| STATEMENT-1: Variance of observations\[2{{x}_{1}},2{{x}_{2}},2{{x}_{3}}\], ....., \[2{{x}_{n}}\] is \[4{{\sigma }^{2}}\]. |
| STATEMENT-2: Arithmetic mean of \[2{{x}_{1}},2{{x}_{2}},2{{x}_{3}}\],......, \[2{{x}_{n}}\] is \[4\overline{x}\]. |
A) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
B) Statement-1 is true, Statement-2 is false.
C) Statement-1 is false, Statement-2 is true.
D) Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation for Statement-1.
Correct Answer: B
Solution :
\[{{\sigma }^{2}}=\sum\limits_{{}}^{{}}{\frac{x_{i}^{2}}{n}-\,{{\left( \sum\limits_{{}}^{{}}{\frac{{{x}_{i}}}{n}} \right)}^{2}}}\] \[\therefore \,\] Variance of \[2{{x}_{1}},\,\,2{{x}_{2}},\,\,2{{x}_{3}},.....2{{x}_{0}}\]. \[=\sum\limits_{{}}^{{}}{\left( \frac{{{(2{{x}_{i}})}^{2}}}{n} \right)\,-{{\left( \sum\limits_{{}}^{{}}{\frac{2{{x}_{i}}}{n}} \right)}^{2}}}\] \[=4\,\left[ \sum\limits_{{}}^{{}}{\frac{2x_{i}^{2}}{n}-\,{{\left( \sum\limits_{{}}^{{}}{\frac{{{x}_{i}}}{n}} \right)}^{2}}} \right]=4{{\sigma }^{2}}\] But arithmetic mean of \[2{{x}_{1}},\,2{{x}_{2}},\,2{{x}_{3}},.....2{{x}_{0}}\] \[=\left[ \frac{2{{x}_{1}}\,+2{{x}_{2}}\,+.......+2{{x}_{0}}}{n} \right]=2\overline{x}\]
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