Mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations a

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13.Statistics

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The mean and variance of $8$ observations are $10$ and $13.5,$ respectively. If $6$ of these observations are $5,7,10,12,14,15,$ then the absolute difference of the remaining two observations is

A

$7$

B

$3$

C

$5$

D

$9$

(JEE MAIN-2020)

Solution

$\bar{x}=10$

$\Rightarrow \bar{x}=\frac{63+a+b}{8}=10 \Rightarrow a+b=17$

since, variance is independent of origin. So, we subtract 10 from each observation.

$So , \sigma^{2}=13.5=\frac{79+( a -10)^{2}+( b -10)^{2}}{8}-(10-10)^{2}$

$\Rightarrow a ^{2}+ b ^{2}-20( a + b )=-171$

$\Rightarrow a ^{2}+ b ^{2}=169 \quad \ldots(2)$

From

$(i) and (ii)$ $; a=12 \& b=5$

Standard 11

Mathematics

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