QuestionThe frequency distribution in the table represents the square footage of a random sample of 522 houses that are owner occupied year round. Approximate the mean and standard deviation square footage.
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Housing Data \begin{tabular}{rc} Square Footage & Frequency \\ \hline $0-499$ & 5 \\ \hline $500-999$ & 20 \\ \hline $1000-1499$ & 38 \\ \hline $1500-1999$ & 116 \\ \hline $2000-2499$ & 125 \\ \hline $2500-2999$ & 83 \\ \hline $3000-3499$ & 49 \\ \hline $3500-3999$ & 50 \\ \hline $4000-4499$ & 30 \\ \hline $4500-4999$ & 6 \\ \hline \end{tabular}

Studdy Solution
Calculate the standard deviation using the midpoints, frequencies, and the mean. The formula for the sample standard deviation is:
$s = \sqrt{\frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i - 1}}$
\begin{align*} s &= \sqrt{\frac{5 \cdot (249.5 - 2434.98)^2 + 20 \cdot (749.5 - 2434.98)^2 + 38 \cdot (1249.5 - 2434.98)^2 + 116 \cdot (1749.5 - 2434.98)^2 + 125 \cdot (2249.5 - 2434.98)^2 + 83 \cdot (2749.5 - 2434.98)^2 + 49 \cdot (3249.5 - 2434.98)^2 + 50 \cdot (3749.5 - 2434.98)^2 + 30 \cdot (4249.5 - 2434.98)^2 + 6 \cdot (4749.5 - 2434.98)^2}{522 - 1}} \\ &= \sqrt{\frac{5 \cdot 4785872.65 + 20 \cdot 2845747.20 + 38 \cdot 1404623.90 + 116 \cdot 471732.50 + 125 \cdot 34461.50 + 83 \cdot 99252.00 + 49 \cdot 661469.80 + 50 \cdot 1710333.70 + 30 \cdot 3294197.00 + 6 \cdot 5354865.20}{521}} \\ &= \sqrt{\frac{23929363.25 + 56914944.00 + 53375608.20 + 54721070.00 + 4307687.50 + 8227896.00 + 32411920.20 + 85516685.00 + 98825910.00 + 32129191.20}{521}} \\ &= \sqrt{\frac{470985275.35}{521}} \\ &\approx \sqrt{904013.83} \\ &\approx 951.32 \end{align*}
The approximate mean square footage is $2434.98$ and the standard deviation is $951.32$.