JEE Mains · Maths · STD 11 - 13. statistics
Let the six numbers \(a_1, a_2, a_3, a_4, a_5, a_6\) be in \(A.P.\) and \(a_1+a_3=10\). If the mean of these six numbers is \(\frac{19}{2}\) and their variance is \(\sigma^2\), then \(8 \sigma^2\) is equal to
- A \(220\)
- B \(210\)
- C \(200\)
- D \(105\)
Answer & Solution
Correct Answer
(B) \(210\)
Step-by-step Solution
Detailed explanation
\(a_1+a_3=10=a_1+d \Rightarrow 5\) \(a_1+a_2+a_3+a_4+a_5+a_6=57\) \(\Rightarrow \frac{6}{2}\left[a_1+a_6\right]=57\) \(\Rightarrow a_1+a_6=19\) \(\Rightarrow 2 a_1+5 d=19 \text { and } a_1+d=5\) \(\Rightarrow a_1=2, d=3\) \(\text { Numbers }: 2,5,8,11,14,17\)…
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