Step-by-step Solution:

We know that the mean is given by: \[ \frac{\sum X}{N} = \mu \] and the variance is given by: \[ \sigma^2 = \frac{1}{N} \sum X^2 - \mu^2. \] The \( n^{\text{th}} \) moment about a number \( m \) is defined as: \[ \frac{1}{N} \sum_{i=1}^{N} (X - m)^k. \] Hence, the first moment about 2 is: \[ \frac{1}{N} \sum_{i=1}^{N} (X - 2) = \frac{1}{N} \sum X - 2 = \mu - 2. \] Given that \( \mu - 2 = 1 \), we have: \[ \mu = 3. \] The second moment about 2 is: \[ \frac{1}{N} \sum_{i=1}^{N} (X - 2)^2 = \frac{1}{N} \sum \left( X^2 - 4X + 4 \right). \] Simplifying, we get: \[ = \frac{1}{N} \sum X^2 - \frac{4}{N} \sum X + 4, \] \[ = \sigma^2 + \mu^2 - 4\mu + 4 = \sigma^2 + 1. \] Given that the second moment about 2 is 16, we have: \[ \sigma^2 + 1 = 16 \Rightarrow \sigma^2 = 15. \]