JEE Mains · Maths · STD 11 - 13. statistics
An online exam is attempted by \(50\) candidates out of which \(20\) are boys. The average marks obtained by boys is \(12\) with a variance \(2 .\) The variance of marks obtained by \(30\) girls is also \(2 .\) The average marks of all \(50\) candidates is \(15 .\) If \(\mu\) is the average marks of girls and \(\sigma^{2}\) is the variance of marks of \(50\) candidates, then \(\mu+\sigma^{2}\) is equal to ...... .
- A \(125\)
- B \(25\)
- C \(60\)
- D \(40\)
Answer & Solution
Correct Answer
(B) \(25\)
Step-by-step Solution
Detailed explanation
\(\sigma_{b}^{2}=2 \quad\) (variance of boys) \(n_{1}=\) no. of boys \(\bar{x}_{b}=12 \quad\quad\quad\quad\quad\quad\quad\quad n_{2}=\) no. of girls \(\sigma_{\mathrm{g}}^{2}=2\) \(\bar{x}_{g}=\frac{50 \times 15-12 \times \sigma_{b}}{30}=\frac{750-12 \times 20}{30}=17=\mu\)…
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