Question

Mean as center of Gravity. Let X be a discrete random variable with a finite number of number of possible values say $x_1,x_2.....x_m$For convenience, set $P_K=P(X=x_k),$for K = 1, 2.....m . Think of a horizontal axis as a seesaw and each $P_K$as a mass placed at point $x_k$on the seesaw. The center of gravity of those 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 $P_K$is proportional to the product of that mass with the signed distance of the point $X_k$From c That is , to ($(x_k-c)$. $P_k$show that the center of gravity equal the mean of the random variable X ( hint: To balance, the total torque acting on the seesaw must be 0)

Short answer

The random variable's mean equals the centre of gravity.

${\mu }_X=c$

Step-by-step solution

Step 1. Given information. 

Set $P_K=P(X=x)$for $k=1,2,3.....m$and consider the possible values of a random variable X to be $x_1,x_2......x_m$. Consider the seesaw below, with each mass $P_K$placed at point $X_K$and the center of gravity at point c. $(x_k-c)p_k$is the torque exerted on the seesaw by mass $p_k$.

A balanced seesaw has a total torque of zero.

Step 2. The center of gravity is equal to the mean of the random variable X.

An event's total probability is:

$\sum _{}P(X=x)=1$

A random variable's expected value is given by:

$\mu =\sum _{}xP(X=x)$

The seesaw is balanced if the total torque acting on it equals:

$$\begin{gathered} \sum _{k‐1}^m(x_k-c)p_k=0 \\ \sum _{k‐1}^m(x_k·p_k-c·p_k)=0 \\ \sum _{k‐1}^m(x_k·p_k-\sum _{k‐1}^{m}c·p_k)=0 \\ {\mu }_x-\sum _{k‐1}^{m}c·p_k=0 \\ {\mu }_x-c\sum _{k‐1}^m·P(X=x_k)=0 \\ {\mu }_x-c·1=0 \\ {\mu }_x-c·1=0 \\ {\mu }_x-c=0 \\ {\mu }_x=c \end{gathered}$$

As a result, the random variable's mean equals the centre of gravity.