JEE Main 2024 — Statistics Question with Solution

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

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

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

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}