## Related questions with answers

The distribution of farm families (as a percent of the population) in a SO-county survey is given in the table.

$\begin{array}{cc} \hline \text { Percent } & \begin{array}{c} \text { Number of } \\ \text { Counties } \end{array} \\ \hline 10-19 & 5 \\ 20-29 & 16 \\ 30-39 & 25 \\ 40-49 & 3 \\ 50-59 & 1 \\ \hline \end{array}$

By using above data solve below statement Determine the standard deviation of the percent of farm families in these $50$ counties.

Solution

VerifiedRecall that Sample Standard Deviation $(s)=\sqrt{\frac{\Sigma(x-\bar{x})^2}{n-1}}$ where $n$ represents the number of scores, $\bar{x}$ is the mean and $x$ is the scores. Using the sample data on the table , we have $\Sigma(x-\bar{x})^2=1088.20$

$\begin{aligned} s&=\sqrt{\frac{\Sigma(x-\bar{x})^2}{n-1}}\\ &=\sqrt{\frac{1088.20}{50-1}} \hspace{1.cm}\text{Substitute the values }\\ &=\sqrt{\frac{1088.20}{49}} \hspace{1.4cm}\text{Subtraction}\\ &=4.71 \hspace{1.8cm}\text{Simplify} \end{aligned}$

Therefore $s=4.71$.

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