 2, . . . ,m. Think of a horizontal axis as a seesaw and each pk as a mass placed at point xk on the seesaw. The center of gravity of these masses is defined to be the point c on the horizontal axis at which a fulcrum could be placed to balance the seesaw.

Relative to the center of gravity, the torque acting on the seesaw by the mass pk is proportional to the product of that mass with the signed distance of the point xk from c, that is, to (xk − c) · pk. Show that the center of gravity equals the mean of the random variable X.

Relative to the center of gravity, the torque acting on the seesaw by the mass pk is proportional to the product of that mass with the signed distance of the point xk from c, that is, to (xk − c) · pk. Show that the center of gravity equals the mean of the random variable X.

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