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13.Statistics
hard
The mean and variance of $7$ observations are $8$ and $16,$ respectively. If five observations are $2, 4, 10,12,14,$ then the absolute difference of the remaining two observations is
A
$2$
B
$4$
C
$3$
D
$1$
(JEE MAIN-2020)
Solution
$\bar{x}=\frac{2+4+10+12+14+x+y}{7}=8$
$x+y=14$
$(\sigma)^{2}=\frac{\sum\left( x _{ i }\right)^{2}}{ n }-\left(\frac{\sum x _{ i }}{ n }\right)^{2}$
$16=\frac{4+16+100+144+196+x^{2}+y^{2}}{7}-8^{2}$
$16+64=\frac{460+x^{2}+y^{2}}{7}$
$560=460+x^{2}+y^{2}$
$x^{2}+y^{2}=100$ (ii)
Clearly by (i) and (ii), $|x-y|=2$
Standard 11
Mathematics
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