- Home
- Standard 11
- Mathematics
If $\mathop \sum \limits_{i = 1}^9 \left( {{x_i} - 5} \right) = 9$ and $\mathop \sum \limits_{i = 1}^9 {\left( {{x_i} - 5} \right)^2} = 45,$ then the standard deviation of the $9$ items ${x_1},{x_2},\;.\;.\;.\;,{x_9}$ is :
$4$
$2$
$3$
$9$
Solution
Given $\sum\limits_{i = 1}^9 {\left( {{x_i} – 5} \right)} = 9 \Rightarrow \sum\limits_{i = 1}^9 {{x_i} = 54\,\,\,…..\left( i \right)} $
Also, $\sum\limits_{i = 1}^9 {{{\left( {{x_i} – 5} \right)}^2}} = 45$
$\sum\limits_{i = 1}^9 {x_i^2} – 10\sum\limits_{i = 1}^9 {{x_i} + 9\left( {25} \right)} = 45\,\,\,\,\,…\left( {ii} \right)$
From $(i)$ and $(ii)$ we get,
$\sum\limits_{i = 1}^9 {x_i^2} = 360$
Since, variance $ = \frac{{\sum {x_i^2} }}{9} – {\left( {\frac{{\sum {{x_i}} }}{9}} \right)^2}$
$ = \frac{{360}}{9} – {\left( {\frac{{54}}{9}} \right)^2} = 40 – 36 = 4$
Standared deviation $ = \sqrt {Variance} = 2$
Similar Questions
Find the mean and variance for the data
| ${x_i}$ | $92$ | $93$ | $97$ | $98$ | $102$ | $104$ | $109$ |
| ${f_i}$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
