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

Moderate

A

$4 / 9$
B

$9 / 4$
C

$3 / 2$
D

none of these

Solution

Let $d_{i} = x_{i} - 8$ but
$\begin{array}{l} σ_{x}^{2} = σ_{d}^{2} = \frac{1}{18} \sum d_{i}^{2} - {(\frac{1}{18} \sum d_{i})}^{2} \\ = \frac{1}{18} \times 45 - {(\frac{9}{18})}^{2} = \frac{5}{2} - \frac{1}{4} = \frac{9}{4} \end{array}$
Therefore $σ_{x} = 3 / 2$ .

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