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 $\overline{X}$ is the mean of the given observations, then we get:
$$\begin{gathered} \Rightarrow \mathrm{Variance}\mathrm{X}=\frac{1}{15}\sum _{i=1}^{15}{\left(x_i-\overline{X}\right)}^2 \\ =4 \end{gathered}$$

Let u_1,u_2,u₃ ... u₁₅ be the new observations such that
$$\begin{gathered} u_{\mathrm{i}}=x_{\mathrm{i}}+9\left(\mathrm{for}i=1,2,3,...,15\right)....\left(1\right) \\ \overline{U}=\frac{1}{n}\sum _{i=1}^{15}{u}_i \\ =\frac{1}{15}\sum _{i=1}^{15}\left(x_i+9\right) \\ =\frac{1}{15}\underset{i=1}{\overset{15}{\sum x_i}}+\frac{9\times 15}{15}\left[\mathrm{as}\sum _{i=1}^{15}9=9\times 15\right] \\ =\overline{\mathrm{X}}+9...\left(2\right) \\ {u}_i-\overline{U}=\left(x_i+9\right)-\left(9+\overline{X}\right)\left[\mathrm{from}\mathrm{eq}\left(1\right)\mathrm{and}\mathrm{eq}\left(2\right)\right] \\ =x_i-\overline{X} \\ \Rightarrow \frac{1}{15}\times {\left(u_i-\overline{U}\right)}^2=\frac{1}{15}{\left(x_i-\overline{X}\right)}^2\left[\mathrm{squaring}\mathrm{both}\text{the}\mathrm{sides}\mathrm{and}\mathrm{then}\mathrm{dividing}\mathrm{by}15\right] \\ \Rightarrow \frac{1}{15}\times \sum _{i=1}^{15}{\left(u_i-\overline{U}\right)}^2=\frac{1}{15}\times \sum _{i=1}^{15}{\left(x_i-\overline{X}\right)}^2 \\ \Rightarrow \frac{1}{15}\times \sum _{i=1}^{15}{\left(u_i-\overline{U}\right)}^2=4 \\ \Rightarrow \mathrm{Variance}\left(U\right)=4 \end{gathered}$$

Thus, variance of the new observation is 4.