JEE Main 2023 — Statistics Question with Solution

$$a_1,a_2,a_3,a_4,a_5,a_6$$ are in AP.

Let

$$a_1=a$$

$$a_2=a+d$$

$$a_3=a+2d$$

$$a_4=a+3d$$

$$a_5=a+4d$$

$$a_6=a+5d$$

Now Mean of $$a_1,a_2,a_3,a_4,a_5$$ and $$a_6$$ is

$$=\frac{a_1+a_2+a_3+a_4+a_5+a_6}{6}=\frac{19}{2}$$

$$\Rightarrow \frac{6a+15d}{6}=\frac{19}{2}$$

$$\Rightarrow 2a+5d=19$$ ...... (1)

Also, given,

$$a_1+a_3=10$$

$$\Rightarrow a+a+2d=10$$

$$\Rightarrow 2a+2d=10$$

$$\Rightarrow a+d=5$$ ..... (2)

From equation (1) and (2), we get $$a=2$$ and $$d=3$$

$$\therefore$$ AP is 2, 5, 8, 11, 14, 17

Now, Variance $$({\sigma }^2)=\frac{\sum x_i^2}{6}-{\left(\overline{x}\right)}^2$$

$$=\frac{2^2+5^2+8^2+{11}^2+{14}^2+{17}^2}{6}-{\left(\frac{19}{2}\right)}^2$$

$$=\frac{105}{4}$$

$$\therefore$$ $$8{\sigma }^2=8\times \frac{105}{4}=210$$