A) 6
B) 7
C) 8
D) None of these
Correct Answer: C
Solution :
[c] Here, \[\bar{x}=\frac{2+4+6+8+10}{5}=6\] Hence, variance\[=\frac{1}{n}\sum{{{({{x}_{i}}-\bar{x})}^{2}}}\] \[=\frac{1}{5}\left\{ {{(2-6)}^{2}}+{{(4-6)}^{2}}+(6-6)+{{(8-6)}^{2}}+{{(10-6)}^{2}} \right\}\] \[=\frac{1}{5}\left\{ (16+4+0+4+16) \right\}=\frac{1}{5}\{40\}=8\]
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