Solution

Total number of items in the distribution = Σ f_i = 3 + 6 + 9 + 13 + 8 + 5 + 4 = 48.

The Mean (x̅) of the given set = \(\rm \dfrac{\sum f_i x_i}{\sum f_i}\).

⇒ x̅ = \(\rm \dfrac{6\times3+7\times6+8\times9+9\times13+10\times8+11\times5+12\times4}{48}=\dfrac{432}{48}\) = 9.

Let's calculate the variance using the formula: \(\rm \sigma^2 =\dfrac{\sum x_i^2}{n}-\bar x^2\).

\(\rm \dfrac{\sum {x_i}^2}{n}=\dfrac{6^2\times3+7^2\times6+8^2\times9+9^2\times13+10^2\times8+11^2\times5+12^2\times4}{48}=\dfrac{4012}{48}\) = 83.58.

∴ σ² = 83.58 - 9² = 83.58 - 81 = 2.58.

And, Standard Deviation (σ) = \(\rm \sqrt{\sigma^2}=\sqrt{Variance}=\sqrt{2.58}\) ≈ 1.607.