Q.For the probability distribution of $X$ (the number of heads in two coin tosses) obtained in the previous example, find the mathematical expectation $E(X)$ and interpret its meaning.

Expectation is linear, so $E(6x)=6E(x)$.

A key property of mathematical expectation is that a constant factor pulls out of the expectation. For any constant $a$,

$E(aX) = a\,E(X).$

With $a=6$:

$E(6X) = 6\,E(X).$

(The factor is not squared — squaring the constant applies to variance, $V(aX)=a^2V(X)$, not to expectation, which rules out option (c).)

$E(X-Y)=E(X)-E(Y)=5-(-2)=7$.

By the linearity of mathematical expectation,

$E(X-Y)=E(X)-E(Y)=5-(-2)=7.$

Expectation is linear and $E(X)$ is a constant, so the mean deviation $E[X-E(X)]=0$.

Let $\mu=E(X)$, which is a constant. Using linearity of expectation:

$E[X-E(X)]=E(X)-E\big(E(X)\big)=E(X)-E(X)=\mu-\mu=0.$

(The expectation of a constant equals the constant itself, so $E(\mu)=\mu$.) This is why the mean of the deviations about the mean is always zero.

Mathematical expectation is $\sum x\,p(x)$; here $10\times\tfrac{1}{10}=1$.

The expected value of a discrete random variable is

$E(X)=\sum x\,p(x).$

Given $x=10$ with $p(x)=\dfrac{1}{10}$:

$E(X)=10\times\frac{1}{10}=1.$