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 $x_{1}, x_{2}, \dots, x_{18}$ is

Easy

A

$\frac{4}{9}$
B

$\frac{9}{4}$
C

$\frac{3}{2}$
D

None of these

Solution

Let, $d_{i} = x_{i} - 8$
$\begin{array}{l} ∴ σ_{x}^{2} = σ_{d}^{2} = \frac{1}{18} {Σd}_{i}^{2} - {(\frac{1}{8} {Σd}_{i})}^{2} \\ = \frac{1}{18} \times 45 - {(\frac{9}{18})}^{2} = \frac{5}{2} - \frac{1}{4} = \frac{9}{4} \Rightarrow σ_{x} = \frac{3}{2} \end{array}$

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