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

H Harsh Kankaria

Given,

The mean and variance of 7 observations are 8 and 16, respectively

Let the remaining two observations be x and y,

Observations: 2, 4, 10, 12, 14, x, y

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

42 + x + y = 56

x + y = 14          -(i)

Now, Variance

$\dpi{100} = \frac{1}{n}\sum_{i=1}^8(x_i - \overline x)^2 = 16$

$\dpi{100} \implies 16 = \frac{1}{7}\left[(-6)^2+ (-4)^2+ 2^2+ 4^2+ 6^2+ x^2+ y^2 -16(x+y)+ 2.8^2 \right ]$

$\dpi{100} \implies 16 = \frac{1}{7}\left[36+16+4+16+36+ x^2+ y^2 -16(14)+ 2(64) \right ]$              (Using (i))

$\dpi{100} \implies 16 = \frac{1}{7}\left[108+x^2+ y^2 -96 \right ] = \frac{1}{7}\left[x^2+ y^2 + 12\right ]$

$\dpi{100} \implies x^2+ y^2 = 112- 12 =100$     -(ii)

Squaring (i), we get

$\dpi{100} x^2+ y^2 +2xy= 196$      (iii)

(iii) - (ii) :

2xy = 96  (iv)

Now, (ii) - (iv):

$\dpi{100} \\ x^2+ y^2 -2xy= 100-96 \\ \implies (x-y)^2 = 4 \\ \implies x-y = \pm 2$         -(v)

Hence, From (i) and (v):

x – y = 2 $\dpi{100} \implies$ x = 8 and y = 6

x – y = -2 $\dpi{100} \implies$ x = 6 and y = 8

Therefore, The remaining observations are 6 and 8. (in no order)

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