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The mean and variance of 5 observations are 3 and 2 respectively. If three of the five observations are 1, 3 and 5, find the values of other two observations. - Mathematics and Statistics

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Sum

The mean and variance of 5 observations are 3 and 2 respectively. If three of the five observations are 1, 3 and 5, find the values of other two observations.

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Solution

`bar"x"` = 3,Var (X) = 2, n = 5, x1 = 1, x2 = 3, x3 = 5 .......[Given]

Let the remaining two observations be x4 and x5.

`bar"x"=(∑"x"_"i")/"n"`

∴ ∑xi = n`bar"x"` = 5 × 3 = 15

Var(X) = `(∑"x"_"i"^2)/"n"-(bar"x")^2`

∴ 2 = `(∑"x"_"i"^2)/5-(3)^2`

∴ `(∑"x"_"i"^2)/5` = 2 + 9

∴ `∑"x"_"i"^2`= 5 × 11

∴ `∑"x"_"i"^2` = 55

Now,

∑xi = 1 + 3 + 5 x4 + x5

∴ 15 = 9 + x4 + x5

∴ x4 + x5 = 15 − 9

∴ x4 + x5 = 6 
∴ x5 = 6 − x4 ...........(i)

∑xi2 = 12 + 32 + 52 + x42 + x52

∴ 55 = 1 + 9 + 25 + x42 + (6 − x4)2 .....[From (i)]

∴ 55 = 35 + x42 + 36 − 12x4 + `"x"_4^2`

∴ 2x42 − 12x4 + 16 = 0

∴ x42 − 6x4 + 8 = 0

∴ x42 − 4x4 − 2x4 + 8 = 0

∴ x4(x4 − 4) −2(x4 − 4) = 0

∴ (x4 − 4)(x4 − 2) = 0

∴ x4 = 4 or x4 = 2

From (i), we get

x5 = 2 or x5 = 4

∴ The two numbers are 2 and 4.

Concept: Measures of Dispersion - Variance and Standard Deviation
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