Question

The mean and variance of $5$ observations are $5$ and $8$ respectively If $3$ observations are $1,3,5$, then the sum of cubes of the remaining two observations is

• A. 1216
• B. 1456
• C. 1072
• D. 1792

Answer 1

The Correct Option is C

$\frac{1+3+5+a+b}{5}=5$
$a+b=16\dots \dots (1)$
${\sigma }^2=\frac{\sum x_1^2}{5}-{\left(\frac{\sum x}{5}\right)}^2$
$8=\frac{1^2+3^2+5^2+a^2+b^2}{5}-25$
$a^2+b^2=130\dots \dots (2)$
$\text{by}(1),(2)$
$a=7,b=9$
$\text{or}a=9,b=7$

Answer 2

Using the mean formula: \[ \frac{1 + 3 + 5 + a + b}{5} = 5 \] \[ a + b = 16 \] Using variance formula: \[ \sigma^2 = \frac{\sum x_i^2}{5} - \left(\frac{\sum x_i}{5}\right)^2 \] \[ 8 = \frac{1^2 + 3^2 + 5^2 + a^2 + b^2}{5} - 25 \] \[ a^2 + b^2 = 130 \] Solving equations: \[ a = 7, b = 9 \quad \text{or} \quad a = 9, b = 7 \] Computing sum of cubes: \[ a^3 + b^3 = 7^3 + 9^3 = 1072 \]