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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:

  1. A 11
  2. B 9
  3. C 10
  4. D 12
Verified Solution

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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