Search any question & find its solution
Question:
Answered & Verified by Expert
The standard deviations of two sets of observations $X=\left\{x_i\right\}$ and $Y=\left\{y_i\right\}$ $(i=1,2, \ldots, 100)$ are respectively 5 and 6 . If $\bar{x}, \bar{y}$ are their means and $\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)=600$, then the standard deviation of $Z=\left\{z_i / z_i=x_i-y_i\right)$ is
Options:
Solution:
2495 Upvotes
Verified Answer
The correct answer is:
7
From the given information, we can say that
$\sqrt{\frac{\sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2}{100}}=5 \text { and } \sqrt{\frac{\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2}{100}}=6$
$\therefore \quad \sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2=2500$ and $\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2=3600$
and it is also given that
$\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)=600$
$\because \quad a^2+b^2-2 a b=(a-b)^2$
$\sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2+\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2-\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)$
$=\sum_{i=1}^{100}\left(\left(x_i-\bar{x}\right)-\left(y_i-\bar{y}\right)\right)^2$
$\Rightarrow \quad \sum_{i=1}^{100}\left[\left(x_i-\bar{x}\right)-\left(y_i-\bar{y}\right)\right]^2=2500+3600-1200$
$\Rightarrow \quad \sum_{i=1}^{100}\left[\left(x_i-y_i\right)-(\bar{x}-\bar{y})\right]^2=4900$
$\Rightarrow \sqrt{\frac{\sum_{i=1}^{100}\left[\left(x_i-y_i\right)-(\bar{x}-\bar{y})\right]^2}{100}}=7$
$\sqrt{\frac{\sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2}{100}}=5 \text { and } \sqrt{\frac{\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2}{100}}=6$
$\therefore \quad \sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2=2500$ and $\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2=3600$
and it is also given that
$\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)=600$
$\because \quad a^2+b^2-2 a b=(a-b)^2$
$\sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2+\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2-\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)$
$=\sum_{i=1}^{100}\left(\left(x_i-\bar{x}\right)-\left(y_i-\bar{y}\right)\right)^2$
$\Rightarrow \quad \sum_{i=1}^{100}\left[\left(x_i-\bar{x}\right)-\left(y_i-\bar{y}\right)\right]^2=2500+3600-1200$
$\Rightarrow \quad \sum_{i=1}^{100}\left[\left(x_i-y_i\right)-(\bar{x}-\bar{y})\right]^2=4900$
$\Rightarrow \sqrt{\frac{\sum_{i=1}^{100}\left[\left(x_i-y_i\right)-(\bar{x}-\bar{y})\right]^2}{100}}=7$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.