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JEE Main 202623 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2026 · 23 January · Evening Shift · Mathematics · Statistics

If the mean and the variance of the data ( Class 4-8 8-12 12-16...

If the mean and the variance of the data (
Class4-88-1212-1616-20
Frequency3\lambda47
) are μ\mu and 19 respectively, then the value of λ+μ\lambda+\mu is

Options

  1. A21
  2. B19
  3. C20
  4. D18

Correct answer

B. 19

Step-by-step solution

Class midpoints: 6,10,14,186, 10, 14, 18.
Frequencies: 3,λ,4,73, \lambda, 4, 7. N=14+λN = 14 + \lambda.
Mean μ=200+10λ14+λ\mu = \frac{200 + 10\lambda}{14 + \lambda}.
∑fixi2=3(36)+100λ+4(196)+7(324)=3160+100λ\sum f_i x_i^2 = 3(36) + 100\lambda + 4(196) + 7(324) = 3160 + 100\lambda.
Variance: 19=3160+100λN−μ219 = \frac{3160+100\lambda}{N} - \mu^2.
Let u=14+λu = 14+\lambda:
19u2=u(1760+100u)−(60+10u)2⇒19u2−560u+3600=019u^2 = u(1760+100u) - (60+10u)^2 \Rightarrow 19u^2 - 560u + 3600 = 0.
u=560±20038u = \frac{560 \pm 200}{38}, so u=20⇒λ=6u = 20 \Rightarrow \lambda = 6.
μ=26020=13\mu = \frac{260}{20} = 13. Hence λ+μ=6+13=19\lambda + \mu = 6 + 13 = 19.
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