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Question
The variance of 20 observations is 5. If each observation is multiplied by 2, find the variance of the resulting observations.
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Solution
Let \[x_1 , x_2 , x_3 , . . . , x_{20}\] be the 20 given observations.
\[\text{ Variance} (X) = 5\]
\[\text{ Variance } (X) = {\frac{1}{20}} \times \sum \left( {x_i - X} \right)^2 = 5 (\text{ Here , is the mean of the given observations } . )\]
Let u1,u2,,u3, ..., u20 be the new observations, such that
\[u_i = 2 x_i (\text{ for } i = 1, 2, 3, . . . , 20) . . . (1)\]
\[\text{ Mean } = \bar{U} = \frac{\sum^{20}_{i = 1} u_i}{n} \]
\[ = \frac{\sum^{20}_{i = 1} 2 x_i}{20} \left[ \text{ substituting} u_i \text{ from eq (1) and taking n as } 20 \right]\]
\[ = 2 \times \frac{\sum^{20}_{i = 1}{ x_i} }{20} \]
\[ = 2 \bar{X}\]
\[u_i - \bar{U} = 2 x_i - 2 \bar{X} (\text{ for } i = 1, 2, . . . , 20)\]
\[ = 2\left( x_i - \bar{X} \right) \]
\[ \left( u_i - \bar{U} \right)^2 = \left( 2\left( x_i - \bar{X} \right) \right)^2 \left(\text{ squaring both the sides } \right)\]
\[ = 4 \left( x_i - \bar{X} \right)^2 \]
\[ \therefore \sum^{20}_{i = 1} \left( u_i - \bar{U} \right)^2 = \sum 4^{20}_{i = 1} \left( x_i - \bar{X} \right)^2 \]
\[\frac{\sum^{20}_{i = 1} \left( u_i - \bar{U} \right)^2}{20} = \frac{\sum 4^{20}_{i = 1} \left( x_i - \bar{X} \right)^2}{20}\]
\[ = 4 \frac{\sum^{20}_{i = 1} \left( x_i - \bar{X} \right)^2}{20}\]
\[\text{ Variance } (U) = 4 \times \text{ Variance }(X)\]
\[ = 4 \times 5 \]
\[ = 20\]
Thus, variance of the new observations is 20.
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