Question

The arithmetic mean and standard deviation of a data of nine numbers are $13$ and $5$ respectively. If $3$ is included as the $10$ th item of the data, then the variance of the data of ten number is

• A. 23.5
• B. 21.5
• C. 31.5
• D. 27

Answer 1

Answer: (C)

$\because \frac{\sum _{i=1}^9{x}_i}{9}=13$
$\Rightarrow \sum _{i=1}^9{x}_i=117.$
and ${\sigma }^2=25=\frac{\sum _{i=1}^9{x}_i^2}{9}-(13{)}^2$
$\Rightarrow \sum _{i=1}^y{x}_i^2=9[25+169]$
Now, after including $10^{\text{th }}$ item as ${}^{\prime }{3}^{\prime }$
New mean $\bar{y}=\frac{\sum _{i=1}^9{x}_i+3}{10}=\frac{117+3}{10}=12$
and new variance $=\frac{\left(\sum _{i=1}^9{x}_i^2+9\right)}{10}-(12{)}^2$
$=\frac{(169+25+1)9}{10}-144=\frac{1755-1440}{10}=31.5$