1. **State the problem:** We need to find the first four central moments and the first three moments about the origin 4 for the data set $\{2,4,6,8\}$. 2. **Recall definitions:** - The $n$th central moment is defined as $\mu_n = \frac{1}{N} \sum_{i=1}^N (x_i - \bar{x})^n$, where $\bar{x}$ is the mean of the data. - The $n$th moment about a point $a$ is $m_n(a) = \frac{1}{N} \sum_{i=1}^N (x_i - a)^n$. 3. **Calculate the mean $\bar{x}$:** $$\bar{x} = \frac{2 + 4 + 6 + 8}{4} = \frac{20}{4} = 5$$ 4. **Calculate central moments:** - First central moment $\mu_1 = \frac{1}{4} \sum (x_i - 5)^1 = 0$ (by definition, mean deviation is zero). - Second central moment $\mu_2 = \frac{1}{4}[(2-5)^2 + (4-5)^2 + (6-5)^2 + (8-5)^2] = \frac{1}{4}[9 + 1 + 1 + 9] = \frac{20}{4} = 5$ - Third central moment $\mu_3 = \frac{1}{4}[(2-5)^3 + (4-5)^3 + (6-5)^3 + (8-5)^3] = \frac{1}{4}[-27 -1 +1 +27] = \frac{0}{4} = 0$ - Fourth central moment $\mu_4 = \frac{1}{4}[(2-5)^4 + (4-5)^4 + (6-5)^4 + (8-5)^4] = \frac{1}{4}[81 + 1 + 1 + 81] = \frac{164}{4} = 41$ 5. **Calculate moments about origin 4:** - First moment about 4: $$m_1(4) = \frac{1}{4}[(2-4) + (4-4) + (6-4) + (8-4)] = \frac{1}{4}[-2 + 0 + 2 + 4] = \frac{4}{4} = 1$$ - Second moment about 4: $$m_2(4) = \frac{1}{4}[(2-4)^2 + (4-4)^2 + (6-4)^2 + (8-4)^2] = \frac{1}{4}[4 + 0 + 4 + 16] = \frac{24}{4} = 6$$ - Third moment about 4: $$m_3(4) = \frac{1}{4}[(2-4)^3 + (4-4)^3 + (6-4)^3 + (8-4)^3] = \frac{1}{4}[-8 + 0 + 8 + 64] = \frac{64}{4} = 16$$ **Final answers:** - Central moments: $\mu_1=0$, $\mu_2=5$, $\mu_3=0$, $\mu_4=41$ - Moments about origin 4: $m_1(4)=1$, $m_2(4)=6$, $m_3(4)=16$