A) 8
B) 5
C) 1
D) 3
Correct Answer: D
Solution :
Given observations are 4, 7, 2, 8, 6, a and mean is 7. We know Mean \[=\frac{4+7+2+8+6+a}{6}\] \[\Rightarrow \]\[7=\frac{4+7+2+8+6+a}{6}\Rightarrow a=15\] Now, given observations can be written in ascending order which is 2,4,6, 7,8,15 Since, No. of observation is even \[\therefore \]Median \[\frac{\left( \frac{\text{6}}{\text{2}} \right)\text{th}\,\text{observation+}\left( \frac{\text{6}}{\text{2}}\text{+1} \right)\text{th}\,\text{observation}}{\text{2}}\] \[\frac{\text{3rd}\,\text{observation}\,\text{+}\,\text{4th}\,\text{observation}}{\text{2}}\]\[=\frac{6+7}{2}=\frac{13}{2}\] Now, Mean deviation \[=\frac{\sum\limits_{i=1}^{6}{\left| {{x}_{i}}-\frac{13}{2} \right|}}{6}\] \[=\frac{\left| 4-\frac{13}{2} \right|+\left| 7-\frac{13}{2} \right|+\left| 2-\frac{13}{2} \right|+\left| 6-\frac{13}{2} \right|+\left| 15-\frac{13}{2} \right|}{6}\] \[=\frac{\frac{5}{2}+\frac{1}{2}+\frac{9}{2}+\frac{3}{2}+\frac{1}{2}+\frac{17}{2}}{6}=\frac{18}{6}=3\]
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