Statistics Question 9
Question 9 - 01 February - Shift 1
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
(1) 1072
(2) 1792
(3) 1216
(4) 1456
Show Answer
Answer: (1)
Solution:
Formula: Arithmetic mean of individual series (ungrouped data), Variance of individual observations (ungrouped data)
$\begin{aligned} & \frac{1+3+5+a+b}{5}=5 \\ & a+b=16 \ldots \ldots(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 \ldots \ldots(2) \\ & \text { by }(1),(2) \\ & a=7, b=9 \\ & \text { or } a=9, b=7 \end{aligned} $
$\therefore$ The sum of cubes of the remaining two observations is 1072.
