QJEEM24Ma250e: JEE Main 2024 Mathematics - Class 11 JEEM24Ma250e Solution: If the mean and variance of five observations are and…

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

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

$$\begin{array}{rl}\mathrm{Var}& =\frac{\sum _{i=1}^4{x}_i^2}{4}-{\left(\frac{\sum _{i=1}^4{x}_i}{4}\right)}^2\\ & =\frac{54}{4}-\frac{49}{4}=\frac{5}{4}\end{array}$$