Question

The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.

Answer 1

Let x_1,,x_2,,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 u_1,u_2,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.