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To calculate the variance of the first 10 natural numbers that are multiples of 3 , we use the formula for variance:$\sigma^2 = \; \frac{1}{n} \sum\limits_{i=1}^{n} (x_i)^2 - \overline{x}^2$Here, let $x_i = 3k$ where $k \in [1,10]$. So the numbers are $3 \times 1, 3 \times 2, \ldots, 3 \times 10$.First, we find the sum of squares of these numbers:$\sigma^2 = \; \frac{1}{n} \times 9 \sum\limits_{k=1}^{10} k^2 - 9(\overline{k})^2$Calculating step-by-step:The formula for the sum of squares of the first $n$ natural numbers is:$\sum\limits_{k=1}^{n} k^2 = \; \frac{n(n+1)(2n+1)}{6}$Substituting $n=10$ :$\sum\limits_{k=1}^{10} k^2 = \; \frac{10 \times 11 \times 21}{6}$The mean $\overline{k}$ of the first 10 natural numbers is:$\overline{k} = \; \frac{1}{10} \sum\limits_{k=1}^{10} k = \; \frac{10 \times 11}{2 \times 10}$Plugging these into the variance formula:$\sigma^2 = \; \frac{9}{10} \times \; \frac{10 \times 11 \times 21}{6} - 9 \left( \; \frac{10 \times 11}{20} \right)^2$This simplifies to: $\sigma^2 = \; \frac{9}{10} \times 385 - 9 \times 30.25$Finally, calculating the result:$\sigma^2 = \; \frac{9}{10} \times 385 - 272.25 = 74.25$Thus, the variance of the first 10 multiples of 3 is 74.25 .