Q.From the bivariate frequency table constructed in Exercise 1, find the marginal distributions of $x$ and $y$, and hence compute the means $\bar{x}$ and $\bar{y}$.

Read the marginal distributions from the totals, then apply $\bar{x} = \sum f_x x / N$ and $\bar{y} = \sum f_y y / N$.

$\bar{x} = (10\cdot3 + 20\cdot4 + 30\cdot3)/10 = 200/10 = 20$

$\bar{y} = (5\cdot4 + 10\cdot3 + 15\cdot3)/10 = 95/10 = 9.5$

Step 1 — Write the marginal distributions (from Exercise 1's totals).

Step 2 — Compute $\bar{x}$ using $\bar{x} = \dfrac{\sum f_x x}{N}$.

Step 3 — Compute $\bar{y}$ using $\bar{y} = \dfrac{\sum f_y y}{N}$.

Independent check. Both weighting totals used $N = 10$ (the same grand total), and each marginal frequency set adds to 10 — so the two means are weighted averages of values that lie within their ranges ($\bar{x} = 20$ sits in $[10, 30]$, $\bar{y} = 9.5$ in $[5, 15]$), a quick plausibility check that neither mean fell outside its data range.

Multiplying a value by the wrong frequency (mixing up $x$-values with $y$-frequencies) is a common error — keep the $x$ marginal and $y$ marginal in separate tables and weight each value only by its own frequency. Dividing by 3 (the number of distinct values) instead of by $N = 10$ (the total frequency) gives a wrong 'unweighted' average.