JEE Mains · Maths · STD 11 - 13. statistics
Let the mean and the variance of \(5\) observations \(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\) be \(\frac{24}{5}\) and \(\frac{194}{25}\) respectively. If the mean and variance of the first \(4\) observation are \(\frac{7}{2}\) and \(a\) respectively, then \(\left(4 a+x_{5}\right)\) is equal to
- A \(13\)
- B \(15\)
- C \(17\)
- D \(18\)
Answer & Solution
Correct Answer
(B) \(15\)
Step-by-step Solution
Detailed explanation
\(\bar{x}=\frac{\sum x_{i}}{5}=\frac{24}{5} \Rightarrow \sum x_{i}=24\) \(\sigma^{2}=\frac{\sum x_{i}^{2}}{5}-\left(\frac{24}{5}\right)^{2}=\frac{194}{25}\) \(\Rightarrow \sum x_{i}^{2}=154\) \(x_{1}+x_{2}+x_{3}+x_{4}=14\) \(\Rightarrow x_{5}=10\)…
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