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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), $|x-y|=2$

Std 11

Mathematics

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