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  • 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.

1.    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.

Answers (1)

best_answer

Given,

The mean and variance of 8 observations are  9  and  9.25, respectively

Let the remaining two observations be x and y,

Observations: 6, 7, 10, 12, 12, 13, x, y.

∴ Mean, \overline X = \frac{6+ 7+ 10+ 12+ 12+ 13+ x+ y}{8} = 9

60 + x + y = 72

x + y = 12          -(i)

Now, Variance

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

\implies 9.25 = \frac{1}{8}\left[(-3)^2+ (-2)^2+ 1^2+ 3^2+ 4^2+ x^2+ y^2 -18(x+y)+ 2.9^2 \right ]

\implies 9.25 = \frac{1}{8}\left[(-3)^2+ (-2)^2+ 1^2+ 3^2+ 4^2+ x^2+ y^2+ -18(12)+ 2.9^2 \right ]              (Using (i))

\implies 9.25 = \frac{1}{8}\left[48+x^2+ y^2 -216+ 162 \right ] = \frac{1}{8}\left[x^2+ y^2 - 6 \right ]

\implies x^2+ y^2 = 80     -(ii)

Squaring (i), we get

x^2+ y^2 +2xy= 144      (iii)

(iii) - (ii) :

2xy = 64  (iv)

Now, (ii) - (iv):

\\ x^2+ y^2 -2xy= 80-64 \\ \implies (x-y)^2 = 16 \\ \implies x-y = \pm 4         -(v)

Hence, From (i) and (v):

x – y = 4 \implies x = 8 and y = 4

x – y = -4 \implies x = 4 and y = 8

Therefore, The remaining observations are 4 and 8. (in no order)

Posted by

HARSH KANKARIA

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