Expectation: $\frac{55}{15}$ ; Variance: $\frac{225}{15} - (\frac {55}{15})^2$

We know that the expectation has the following format:

$$E[x] = \sum_{\forall x} f_{X}(x) x$$

In this case, we're given a sort of strange function for $f_{x}(x)$, namely that the probability is increasing for each of the value. We can name each of these values out:

$$\begin{aligned}P(X=1) &= \frac{1}{15} \P(X=2) &= \frac{2}{15} \P(X=3) &= \frac{3}{15} = \frac{1}{5} \P(X=4) &= \frac{4}{15} \P(X=5) &= \frac{5}{15} = \frac{1}{3}\end{aligned}$$

And so, using the definition of expectation, we can find that

$$\begin{aligned}E[X] &= (1) \cdot \frac{1}{15} + (2) \cdot \frac{2}{15} + (3) \cdot \frac{3}{15} + (4) \cdot \frac{4}{15} + (5) \cdot \frac{5}{15} \&= \frac{1}{15} + \frac{4}{15} + \frac{9}{15} + \frac{16}{15} + \frac{25}{15} \&= \frac{55}{15}\end{aligned}$$

To figure out the variance, we can use the alternative form of variance: $V[X] = E[X^2] - E[X]^2$. We've already found that $E[X] = \frac{55}{15}$, so we can quickly note that $E[X]^2 = \frac{55}{15}^2 = \frac{3025}{225}$.

What is left is to find $E[X^2]$, which, using $LOTUS$ is:

$$\begin{aligned}E[X^2] &= (1^2) \cdot \frac{1}{15} + (2^2) \cdot \frac{2}{15} + (3^2) \cdot \frac{3}{15} + (4^2) \cdot \frac{4}{15} + (5^2) \cdot \frac{5}{15} \&= \frac{1}{15} + \frac{8}{15} + \frac{27}{15} + \frac{64}{15} + \frac{125}{15} \&= \frac{225}{15}\end{aligned}$$

As a consequence, the variance of this random variable is: $\frac{225}{15} - (\frac{55}{15})^2$.