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13.Statistics
medium
If the mean and the variance of $6,4, a, 8, b, 12,10, 13$ are $9$ and $9.25$ respectively, then $a+b+a b$ is equal to :
A$105$
B$103$
C$100$
D$106$
(JEE MAIN-2025)
Solution
$\because \text { mean }=9$
$\therefore 53+ a + b =72$
$\Rightarrow a + b =19$
$\because \sigma^2=\frac{37}{4} \text { and }(\overline{ X })^2+\sigma^2=\frac{\sum x _1^2}{N}$
$\Rightarrow 81+\frac{37}{4}=\frac{529+ a ^2+ b ^2}{8}$
$\Rightarrow 648+74=529+ a ^2+ b ^2$
$\Rightarrow a ^2+ b ^2=193$
$\because a + b =19 \Rightarrow a ^2+ b ^2+2 ab =361$
$\Rightarrow 2 ab =168$
$\Rightarrow ab =84$
$\therefore a + b + ab =103$
$\therefore 53+ a + b =72$
$\Rightarrow a + b =19$
$\because \sigma^2=\frac{37}{4} \text { and }(\overline{ X })^2+\sigma^2=\frac{\sum x _1^2}{N}$
$\Rightarrow 81+\frac{37}{4}=\frac{529+ a ^2+ b ^2}{8}$
$\Rightarrow 648+74=529+ a ^2+ b ^2$
$\Rightarrow a ^2+ b ^2=193$
$\because a + b =19 \Rightarrow a ^2+ b ^2+2 ab =361$
$\Rightarrow 2 ab =168$
$\Rightarrow ab =84$
$\therefore a + b + ab =103$
Standard 11
Mathematics
