JEE Mains · Maths · STD 11 - 13. statistics
Let the mean and variance of 8 numbers -10, -7, -1, x, y, 9, 2, 16 be \( \frac{7}{2} \) and \( \frac{293}{4} \) respectively. Then the mean of 4 numbers x, y, \( x+y+1 \), \( |x-y| \) is:
- A 11
- B 9
- C 10
- D 12
Answer & Solution
Correct Answer
(A) 11
Step-by-step Solution
Detailed explanation
Mean \(=\frac{-18+x+y+2+9+16}{8}=\frac{7}{3}\) \(=\frac{x+y+9}{8}=\frac{7}{2} \Rightarrow x+y+9=28\) ....(1) \( Variance = \frac{\sum z_{i}^{2}}{8} - (\mu)^{2} = \frac{293}{4} \)…
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