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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:
$12$
$24$
$16$
$32$
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 $(1)$ $a^{2}+b^{2}+2 a b=144$
$80+2 a b=144 \therefore 2 a b=64$
$(a-b)^{2}=a^{2}+b^{2}-2 a b=80-64=16$
Similar Questions
Find the mean and variance for the data
${x_i}$ | $92$ | $93$ | $97$ | $98$ | $102$ | $104$ | $109$ |
${f_i}$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
The data is obtained in tabular form as follows.
${x_i}$ | $60$ | $61$ | $62$ | $63$ | $64$ | $65$ | $66$ | $67$ | $68$ |
${f_i}$ | $2$ | $1$ | $12$ | $29$ | $25$ | $12$ | $10$ | $4$ | $5$ |