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Is this the correct calculation of the variance of $\{57; 53; 58; 65; 48; 50; 66; 51\}$?

$$\begin{array}{cc}
\textbf{Variance} & = & \frac{\sum{(X - \bar{X})^2}}{n} \\
&=& \frac{320}{8} \\
&=& 40
\end{array}$$

$$\begin{array}{cc}
\textbf{Standard deviation} & = & \sqrt{\mbox{variance}} \\
&=& \sqrt{\frac{\sum{(X - \bar{X})^2}}{n}} \\
&=& \sqrt{\frac{320}{8}} \\
&=& \sqrt{40} \\
&=& 6.32 \\
\end{array}$$
in Statistics Questions by Diamond (41.5k points) | 46 views

1 Answer

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Mean = (57 + 53 + 58 + 65 + 48 + 50 + 66 + 51) / 8 = 448/8 = 56.

Variance = (Sum((xi - mean)^2)) / (n - 1)

    = (((57-56)^2) + ((53-56)^2) + ((58-56)^2) + ((65-56)^2) + ((48-56)^2) + ((50-56)^2) + ((66-56)^2) + ((51-56)^2))) / (8 - 1)

    = (320/7)

    = 45.7

Standard deviation = √ Variance

    = √ 45.7

    = 6.76

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