Question

The mean deviation from the median for the following data is

\( x_i \)	2	9	8	3	5	7
\( f_i \)	5	3	1	6	6	1

• A. 2
• B. \( \frac{8}{3} \)
• C. \( \frac{9}{2} \)
• D. 9

Hint:

Remember to sort the data by \( x_i \) values first. Finding the median from unsorted cumulative frequencies will lead to an incorrect answer.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

We need to calculate the Mean Deviation from the Median. This involves sorting the data, finding the cumulative frequency to identify the median, finding the absolute deviation of each observation from the median, and then computing the weighted mean of these deviations.

Step 2: Key Formula or Approach:

\[ \text{Mean Deviation} = \frac{\sum f_i |x_i - \text{Median}|}{\sum f_i} \]

Step 3: Detailed Explanation:

First, arrange the table in ascending order of \( x_i \): \[ \begin{array}{|c|c|c|} \hline x_i & f_i & \text{Cumulative Freq (CF)} \\ \hline 2 & 5 & 5 \\ 3 & 6 & 11 \\ 5 & 6 & 17 \\ 7 & 1 & 18 \\ 8 & 1 & 19 \\ 9 & 3 & 22 \\ \hline \end{array} \] Total observations \( N = \sum f_i = 22 \). Since \( N \) is 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. \( \frac{N}{2} = 11 \), \( \frac{N}{2}+1 = 12 \). From the CF column: - The 11th term is 3. - The 12th term falls in the next class, so it is 5. \[ \text{Median } (M) = \frac{3 + 5}{2} = 4 \] Now, compute \( |x_i - 4| \) and \( f_i |x_i - 4| \): \[ \begin{array}{|c|c|c|c|} \hline x_i & f_i & |x_i - 4| & f_i |x_i - 4| \\ \hline 2 & 5 & 2 & 10 \\ 3 & 6 & 1 & 6 \\ 5 & 6 & 1 & 6 \\ 7 & 1 & 3 & 3 \\ 8 & 1 & 4 & 4 \\ 9 & 3 & 5 & 15 \\ \hline \end{array} \] Sum of deviations: \( \sum f_i |x_i - 4| = 10 + 6 + 6 + 3 + 4 + 15 = 44 \). Calculate Mean Deviation: \[ \text{M.D.} = \frac{44}{22} = 2 \]

Step 4: Final Answer:

The mean deviation is 2.