Question

In an experiment with $250$ trials, the first $150$ trials has mean $16$ and standard deviation $4.$ The overall experiment has mean $15.6$ and variance $\mathrm{13.44.}$The standard deviation of the next $100$ trials of given experiment is

• A. $3$
• B. $4$
• C. $6$
• D. $7$

Answer 1

The correct option is A $3$

Given first $150$ trials with mean $\overline{x}=16$ and standard deviation ${\sigma }_x=4$
overall $250$ trials with mean $\overline{z}=15.6$ and standard deviation ${\sigma }_z^2=13.44$

Let the next $100$ trials has mean $\overline{y}$ and standard deviation ${\sigma }_y$
$\therefore \overline{z}=\frac{\sum x_i+\sum y_i}{250}$
$\because \sum x_i=150\times 16=2400$
$\therefore 15.6=\frac{2400+\sum y_i}{250}\Rightarrow \sum y_i=1500$
Therefore, mean $\overline{y}=\frac{1500}{100}=15$

${\sigma }_x^2=\frac{\sum x_i^2}{150}-(\frac{\sum x_i}{150}{)}^2$
$\Rightarrow 16=\frac{\sum x_i^2}{150}-(\frac{2400}{150}{)}^2$
$\therefore \sum x_i^2=40800$


${\sigma }_z^2=\frac{\sum x_i^2+\sum y_i^2}{250}-(\frac{\sum x_i+\sum y_i}{250}{)}^2$
$\Rightarrow 13.44=\frac{40800+\sum y_i^2}{250}-(\frac{2400+1500}{250}{)}^2$
$\Rightarrow \sum y_i^2=23400$

${\sigma }_y^2=\frac{\sum y_i^2}{100}-(\frac{\sum y_i}{100}{)}^2\Rightarrow {\sigma }_y^2=\frac{23400}{100}-(15{)}^2=9$
Hence the standard deviation ${\sigma }_y=\sqrt{9}=3$