JEE Main 2022 — Statistics Question with Solution

Mean $$(\overline{x})=\frac{x_1+x_2+x_3+x_4+x_5}{5}$$

Given, $$\frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5}$$

$$\Rightarrow x_1+x_2+x_3+x_4+x_5=24$$ ...... (1)

Now, Mean of first 4 observation

$$=\frac{x_1+x_2+x_3+x_4}{4}$$

Given, $$=\frac{x_1+x_2+x_3+x_4}{4}=\frac{7}{2}$$

$$\Rightarrow x_1+x_2+x_3+x_4=14$$ ...... (2)

From equation (1) and (2), we get

$$14+x_5=24$$

$$\Rightarrow x_5=10$$

Now, variance of first 5 observation

$$=\frac{\sum x_i^2}{n}-{\left(\overline{x}\right)}^2$$

$$=\frac{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2}{5}-{\left(\frac{24}{5}\right)}^2$$

Given,

$$\frac{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2}{5}-{\left(\frac{24}{5}\right)}^2=\frac{194}{24}$$

$$\Rightarrow x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=5\left(\frac{194}{25}+\frac{576}{25}\right)$$

$$\Rightarrow x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=154$$

$$\Rightarrow x_1^2+x_2^2+x_3^2+x_4^2+(10{)}^2=154$$

$$\Rightarrow x_1^2+x_2^2+x_3^2+x_4^2=54$$

Now, variance of first 4 observation

$$=\frac{x_1^2+x_2^2+x_3^2+x_4^2}{4}-{\left(\frac{7}{2}\right)}^2$$

Given,

$$\frac{x_1^2+x_2^2+x_3^2+x_4^2}{4}-{\left(\frac{7}{2}\right)}^2=a$$

$$\Rightarrow \frac{54}{4}-\frac{49}{4}=a$$

$$\Rightarrow a=\frac{5}{4}$$

$$\therefore$$ $$4a+x_5$$

$$=4\times \frac{5}{4}+10=15$$