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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.
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Solution
\[\text{ Mean } = \bar{X} = 8 \]
\[n = 6 \]
\[\sigma = S . D = 4\]
\[\text{ If } x_{1,} x_{2 . . . .} x_6 \text{ are the given observations } \]
\[ X = \frac{1}{n} \times \sum^6_{i = 1} x_i \]
\[ \Rightarrow 8 = \frac{1}{6} \times \sum^6_{i = 1} x_i \]
\[\text{ Let } u_{1,} u_2 . . . . . u_6 \text{ be the new observations } \]
\[ \Rightarrow u_i = 3 x_i \left( \text{ for } i = 1, 2, 3 . . . 6 \right)\]
\[ \Rightarrow \text{ Mean of new observations } = \bar{U} = \frac{1}{n} \times \sum ^6_{i = 1} u_i \]
\[ = \frac{1}{6} \times \sum ^6 _{i = 1} 3x_i \]
\[ = 3 \times \frac{1}{6} \times \sum^ 6_{i = 1} x_i \]
\[ = 3 \bar{X} \]
\[ = 3 \times 8\]
\[ = 24\]
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.
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