Question:
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:
Correct Option: 1
Solution:
$\bar{x}=\frac{2+4+10+12+14+x+y}{7}=8$
$x+y=14$......(i)
$(\sigma)^{2}=\frac{\sum\left(\mathrm{x}_{\mathrm{i}}\right)^{2}}{\mathrm{n}}-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}\right)^{2}$
$16=\frac{4+16+100+144+196+x^{2}+y^{2}}{7}-8^{2}$
$16+64=\frac{460+x^{2}+y^{2}}{7}$
$560=460+x^{2}+y^{2}$
$x^{2}+y^{2}=100$.........(ii)
Clearly by (i) and (ii), $|x-y|=2$
Ans. 1