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Question
The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.
Solution
Let x1,,x2,,x3 , ..., x15 be the given observations.
Variance X is given as 4.
If \[\bar{ X} \] is the mean of the given observations, then we get:
\[\Rightarrow \text{ Variance } X = \frac{1}{15} \sum^{15}_{i = 1} \left( x_i - X \right)^2 \]
\[ = 4\]
Let u1,u2,u3 ... u15 be the new observations such that
\[u_i = x_i + 9 \left( for i = 1, 2 , 3, . . . , 15 \right) . . . . (1) \]
\[ \bar{U} = \frac{1}{n} \sum^{15}_{i = 1} u_i \]
\[ = \frac{1}{15} \sum^{15}_{i = 1} \left( x_i + 9 \right) \]
\[ = \frac{1}{15} \sum^{15}_{i = 1} x_i+ \frac{9 \times 15}{15} \left[ \text{as } \sum^{15}_{i = 1} 9 = 9 \times 15 \right]\]
\[ = X + 9 . . . (2)\]
\[ u_i - \bar{U} = \left( x_i + 9 \right) - \left( 9 + \bar{X} \right) \left[\text{ from eq (1) and eq (2) } \right]\]
\[ = x_i - \bar{X}\]
\[ \Rightarrow \frac{1}{15} \times \left( u_i - \bar{U} \right)^2 = \frac{1}{15} \left( x_i - \bar{X} \right)^2 \left[ \text{ squaring both thesides and then dividing by 15} \right]\]
\[ \Rightarrow \frac{1}{15} \times \sum^{15}_{i = 1} \left( u_i - \bar{U} \right)^2 = \frac{1}{15} \times \sum^{15}_{i = 1} \left( x_i - \bar{X} \right)^2 \]
\[ \Rightarrow \frac{1}{15} \times \sum^{15}_{i = 1} \left( u_i - \bar{U} \right)^2 = 4\]
\[ \Rightarrow \text{ Variance } \left( U \right) = 4 \]
Thus, variance of the new observation is 4.
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