The mean of five observations is 5 and their variance is 9.20 . If...

Let, the two unknown terms are $x \text{ and } y .$

The mean is defined as $\mu = \dfrac{∑ {x}_{i}}{N}$

$⇒ 5 = \dfrac{1 + 3 + 8 + x + y}{5}$

$⇒ 25 = 12 + x + y$

$⇒ x + y = 13 . . . \left(1\right)$

And, the variance is defined as ${\sigma}^{2} = \dfrac{∑ {\left({x}_{i} - \mu\right)}^{2}}{N} = \dfrac{∑ {x}_{i}^{2}}{N} - {\mu}^{2}$

$⇒ 9.2 = \dfrac{1 + 9 + 64 + {x}^{2} + {y}^{2}}{5} - 25$

$⇒ 34.2 \times 5 = 74 + {x}^{2} + {y}^{2}$

$⇒ 171 = 74 + {x}^{2} + {y}^{2}$

$⇒ 97 = {x}^{2} + {y}^{2} . . . \left(2\right)$

Put the value of $y$ from equation \left(1\right)$and put in \left(2\right),$ to get

$97 = {x}^{2} + {\left(13 - x\right)}^{2}$

$⇒ 97 = {x}^{2} + 169 - 26x + {x}^{2}$

$⇒ 2{x}^{2} - 26x + 72 = 0$

$⇒ {x}^{2} - 13x + 36 = 0$

$⇒ \left(x - 4\right)\left(x - 9\right) = 0$

$⇒ x = 4$ or $x = 9$

And, $y = 9$ or $y = 4$ respectively.

Hence, the ratio of $x: y$ is $4: 9$ or $9: 4 .$