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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.
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Solution
Let those two numbers be x and y.
`overline x = 8 = (2 + 4 + 10 + 12 + 14 + x + y)/7`
or 56 = 42 + x + y or x + y = 56 − 42 = 14 ...(i)
σ2 = `1/n^2 [nsumx_i^2 - (sumx_i)^2]`
`[overline x = (sumx_i)/n ∴ sumx_i = n overline x = 7 xx 8 = 56]`
`σ^2 = 16 = 1/49 [7 xx sumx_i^2 - (56)^2]`
∴ `7sumx_i^2` = 49 × 16 + 56 × 56
or `sumx_i^2` = 7 × 16 + 8 × 56
= 560
or 22 + 42 + 102 + 122 + 142 + x2 + y2
= 560
460 + x2 + y2 = 560
x2 + y2 = 560 – 460 = 100 ...(ii)
From equations (i) and (ii),
x2 + (14 – x)2 = 100
or 2x2 – 28x + 196 – 100 = 0
or x2 – 14x + 48 = 0
∴ (x – 6)(x – 8) = 0
∴ x = 6 or 8
∴ y = 8 or 6
∴ Those two numbers are 6 and 8.
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