Given: Mean of 5 observations = 4.4

and Variance of 5 observations = 8.24

Let the other two observations be x and y

∴, our observations are 1, 2, 6, x and y

Now, we know that,

⇒ 5 × 4.4 = 9 + x + y

⇒ 22 – 9= x + y

⇒ 13 = x + y

or x + y = 13 …(i)

Also,

Variance = 8.24

So,

⇒ 41.2 = 19.88 + (x² + 19.36 – 8.8x) + (y² + 19.36 – 8.8y)

⇒ 41.2 – 19.88 = x² + 19.36 – 8.8x + y² + 19.36 – 8.8y

⇒ 21.32 = x² + y² + 38.72 – 8.8(x + y)

⇒ x² + y² + 38.72 – 8.8(13) – 21.32 = 0 [from (i)]

⇒ x² + y² + 17.4 – 114.4 = 0

⇒ x² + y² – 97 = 0

⇒ x² + y² = 97 …(ii)

From eq. (i)

x + y = 17.4

Squaring both the sides, we get

(x + y)² = (13)²

⇒ x² + y² + 2xy = 169

⇒ 97 + 2xy = 169 [from (ii)]

⇒ 2xy = 169 – 97

⇒ 2xy = 72

⇒ xy = 36

…(iii)

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

x + y = 13

⇒ y² + 36 = 13y

⇒ y² – 13y + 36 = 0

⇒ y² – 4y – 9y + 36 = 0

⇒ y(y – 4) – 9(y – 4)= 0

⇒ (y – 4)(y – 9) = 0

⇒ y – 4 = 0 and y – 9 = 0

⇒ y = 4 and y = 9

For y = 4

Hence, x = 9, y = 4 are the remaining two observations

For y = 9

Hence, x = 4, y = 9 are the remaining two observations

Thus, remaining two observations are 4 and 9.