Question

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

 Percent  Number of  Counties 101952029163039254049350591\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 5050 counties.

Solution

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Answered 1 year ago
Answered 1 year ago

Recall that Sample Standard Deviation (s)=Σ(xxˉ)2n1(s)=\sqrt{\frac{\Sigma(x-\bar{x})^2}{n-1}} where nn represents the number of scores, xˉ\bar{x} is the mean and xx is the scores. Using the sample data on the table , we have Σ(xxˉ)2=1088.20\Sigma(x-\bar{x})^2=1088.20

s=Σ(xxˉ)2n1=1088.20501Substitute the values =1088.2049Subtraction=4.71Simplify\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=4.71s=4.71.

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