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  • Let x1, x2, x3, x4, x5 be the observations with mean m and standard deviation s. The standard deviation of the observations kx1, kx2, kx3, kx4, kx5 is A. k + s

Let x1, x2, x3, x4, x5 be the observations with mean m and standard deviation s.

The standard deviation of the observations kx1, kx2, kx3, kx4, kx5 is
A. k + s

B.\frac{s}{k}

C. ks
D. s

 

Answers (1)

\\ \text{Given observations are }x_{1},~x_{2},~x_{3},~x_{4},~x_{5}~~~ \\ \\ \text{ So the mean of the 5 observations is} \\ \\ m=\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}}{5}=\frac{ \sum x_{i}}{5} \\ \\ \sum x_{i}=5m=x_{1}+x_{2}+x_{3}+x_{4}+x_{5} \\ \\ \sigma =\sqrt {\frac{ \sum \left( x_{i} \right) ^{2}~}{5}- \left( \frac{ \sum x_{i}}{5} \right) ^{2}}=\sqrt {\frac{ \sum \left( x_{i} \right) ^{2}~}{5}-m^{2}} \\ \\ m_{1}=\frac{kx_{1}+kx_{2}+kx_{3}+kx_{4}+kx_{5}}{5}=\frac{k \left( x_{1}+x_{2}+x_{3}+x_{4}+x_{5} \right) }{5}=km \\
\\ \sigma _{1}=\sqrt {\frac{ \sum \left( kx_{i} \right) ^{2}~}{5}- \left( \frac{ \sum kx_{i}}{5} \right) ^{2}} \\ \\ =\sqrt {\frac{ \sum k^{2} \left( x_{i} \right) ^{2}~}{5}- \left( m_{1} \right) ^{2}} \\ \\ =\sqrt {\frac{k^{2} \sum \left( x_{i} \right) ^{2}~}{5}-k^{2}m^{2}} \\ \\ =\sqrt {k^{2} \left( \frac{ \sum \left( x_{i} \right) ^{2}~}{5}-m^{2} \right) } \\ \\ =k\sqrt {\frac{ \sum \left( x_{i} \right) ^{2}~}{5}-m^{2}} \\ \\ \sigma _{1}=k \sigma ~ \\ \\ \text{ Hence, the standard deviation of new set of observations k}x_{1}\text{, k}x_{2}\text{, k}x_{3}\text{, k}x_{4}\text{, k}x_{5} \: \: \:\ \ is\: \: \: \ \ k \sigma \\

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