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JEE Mains · Maths · STD 11 - 13. statistics

Let the mean and variance of the frequency distribution
\(\mathrm{x}\) \(\mathrm{x}_{1}=2\) \(\mathrm{x}_{2}=6\) \(\mathrm{x}_{3}=8\) \(\mathrm{x}_{4}=9\)
\(\mathrm{f}\) \(4\) \(4\) \(\alpha\) \(\beta\)
be \(6\) and \(6.8\) respectively. If \(x_{3}\) is changed from \(8\) to \(7 ,\) then the mean for the new data will be:

  1. A \(\frac{16}{3}\)
  2. B \(4\)
  3. C \(\frac{17}{3}\)
  4. D \(5\)
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Answer & Solution

Correct Answer

(C) \(\frac{17}{3}\)

Step-by-step Solution

Detailed explanation

\(\text { Given } 32+8 \alpha+9 \beta=(8+\alpha+\beta) \times 6\) \(\Rightarrow 2 \alpha+3 \beta=16 \quad \ldots \text { (i) }\) \(\text { Also, } 4 \times 16+4 \times \alpha+9 \beta=(8+\alpha+\beta) \times 6.8\) \(\Rightarrow 640+40 \alpha+90 \beta=544+68 \alpha+68 \beta\)…
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