The mean and variance of 7 observations are 8 | vedclass

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Question

$\bar{x}=\frac{2+4+10+12+14+x+y}{7}=8$

$x+y=14$

$(\sigma)^{2}=\frac{\sum\left( x _{ i }\right)^{2}}{ n }-\left(\frac{\sum x _{ i }}{ n }\right)^

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\bar{x}=\frac{2+4+10+12+14+x+y}{7}=8$

$x+y=14$

$(\sigma)^{2}=\frac{\sum\left( x _{ i }\right)^{2}}{ n }-\left(\frac{\sum x _{ i }}{ n }\right)^{2}$

$16=\frac{4+16+100+144+196+x^{2}+y^{2}}{7}-8^{2}$

$16+64=\frac{460+x^{2}+y^{2}}{7}$

$560=460+x^{2}+y^{2}$

$x^{2}+y^{2}=100$      (ii)

Clearly by (i) and (ii), $|x-y|=2$

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

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