Question

The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Answer 1

Let the remaining two observations be x and y.

Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.

$Mean\,\bar{x}\frac{6+7+10+12+13+x+y}{8}=9$

$⇒60+x+y+72$

$⇒x+y=12$…….(1)

$varience=9.25=\frac{1}{n}\sum_{i=1}^8(x_i-\bar{x})^2$

$9.25=\frac{1}{8}[(-3)^2+(-2)^2+(1)^1+(3)^2+(3)^2+(4)^2+x^2+y^2-2×9(x+y)+2×(9)^2]$

$9.25=\frac{1}{8}[9+4+1+9+9+16+x^2+y^2-18(12)+162].........[Using(1)]$

$9.25=\frac{1}{8}[48+x^2+y^2-216+162]$

$9.25=\frac{1}{8}[x^2+y^2-6]$

$⇒x^2+y^2=80$ ……(2)

From (1), we obtain

$x ^2 + y^2 + 2xy = 144 -(3)$

From (2) and (3), we obtain $2xy = 64 -(4)$

Subtracting (4) from (2), we obtain

$x 2 + y 2 -2xy = 80-64 = 16$

$⇒ x-y = ± 4(5)$

Therefore, from (1) and (5), we obtain

x = 8 and y = 4, when x-y = 4

x = 4 and y = 8, when x-y = 4

Thus, the remaining observations are 4 and 8.