First slide

Introduction to statistics

Question

If $\sum_{i = 1}^{9} (x_{i} - 5) = 9$ and $\sum_{i = 1}^{9} {(x_{i} - 5)}^{2} = 45$ , then the standard deviation of the $9$ items $x_{i}, x_{2}, \dots, x_{9}$ is

Moderate

A

4
B

2
C

3
D

9

Solution

We have,
$∴ \sum_{i = 1}^{9} (x_{i} - 5) = 9$ and $\sum_{i = 1}^{9} {(x_{i} - 5)}^{2} = 45$
$\Rightarrow \sum_{i = 1}^{9} x_{i} - 45 = 9$ and $\sum_{i = 1}^{9} ({x_{i}}^{2} - 10 x_{i} + 25) = 45$

$\Rightarrow \sum_{i = 1}^{9} x_{i} = 54 and \sum_{i = 1}^{9} x_{i}^{2} - 10 \sum_{i = 1}^{9} x_{i} + 25 \times 9 = 45$
$\Rightarrow \sum_{i = 1}^{9} x_{i} = 54$ and $\sum_{i = 1}^{9} {x_{i}}^{2} - 10 \times 54 + 225 = 45$
$\Rightarrow \sum_{i = 1}^{9} x_{i} = 54$ and $\sum_{i = 1}^{9} x_{i}^{2} = 360$
$∴$ Variance $= \frac{1}{9} \sum_{i = 1}^{9} x_{i}^{2} - {(\frac{1}{9} \sum_{i = 1}^{9} x_{i})}^{2}$
$\Rightarrow$ Variance $= \frac{360}{9} - {(\frac{54}{9})}^{2} = 40 - 36 = 4$
Hence, standard deviation $= \sqrt{4} = 2 .$

ALITER Let the variable $U$ take values $u_{1}, u_{2}, \dots, u_{9}$ such that $u_{i} = x_{i} - 5, i = 1, 2, \dots, 9 .$ Then
$\bar{U} = \bar{X} - 5$ and $Var (U) = Var (X)$
Now,
$Var (U) = \frac{1}{9} \sum_{i = 1}^{9} u_{i}^{2} - {(\frac{1}{9} \sum_{i = 1}^{9} u_{i})}^{2}$
$= \frac{1}{9} \sum_{i = 1}^{9} {(x_{i} - 5)}^{2} - {\{\frac{1}{9} \sum_{i = 1}^{9} (x_{i} - 5)\}}^{2} = \frac{45}{9} - {(\frac{9}{9})}^{2} = 5 - 1 = 4$
$∴ σ_{U} = \sqrt{Var (U)} = 2$

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