If the mean and variance of five observations are [(24)/(5)] and [(194)/(25)] respectively…

Question

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

Accepted Answer

$$\bar{X}=\frac{24}{5} ; \sigma^2=\frac{194}{25}$$

Let first four observation be $$\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4$$

Here, $$\frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5}$$. ..... (1)

Also, $$\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4}{4}=\frac{7}{2}$$

$$\Rightarrow \mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4=14$$

Now from eqn -1

$$\mathrm{x}_5=10$$

Now, $$\sigma^2=\frac{194}{25}$$

$$\begin{aligned}
& \frac{\mathrm{x}_1^2+\mathrm{x}_2^2+\mathrm{x}_3^2+\mathrm{x}_4^2+\mathrm{x}_5^2}{5}-\frac{576}{25}=\frac{194}{25} \
& \Rightarrow \mathrm{x}_1^2+\mathrm{x}_2^2+\mathrm{x}_3^2+\mathrm{x}_4^2=54
\end{aligned}$$

Now, variance of first 4 observations

$$\begin{aligned}
\operatorname{Var} & =\frac{\sum_\limits{i=1}^4 x_i^2}{4}-\left(\frac{\sum_\limits{i=1}^4 x_i}{4}\right)^2 \
& =\frac{54}{4}-\frac{49}{4}=\frac{5}{4}
\end{aligned}$$

Solutions · 1