Calculate Standard Deviation from Sums of Squared Deviations

To find the standard deviation of the $10$ observations $x_1, x_2, \dots, x_{10}$, we start by expanding the given summations.

Given:

1. $\sum_{i=1}^{10} (x_i + 2)^2 = 180$
2. $\sum_{i=1}^{10} (x_i - 1)^2 = 90$

Expanding equation (1): $\sum_{i=1}^{10} (x_i^2 + 4x_i + 4) = 180$ $\sum_{i=1}^{10} x_i^2 + 4\sum_{i=1}^{10} x_i + \sum_{i=1}^{10} 4 = 180$ $\sum_{i=1}^{10} x_i^2 + 4\sum_{i=1}^{10} x_i + 40 = 180$ $\sum_{i=1}^{10} x_i^2 + 4\sum_{i=1}^{10} x_i = 140 \quad \text{--- (Equation 1)}$

Expanding equation (2): $\sum_{i=1}^{10} (x_i^2 - 2x_i + 1) = 90$ $\sum_{i=1}^{10} x_i^2 - 2\sum_{i=1}^{10} x_i + \sum_{i=1}^{10} 1 = 90$ $\sum_{i=1}^{10} x_i^2 - 2\sum_{i=1}^{10} x_i + 10 = 90$ $\sum_{i=1}^{10} x_i^2 - 2\sum_{i=1}^{10} x_i = 80 \quad \text{--- (Equation 2)}$

Subtracting Equation 2 from Equation 1: $\left(\sum_{i=1}^{10} x_i^2 + 4\sum_{i=1}^{10} x_i\right) - \left(\sum_{i=1}^{10} x_i^2 - 2\sum_{i=1}^{10} x_i\right) = 140 - 80$ $6\sum_{i=1}^{10} x_i = 60$ $\sum_{i=1}^{10} x_i = 10$

Substituting $\sum_{i=1}^{10} x_i = 10$ back into Equation 2: $\sum_{i=1}^{10} x_i^2 - 2(10) = 80$ $\sum_{i=1}^{10} x_i^2 = 100$

Now, the variance ($\sigma^2$) for $N = 10$ observations is calculated using the formula: $\sigma^2 = \frac{1}{N} \sum_{i=1}^{10} x_i^2 - \left(\frac{1}{N} \sum_{i=1}^{10} x_i\right)^2$ $\sigma^2 = \frac{100}{10} - \left(\frac{10}{10}\right)^2$ $\sigma^2 = 10 - 1 = 9$

Taking the square root to find the standard deviation ($\sigma$): $\sigma = \sqrt{9} = 3$

Thus, the correct option is D.