If standard deviation of 1,2,3,4,⋯,10 is σ, then standard deviation of 11,12,⋯20 is

x¯=1+2+⋯+1010=5.5x¯1=11+12+⋯+2010=15.5σ=(1−5.5)2+(2−5.5)2+⋯+(10−5.5)210,SD=(11−15.5)2+(12−15.5)2+⋯+(20−15.5)210σ=(4.5)2+(3.5)2+⋯+(4.5)210, SD=σ

If standard deviation of 1,2,3,4,⋯,10 is σ, then standard deviation of 11,12,⋯20 is

x¯=1+2+⋯+1010=5.5x¯1=11+12+⋯+2010=15.5σ=(1−5.5)2+(2−5.5)2+⋯+(10−5.5)210,SD=(11−15.5)2+(12−15.5)2+⋯+(20−15.5)210σ=(4.5)2+(3.5)2+⋯+(4.5)210, SD=σ

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If standard deviation of $1, 2, 3, 4, \dots, 10$ is $σ$ , then standard deviation of $11, 12, \dots 20$ is

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$σ + 10$

$10 σ$

$5 σ$

$σ$

answer is C.

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Detailed Solution

$\begin{array}{l} \bar{x} = \frac{1 + 2 + \dots + 10}{10} = 5.5 \\ {\bar{x}}_{1} = \frac{11 + 12 + \dots + 20}{10} = 15.5 \\ σ = \sqrt{\frac{{(1 - 5.5)}^{2} + {(2 - 5.5)}^{2} + \dots + {(10 - 5.5)}^{2}}{10}}, S D = \sqrt{\frac{{(11 - 15.5)}^{2} + {(12 - 15.5)}^{2} + \dots + {(20 - 15.5)}^{2}}{10}} \\ σ = \sqrt{\frac{{(4.5)}^{2} + {(3.5)}^{2} + \dots + {(4.5)}^{2}}{10}}, S D = σ \end{array}$