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JEE Mains · Maths · STD 11 - 13. statistics

माना बंटन
\(X_i\) \(0\) \(1\) \(2\) \(3\) \(4\) \(5\)
\(f_i\) \(k+2\) \(2k\) \(K^{2}-1\) \(K^{2}-1\) \(K^{2}-1\) \(k-3\)
जहाँ \(\sum \mathrm{f}_{\mathrm{i}}=62\) है, का माध्य \(\mu\) तथा मानक विचलन \(\sigma\) हैं। यदि \([\mathrm{x}]\) महत्तम पूर्णांक \(\leq \mathrm{x}\) है, तो \(\left[\mu^2+\sigma^2\right]\) बराबर है

  1. A \(8\)
  2. B \(7\)
  3. C \(6\)
  4. D \(9\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(8\)

Step-by-step Solution

Detailed explanation

\(\sum f _{ i }=62\) \(3 k ^2+16 k -12 k -64=0\) \(k =\text { or }-\frac{16}{3}(\text { rejected) }\) \(\mu=\frac{\sum f _{ i } x _{ i }}{\sum f _{ i }}\) \(\mu=\frac{8+2(15)+3(15)+4(17)+5}{62}=\frac{156}{62}\)…
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