JEE Mains · Maths · STD 11 - 13. statistics
Consider a set of \(3 n\) numbers having variance \(4.\) In this set, the mean of first \(2 n\) numbers is \(6\) and the mean of the remaining \(n\) numbers is \(3.\) A new set is constructed by adding \(1\) into each of first \(2 n\) numbers, and subtracting \(1\) from each of the remaining \(n\) numbers. If the variance of the new set is \(k\), then \(9 k\) is equal to .... .
- A \(76\)
- B \(68\)
- C \(82\)
- D \(56\)
Answer & Solution
Correct Answer
(B) \(68\)
Step-by-step Solution
Detailed explanation
Let number be \(a _{1}, a _{2}, a _{3}, \ldots \ldots a _{2 n }, b _{1}, b _{2}, b _{3} \ldots b _{ n }\) \(\sigma^{2}=\frac{\sum a^{2}+\sum b^{2}}{3 n}-(5)^{2}\) \(\Rightarrow \sum a^{2}+\sum b^{2}=87 n\) Now, distribution becomes…
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