Given: Mean of 5 observations = 6

and Variance of 5 observations = 4

Let the other two observations be x and y

∴, our observations are 5, 7, 9, x and y

Now, we know that,

⇒ 6 × 5 = 21 + x + y

⇒ 30 – 21 = x + y

⇒ 9 = x + y

or x + y = 9 …(i)

Also,

Variance = 4

So,

⇒ 20 = 11 + (x² + 36 – 12x) + (y² + 36 – 12y)

⇒ 20 – 11 = x² + 36 – 12x + y² + 36 – 12y

⇒ 9 = x² + y² + 72 – 12(x + y)

⇒ x² + y² + 72 – 12(9) – 9 = 0 [from (i)]

⇒ x² + y² + 63 – 108 = 0

⇒ x² + y² – 45 = 0

⇒ x² + y² = 45 …(ii)

From eq. (i)

x + y = 9

Squaring both the sides, we get

(x + y)² = (9)²

⇒ x² + y² + 2xy = 81

⇒ 45 + 2xy = 81 [from (ii)]

⇒ 2xy = 81 – 45

⇒ 2xy = 36

⇒ xy = 18

…(iii)

Putting the value of x in eq. (i), we get

x + y = 9

⇒ y² + 18 = 9y

⇒ y² – 9y +18 = 0

⇒ y² – 6y – 3y + 18 = 0

⇒ y(y – 6) – 3(y – 6)= 0

⇒ (y – 3)(y – 6) = 0

⇒ y – 3 = 0 and y – 6 = 0

⇒ y = 3 and y = 6

For y = 3

Hence, x = 6, y = 3 are the remaining two observations

For y = 6

Hence, x = 3, y = 6 are the remaining two observations

Thus, remaining two observations are 3 and 6.