JEE MAIN - Mathematics (2023 - 1st February Evening Shift - No. 1)
Let $$9=x_{1} < x_{2} < \ldots < x_{7}$$ be in an A.P. with common difference d. If the standard deviation of $$x_{1}, x_{2}..., x_{7}$$ is 4 and the mean is $$\bar{x}$$, then $$\bar{x}+x_{6}$$ is equal to :
$$2\left(9+\frac{8}{\sqrt{7}}\right)$$
25
$$18\left(1+\frac{1}{\sqrt{3}}\right)$$
34
Explanation
$\begin{aligned} & \text { Mean } \Rightarrow \bar{x}=\frac{\sum\limits_{i=1}^7 x_i}{7}=\frac{\frac{7}{2}[2 a+6 d]}{7}=a+3 d=x_4 \\\\ & \text { Variance }=\frac{\sum\limits_{i=1}^7\left(x_i-\bar{x}\right)^2}{7}=(4)^2 \Rightarrow \frac{\sum\limits_{i=1}^7\left(x_i-x_4\right)^2}{7}=16 \\\\ & \Rightarrow \frac{(3 d)^2+(2 d)^2+d^2+0+d^2+(2 d)^2+(3 d)^2}{7}=16 \\\\ & =4 d^2=16 \Rightarrow d=2 \\\\ & \Rightarrow \bar{x}=9+3(2)=15 \\\\ & x_6=a+5 d=9+5(2)=19 \Rightarrow \bar{x}+x_6=34\end{aligned}$
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