This is bivariate data — a pair $(x,y)$ for each month — so we first find the means, then average the products of deviations from those means.

$\bar{x}=3$, $\bar{y}=4$, $\sum(x-\bar{x})(y-\bar{y})=6$

$\text{Cov}(x,y) = 6/5 = 1.2$

Since each month contributes a pair of values $(x,y)$, this is bivariate data, and covariance is found using $\text{Cov}(x,y) = \dfrac{1}{n}\sum(x_i-\bar{x})(y_i-\bar{y})$.

Step 1 — Find the means. $\bar{x} = \dfrac{1+2+3+4+5}{5} = \dfrac{15}{5} = 3$. $\bar{y} = \dfrac{2+4+5+4+5}{5} = \dfrac{20}{5} = 4$.

Step 2 — Tabulate the deviations and their product.

$\sum(x-\bar{x})(y-\bar{y}) = 4+0+0+0+2 = 6$.

Step 3 — Divide by $n$.

Independent check (direct method). $\sum xy = 1(2)+2(4)+3(5)+4(4)+5(5) = 2+8+15+16+25 = 66$. Then $\dfrac{1}{n}\sum xy - \bar{x}\bar{y} = \dfrac{66}{5} - (3)(4) = 13.2 - 12 = 1.2$ — matches exactly.

Forgetting to divide the sum of cross-products by $n$ (stopping at 6 instead of 1.2) is the most common slip. A sign error in either deviation column (mixing up $x-\bar{x}$ with $\bar{x}-x$) also silently flips the sign of individual products, so it helps to double-check that the deviations sum to zero in each column ($\sum(x-\bar{x})=0$ and $\sum(y-\bar{y})=0$) before multiplying.