Question

The mean and standard deviation of 6 observations are 8 and 4 respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Answer 1

\[\text{ Mean }  = \bar{X} = 8 \]

\[n = 6 \]

\[\sigma = S . D = 4\]

\[\text{ If }  x_{1,} x_{2 . . . .} x_6 \text{  are the given observations } \]

\[ X = \frac{1}{n} \times \sum^6_{i = 1} x_i \]

\[ \Rightarrow 8 = \frac{1}{6} \times \sum^6_{i = 1} x_i \]

\[\text{ Let }  u_{1,} u_2 . . . . . u_6 \text{ be the new observations } \]

\[ \Rightarrow u_i = 3 x_i \left( \text{ for }  i = 1, 2, 3 . . . 6 \right)\]

\[ \Rightarrow \text{ Mean of new observations } = \bar{U} = \frac{1}{n} \times \sum ^6_{i = 1} u_i \]

\[ = \frac{1}{6} \times \sum ^6 _{i = 1} 3x_i \]

\[ = 3 \times \frac{1}{6} \times \sum^ 6_{i = 1} x_i \]

\[ = 3 \bar{X} \]

\[ = 3 \times 8\]

\[ = 24\]

Thus, mean of the new observations is 24.

\[SD = \sigma_x = 4\]

\[ \sigma_x^2 = \text{ Variance } X\]

\[ \therefore \text{ Variance }  X = 16\]

\[ \Rightarrow \frac{1}{6} \sum^6_{i = 1} \left( x_i - \bar{X} \right)^2 = 16 . . . \left( 1 \right)\]

\[\text{ Variance }  \left( U \right) = \sigma_u^2 = \frac{1}{6} \sum^6_{i = 1} \left( u_i - \bar{U} \right)^2 \]

\[ = \frac{1}{6} \times \sum^6_{i = 1} \left( 3 x_i - 3 \bar{X} \right)^2 \]

\[ = 3^2 \times \frac{1}{6} \sum^6_{i = 1} \left( x_i - \bar{X} \right)^2 \]

\[ = 9 \times 16\]

\[ \sigma_u = \sqrt{ \text{ Variance } \left( U \right)}\]

\[ = \sqrt{9 \times 16}\]

\[ = 12\]

Thus, standard deviation of the new observations is 12.