If mean and variance of following data: 6,10,7,13, a, 12, b, 12 are 9 and frac{37}{4} respectively,

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13.Statistics

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If the mean and variance of the following data:

$6,10,7,13, a, 12, b, 12$ are 9 and $\frac{37}{4}$ respectively, then $(a-b)^{2}$ is equal to:

$12$

$24$

$16$

$32$

(JEE MAIN-2021)

Solution

$\text { Mean }=\frac{6+10+7+13+a+12+b+12}{8}=9$

$60+a+b=72$

$a+b=12$

$\text { veriance }=\frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)=\frac{37}{4}$

$\sum x_{i}^{2}=6^{2}+10^{2}+7^{2}+13^{2}+a^{2}+b^{2}+12^{2}+12^{2}$

$=a^{2}+b^{2}+642$

$\frac{a^{2}+b^{2}+642}{8}-(9)^{2}=\frac{37}{4}$

$\frac{a^{2}+b^{2}}{8}+\frac{321}{4}-81=\frac{37}{4}$

$\frac{a^{2}+b^{2}}{8}=81+\frac{37}{4}-\frac{321}{4}$

$\frac{a^{2}+b^{2}}{8}=81-71$

$\therefore a^{2}+b^{2}=80$

From $(1)$ $a^{2}+b^{2}+2 a b=144$

$80+2 a b=144 \therefore 2 a b=64$

$(a-b)^{2}=a^{2}+b^{2}-2 a b=80-64=16$

Standard 11

Mathematics

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