13.Statistics
medium

જો માહિતી $65,68,58,44,48,45,60, \alpha, \beta, 60$ જ્યાં $\alpha>\beta$ નો મધ્યક અને વિચરણ અનુક્રમે $56$ અને $66.2$ હોય, તો $\alpha^2+\beta^2=$.............................

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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