The arithmetic mean and standard deviation of a data of nine numbers are 13 and 5 respectively. If 3 is…
The arithmetic mean and standard deviation of a data of nine numbers are 13 and 5 respectively. If 3 is included as the 10th item of the data, then the variance of the data of ten number is
23.5
21.5
31.5
27
Solution
$\because \frac{\sum_{i=1}^9 x_i}{9}=13 \Rightarrow \sum_{i=1}^9 x_i=117$.
$\begin{array}{rlrl} & & & \\ \text { and } & & \sigma^2=25=\frac{\sum_{i=1}^9 x_i^2}{9}-(13)^2 \\ \Rightarrow & \quad \sum_{i=1}^9 x_i^2=9[25+169]\end{array}$
Now, after including $10^{\text {th }}$ item as ' 3 '
$
\begin{gathered}
\text { New mean } \bar{y}=\frac{\sum_{i=1}^9 x_i+3}{10}=\frac{117+3}{10}=12 \\
\text { 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
\end{gathered}
$