Statistics Part 2

Concept: The Mean Deviation about the Median ($\text{MD}_{\text{median}}$) is the average of the absolute deviations of each observation from the median value. To find the median, the data must first be sorted.Formula:$\text{MD}_{\text{median}} = \frac{\Sigma |x_i -M|}{N}$Where $M$ is the median and $N$ is the number of observations.Solution:Let the 10 observations be $x_1, x_2, \dots, x_{10}$.We are given 8 observations: $\{2, 3, 5, 10, 11, 13, 15, 21\}$.Let the two unknown observations be $a$ and $b$.1. Find the sum of all observations ($\sum x_i$):Given Mean ($\overline{x}$) = 9 and $N=10$.$\sum x_i = N \times \overline{x} = 10 \times 9 = 90$2. Find the sum of the unknown observations ($a+b$):Sum of the 8 known observations $= 2+3+5+10+11+13+15+21 = 80$.$a + b = \sum x_i - 80 = 90 - 80 = 10$*(Note: The variance information is inconsistent with the other given values if $a$ and $b$ are real numbers. We must assume the intended values for $a$ and $b$ are simple integers that satisfy $a+b=10$ and allow the problem to be solved, such as $a=4$ and $b=6$.)*3. List and Sort all 10 observations:Assuming $a=4$ and $b=6$, the 10 observations in ascending order are:$\{2, 3, 4, 5, 6, 10, 11, 13, 15, 21\}$4. Calculate the Median ($M$):Since $N=10$ (even), the median is the average of the $\left(\frac{N}{2}\right)^{\text{th}}$ and $\left(\frac{N}{2} + 1\right)^{\text{th}}$ terms (the 5th and 6th terms).$M = \frac{x_{(5)} + x_{(6)}}{2} = \frac{6 + 10}{2} = 8$5. Calculate the Mean Deviation about the Median ($\text{MD}_{\text{median}}$):$\text{MD}_{\text{median}} = \frac{\Sigma |x_i - 8|}{10}$$\sum|x_i - 8| = |2-8| + |3-8| + |4-8| + |5-8| + |6-8| + |10-8| + |11-8| + |13-8| + |15-8| + |21-8|$$\sum|x_i - 8| = 6 + 5 + 4 + 3 + 2 + 2 + 3 + 5 + 7 + 13$$\sum|x_i - 8| = 50$$\text{MD}_{\text{median}} = \frac{50}{10} = 5$ Answer: C. 5