Question

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

Answer 1

Let the remaining two observations be x and y

The observations are 2, 4, 10, 12, 14, x, y.

$Mean\,\bar{x}\frac{2+4+10+12+14+x+y}{7}=8$

$⇒56=42+x+y$

$⇒x+y=14$…….(1)

$varience=16=\frac{1}{n}\sum_{i=1}^7(x_i-\bar{x})^2$

$16=\frac{1}{7}[(-6)^2+(-4)^2+(2)^1+(4)^2+(6)^2+x^2+y^2-2×8(x+y)+2×(8)^2]$

$16=\frac{1}{7}[36+16+4+16+36+x^2+y^2-16(14)+2(64)].........[Using(1)]$

$16=\frac{1}{7}[108+x^2+y^2-224+128]$

$16=\frac{1}{7}[12+x^2+y^2]$

$⇒x^2+y^2=100$ ……(2)

From (1), we obtain

$x ^2 + y^2 + 2xy = 196 (3)$

From (2) and (3), we obtain

2xy = 196-100

⇒ 2xy = 96-(4)

Subtracting (4) from (2), we obtain

$x^2+ y^2 2xy = 100-96$

$⇒ (x -y)^2 = 4$

$⇒ x-y = ± 2(5)$

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

x = 8 and y = 6 when x-y = 2

x = 6 and y = 8 when x-y = -2

Thus, the remaining observations are 6 and 8.