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JEE Mains · Maths · STD 11 - 13. statistics

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वर्ग4-88-1212-1616-20
बारंबारता3\(\lambda\)47
क्रमशः \(\mu\) और 19 हैं, तो \(\lambda+\mu\) का मान ___ है।

  1. A 18
  2. B 21
  3. C 20
  4. D 19
Verified Solution

Answer & Solution

Correct Answer

(D) 19

Step-by-step Solution

Detailed explanation

\(\mu=\frac{\sum f_i x_i}{\sum f_i}=\frac{18+10 \lambda+56+126}{14+\lambda}\) \(=\frac{200+10 \lambda}{\lambda+14}=10+\left(\frac{60}{\lambda+14}\right)\) \(\lambda+14\) is multiply of \(60 \Rightarrow \lambda=1\) or 6 or 16. \(\sigma^2=\frac{\sum x_1^2}{\lambda+14}-(\mu)^2\)…
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