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Question

The mean and variance of 7 observation are 8 and 16 respectively. If five of the observation are 2,4,10,12 and 14. Find the remaining two observations.

Solution
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Let the remaining two observation be x and y
The observation are 2,4,10,12,14,x,y
Mean =¯x=2+4+10+12+14+x+y7=8
56=42+x+y
x+y=14 ......(1)
Variance =16=1n7i=l(Xi¯X)2
16=17[(6)2+(4)2+(2)2+(4)2+(6)2+x2+y22×8(x+y)+2×(8)2]
16=17[36+16+4+6+36+x2+y216(14)+2(64)] ...........[using (1)]
16=17[108+x2+y2224+128]
16=17[12+x2+y2]
x2+y2=11212=100
x2+y2=100 ..........(2)
From (1), we obtain
x2+y2+2xy=196 ....(3)
From (2) and (3), we obtain
2xy=196100
2xy=96 ...........(4)
subtracting (4) from (2), we obtain
x2+y22xy=10096
(xy)2=4
xy=±2 ............(5)
Therefore, from (1) and (5) we obtain
x=8 and y=6 when xy=2
x=6 and y=8 when xy=2
Thus the remaining observation are 6 and 8

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