JEE Mains · Maths · STD 11 - 13. statistics
Let \(9 < x_1 < x_2 < \ldots < x_7\) be in an \(A.P.\) with common difference \(d\). If the standard deviation of \(x_1, x_2 \ldots\), \(x _7\) is \(4\) and the mean is \(\overline{ x }\), then \(\overline{ x }+ x _6\) is equal to:
- A \(18\left(1+\frac{1}{\sqrt{3}}\right)\)
- B \(34\)
- C \(2\left(9+\frac{8}{\sqrt{7}}\right)\)
- D \(25\)
Answer & Solution
Correct Answer
(B) \(34\)
Step-by-step Solution
Detailed explanation
\(9=x_1 < x_2 < \ldots \ldots < x_7\) \(9,9+d, 9+2 d, \ldots \ldots .9+6 d\) \(0, d, 2 d, \ldots \ldots \cdot 6\) \(\bar{x}_{\text {new }}=\frac{21 d }{7}=3 d\) \(16=\frac{1}{7}\left(0^2+1^2+\ldots \ldots+6^2\right) d^2-9 d^2\)…
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