JEE Main 2024 — 29 January, Shift 2 — Question 4

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

Let first four observation be

$\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \mathrm{x}_{4}$

Here, $\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}}{5}=\frac{24}{5} \ldots .$. .(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, $\sigma^{2}=\frac{194}{25}$

$\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}}{5}-\frac{576}{25}=\frac{194}{25}$

$\Rightarrow \mathrm{x}_{1}^{2}+\mathrm{x}_{2}^{2}+\mathrm{x}_{3}^{2}+\mathrm{x}_{4}^{2}=54$

Now, variance of first 4 observations

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