Mathematics · Probability

JEE Main 2024 — 29 January, Shift 1 — Question 26

If the mean and variance of the data 65,68,58,4465,68,58,44, 48,45,60,α,β,6048,45,60, \alpha, \beta, 60 where α>β\alpha>\beta are 56 and 66.2 respectively, then α2+β2\alpha^{2}+\beta^{2} is equal to

Answer: 6344

Numerical answer — enter this value.

Step-by-step solution

Sol. x‾=56\overline{\mathrm{x}}=56

σ2=66.2\sigma^{2}=66.2

⇒α2+β2+2567810−(56)2=66.2\Rightarrow \frac{\alpha^{2}+\beta^{2}+25678}{10}-(56)^{2}=66.2

∴α2+β2=6344\therefore \alpha^{2}+\beta^{2}=6344

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Probability
Topic
Mean, variance, expected values of distributions
If the mean and variance of the data 65,68,58,44 , 48,45,60, α, β, 60… | JEE Main 2024 PYQ with Solution · DhiX AI