Q.The probability mass function of a random variable is defined as

[!FORMULA] $\begin{array}{c|ccccc} x & -2 & -1 & 0 & 1 & 2\\\hline f(x) & k & 2k & 3k & 4k & 5k\end{array}$

Then $E(X)$ is equal to

Computes mean and variance of the given uniform-type density on $(0,l)$ using $E[X]$ and $E[X^2]-\left(E[X]\right)^2$.

1. $\text{Mean}=E[X]=\displaystyle\int_0^lx\,f(x)\,dx=\int_0^lx\cdot\dfrac1l\,dx=\dfrac1l\left[\dfrac{x^2}2\right]_0^l=\dfrac1l\cdot\dfrac{l^2}2=\dfrac l2$.
2. $E[X^2]=\displaystyle\int_0^lx^2\,f(x)\,dx=\int_0^lx^2\cdot\dfrac1l\,dx=\dfrac1l\left[\dfrac{x^3}3\right]_0^l=\dfrac1l\cdot\dfrac{l^3}3=\dfrac{l^2}3$. …

A constant has no spread around its mean, so its variance is identically zero, regardless of the constant's value.

1. Variance measures the expected squared deviation from the mean: $\text{Var}(X)=E[(X-E[X])^2]$. …

Since $\text{Var}(X)=E(X^2)-[E(X)]^2$, with $\text{Var}(X)=4$ and $E(X)=2$, we get $E(X^2)=8$.

1. The variance formula is $\text{Var}(X)=E(X^2)-[E(X)]^2$.
2. Given $\text{Var}(X)=4$ and $E(X)=2$. …

Solving the two expectation equations gives $C=-2$.

1. By linearity of expectation, $E(X+C)=E(X)+C=8$ and $E(X-C)=E(X)-C=12$.
2. Subtracting the second equation from the first: $(E(X)+C)-(E(X)-C)=8-12\Rightarrow 2C=-4$.
3. So $C=-2$. …