Question

Let $X=\{11,12,13, \ldots ., 40,41\}$ and $Y=\{61,62$, $63, \ldots \ldots, 90,91\}$ be the two sets of observations. If $\overline{ x }$ and $\overline{ y }$ are their respective means and $\sigma^2$ is the variance of all the observations in $X \cup Y$, then $\left|\overline{ x }+\overline{ y }-\sigma^2\right|$ is equal to

Answer 1

$ \overline{ x }=\frac{\displaystyle\sum_{ i =11}^{41} i }{31}=\frac{11+41}{2}=26 \quad(31 \text { elements }) $
$ \overline{ y }=\frac{\displaystyle\sum_{ j =61}^{91} j }{31}=\frac{61+91}{2}=76 \quad(31 \text { elements) }$
Combined mean, $ \mu=\frac{31 \times 26+31 \times 76}{31+31} $
$=\frac{26+76}{2}=51 $
$ \sigma^2=\frac{1}{62} \times\left(\displaystyle\sum_{i=1}^{31}\left(x_i-\mu\right)^2+\displaystyle\sum_{i=1}^{31}\left(y_i-\mu\right)^2\right)=705 $
Since, $x_i \in X$ are in A.P. with 31 elements & common difference 1 , same is $y _{ i } \in y$, when written in increasing order.
$ \therefore \displaystyle\sum_{ i =1}^{31}\left( x _{ i }-\mu\right)^2=\displaystyle \sum_{ i =1}^{31}\left( y _{ i }-\mu\right)^2 $
$=10^2+11^2+\ldots . .+40^2 $
$ =\frac{40 \times 41 \times 81}{6}-\frac{9 \times 10 \times 19}{6}=21855$
$ \therefore\left|\overline{ x }+\overline{ y }-\sigma^2\right|=|26+76-705|=603$