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.

#### Solution

`bar"x"` = 3,Var (X) = 2, n = 5, x_{1 }= 1, x_{2} = 3, x_{3} = 5 .......[Given]

Let the remaining two observations be x_{4} and x_{5}.

`bar"x"=(∑"x"_"i")/"n"`

∴ ∑x_{i} = 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,

∑x_{i }= 1 + 3 + 5 x_{4 }+ x_{5}

∴ 15 = 9 + x_{4} + x_{5}

∴ x_{4 }+ x_{5} = 15 − 9

∴ x_{4 }+ x_{5} = 6

∴ x_{5 }= 6 − x_{4} ...........(i)

∑x_{i}^{2 }= 1^{2} + 3^{2} + 5^{2} + x_{4}^{2} + x_{5}^{2}

∴ 55 = 1 + 9 + 25 + x_{4}^{2 }+ (6 − x_{4})^{2} .....[From (i)]

∴ 55 = 35 + x_{4}^{2 }+ 36 − 12x_{4} + `"x"_4^2`

∴ 2x_{4}^{2 }− 12x_{4 }+ 16 = 0

∴ x_{4}^{2 }− 6x_{4} + 8 = 0

∴ x_{4}^{2 }− 4x_{4 }− 2x_{4} + 8 = 0

∴ x_{4}(x_{4} − 4) −2(x_{4} − 4) = 0

∴ (x_{4} − 4)(x_{4 }− 2) = 0

∴ x_{4 }= 4 or x_{4} = 2

From (i), we get

x_{5} = 2 or x_{5 }= 4

∴ The two numbers are 2 and 4.