First slide

Introduction to statistics

Question

$If \sum_{i = 1}^{18} (x_{i} - 8) = 9 and \sum_{i = 1}^{18} {(x_{i} - 8)}^{2} = 45, then the standard deviation of x_{1}, x_{2}, \dots \dots \dots, x_{18} is$

Difficult

A

$\frac{4}{9}$
B

$\frac{9}{4}$
C

$\frac{3}{2}$
D

$\frac{2}{3}$

Solution

[SolutionStep1]:
$Given that \sum_{i = 1}^{18} (x_{i} - 8) = 9 and \sum_{i = 1}^{18} {(x_{i} - 8)}^{2} = 45,$
$Let d_{i} = x_{i} - 8$
$But σ_{x}^{2} = σ_{d}^{2} = \frac{1}{18} \sum d_{i}^{2} - {(\frac{1}{18} \sum d_{i})}^{2}$
$= \frac{1}{18} \times 45 - {(\frac{9}{18})}^{2}$
$= \frac{5}{2} - \frac{1}{4} = \frac{9}{4}$
$Standard deviation, σ_{x} = \frac{3}{2}$

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