Assume, the two remaining observations are x and yxˉ=72+4+10+12+14+x+y=8
⇒42+x+y=56
⇒x+y=14……(1)
σ2=nΣx12−(nΣx1)2
16=74+16+100+144+196+x2+y2−(8)2
⇒16+64=7460+x2+y2
⇒560=460+x2+y2
⇒x2+y2=100……(2)
⇒xy=48, solving equations (1)&(2)
(x−y)2=(x+y)2−4xy=4
∴∣x−y∣=2