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), $|x-y|=2$

${x_i}$	$4$	$8$	$11$	$17$	$20$	$24$	$32$
${f_i}$	$3$	$5$	$9$	$5$	$4$	$3$	$1$

Class:	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$2$	$3$	$x$	$5$	$4$

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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