13.Statistics
medium

If the mean and variance of the data $65,68,58,44$, $48,45,60, \alpha, \beta, 60$ where $\alpha>\beta$ are $56$ and $66.2$ respectively, then $\alpha^2+\beta^2$ is equal to

A

$6435$

B

$6798$

C

$6344$

D

$4312$

(JEE MAIN-2024)

Solution

$ \overline{\mathrm{x}}=56 $

$ \sigma^2=66.2 $

$ \Rightarrow \frac{\alpha^2+\beta^2+25678}{10}-(56)^2=66.2 $

$ \therefore \alpha^2+\beta^2=6344$

Standard 11
Mathematics

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