Question

Mean deviation from the mean for the ungrouped data \[ 8,\,7,\,15,\,12,\,12,\,60,\,15,\,6,\,65,\,4,\,6,\,18 \] is

• A. \(14.5\)
• B. \(11.16\)
• C. \(12.6\)
• D. \(13.4\)

Hint:

For ungrouped data: \[ \text{Mean Deviation about Mean} = \frac{\sum |x-\bar x|}{n}. \] Always take absolute values before adding deviations; otherwise positive and negative deviations cancel out.

Answer 1

The Correct Option is A

Concept: The mean deviation about the mean is defined as \[ \text{M.D.} = \frac{\sum |x_i-\bar x|}{n}, \] where \[ \bar x = \frac{\sum x_i}{n} \] is the arithmetic mean and \(n\) is the total number of observations. The procedure consists of:
• Finding the arithmetic mean.
• Computing each absolute deviation from the mean.
• Adding all absolute deviations.
• Dividing by the number of observations.

Step 1: Calculate the arithmetic mean. Given observations: \[ 8,7,15,12,12,60,15,6,65,4,6,18. \] Number of observations: \[ n=12. \] Sum of observations: \[ 8+7+15+12+12+60+15+6+65+4+6+18 = 228. \] Hence \[ \bar x = \frac{228}{12} = 19. \]

Step 2: Find absolute deviations from the mean. \[ |8-19|=11 \] \[ |7-19|=12 \] \[ |15-19|=4 \] \[ |12-19|=7 \] \[ |12-19|=7 \] \[ |60-19|=41 \] \[ |15-19|=4 \] \[ |6-19|=13 \] \[ |65-19|=46 \] \[ |4-19|=15 \] \[ |6-19|=13 \] \[ |18-19|=1. \]

Step 3: Add all absolute deviations. \[ 11+12+4+7+7+41+4+13+46+15+13+1 = 174. \] Thus \[ \sum |x_i-\bar x| = 174. \]

Step 4: Compute the mean deviation. \[ \text{M.D.} = \frac{174}{12}. \] \[ = 14.5. \]

Step 5: Verify the result. The large values \(60\) and \(65\) contribute significantly to the deviations, increasing the overall mean deviation. The calculation is therefore consistent with the answer being substantially larger than \(10\). \[ \text{Mean Deviation} = 14.5. \] Conclusion: \[ \boxed{14.5} \]