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Let $X=\{11,12,13, \ldots ., 40,41\}$ and $Y=\{61,62$, $63, \ldots ., 90,91\}$ be the two sets of observations. If $\bar{x}$ and $\bar{y}$ 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 $.................$.
$603$
$604$
$605$
$606$
Solution
$\overline{ x }=\frac{\sum \limits_{ i =11}^{41} i }{31}=\frac{11+41}{2}=26 \quad(31 \text { elements) }$
$\overline{ y }=\frac{\sum \limits_{ j =61}^{91} j }{31}=\frac{61+91}{2}=76 \quad \text { (31 elements) }$
$\text { Combined mean, }$
$\mu =\frac{31 \times 26+31 \times 76}{31+31}$
$=\frac{26+76}{2}=51$
$\sigma^2=\frac{1}{62} \times\left(\sum_{i=1}^{31}\left(x_i-\mu\right)^2+\sum_{i=1}^{31}\left(y_i-\mu\right)^2\right)=705$
Since, $x _{ i } \in X$ are in $A.P.$ with $31$ elements and common difference $1$,same is $y _{ i } \in y$, when written
in increasing order.
$\therefore \sum \limits_{i=1}^{31}\left(x_i-\mu\right)^2=\sum \limits_{i=1}^{31}\left(y_i-\mu\right)^2$
$=10^2+11^2+\ldots . .+40^2$
$=\frac{40 \times 41 \times 81}{6}-\frac{9 \times 10 \times 19}{6}=21855$
$\therefore \left|\bar{x}+\bar{y}-\sigma^2\right|=|26+76-705|=603$
Similar Questions
Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution
| $X_i$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $f_i$ | $k+2$ | $2k$ | $K^{2}-1$ | $K^{2}-1$ | $K^{2}-1$ | $k-3$ |
where $\sum f_i=62$. if $[x]$ denotes the greatest integer $\leq x$, then $\left[\mu^2+\sigma^2\right]$ is equal $………$.
