If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of the first four observations is $\frac{7}{2}$, then the variance of the first four observations in equal to

<p>$\bar{X}=\frac{24}{5} ; \sigma^2=\frac{194}{25}$</p> <p>Let first four observation be $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4$</p> <p>Here, $\frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5}$. ..... (1)</p> <p>Also, $\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4}{4}=\frac{7}{2}$</p> <p>$\Rightarrow \mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4=14$</p> <p>Now from eqn -1</p> <p>$\mathrm{x}_5=10$</p> <p>Now, $\sigma^2=\frac{194}{25}$</p> <p>$$\begin{aligned} & \frac{\mathrm{x}_1^2+\mathrm{x}_2^2+\mathrm{x}_3^2+\mathrm{x}_4^2+\mathrm{x}_5^2}{5}-\frac{576}{25}=\frac{194}{25} \\ & \Rightarrow \mathrm{x}_1^2+\mathrm{x}_2^2+\mathrm{x}_3^2+\mathrm{x}_4^2=54 \end{aligned}$$</p> <p>Now, variance of first 4 observations</p> <p>$$\begin{aligned} \operatorname{Var} & =\frac{\sum_\limits{i=1}^4 x_i^2}{4}-\left(\frac{\sum_\limits{i=1}^4 x_i}{4}\right)^2 \\ & =\frac{54}{4}-\frac{49}{4}=\frac{5}{4} \end{aligned}$$</p>