Math  /  Data & Statistics

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 04990-499 & 5 \\ \hline 500999500-999 & 20 \\ \hline 100014991000-1499 & 38 \\ \hline 150019991500-1999 & 116 \\ \hline 200024992000-2499 & 125 \\ \hline 250029992500-2999 & 83 \\ \hline 300034993000-3499 & 49 \\ \hline 350039993500-3999 & 50 \\ \hline 400044994000-4499 & 30 \\ \hline 450049994500-4999 & 6 \\ \hline \end{tabular}

Studdy Solution

STEP 1

1. The square footage ranges are represented by their midpoints for calculation purposes.
2. The mean and standard deviation are approximated using these midpoints and the frequencies given.
3. The formula for the sample mean (xˉ\bar{x}) and the sample standard deviation (ss) for grouped data are used.

STEP 2

1. Determine the midpoints for each square footage range.
2. Calculate the mean square footage using the midpoints and frequencies.
3. Calculate the standard deviation using the midpoints, frequencies, and the mean.

STEP 3

Determine the midpoints for each square footage range.
\begin{align*} & \text{Midpoint for } 0-499: \frac{0+499}{2} = 249.5 \\ & \text{Midpoint for } 500-999: \frac{500+999}{2} = 749.5 \\ & \text{Midpoint for } 1000-1499: \frac{1000+1499}{2} = 1249.5 \\ & \text{Midpoint for } 1500-1999: \frac{1500+1999}{2} = 1749.5 \\ & \text{Midpoint for } 2000-2499: \frac{2000+2499}{2} = 2249.5 \\ & \text{Midpoint for } 2500-2999: \frac{2500+2999}{2} = 2749.5 \\ & \text{Midpoint for } 3000-3499: \frac{3000+3499}{2} = 3249.5 \\ & \text{Midpoint for } 3500-3999: \frac{3500+3999}{2} = 3749.5 \\ & \text{Midpoint for } 4000-4499: \frac{4000+4499}{2} = 4249.5 \\ & \text{Midpoint for } 4500-4999: \frac{4500+4999}{2} = 4749.5 \\ \end{align*}

STEP 4

Calculate the mean square footage using the midpoints and frequencies. The formula for the mean is:
xˉ=(fixi)fi\bar{x} = \frac{\sum (f_i \cdot x_i)}{\sum f_i}
Where fi f_i is the frequency and xi x_i is the midpoint.
\begin{align*} \bar{x} &= \frac{(5 \cdot 249.5) + (20 \cdot 749.5) + (38 \cdot 1249.5) + (116 \cdot 1749.5) + (125 \cdot 2249.5) + (83 \cdot 2749.5) + (49 \cdot 3249.5) + (50 \cdot 3749.5) + (30 \cdot 4249.5) + (6 \cdot 4749.5)}{522} \\ &= \frac{1247.5 + 14990 + 47481 + 202542 + 281187.5 + 228004.5 + 159225.5 + 187475 + 127485 + 28497}{522} \\ &= \frac{1271135}{522} \\ &\approx 2434.98 \end{align*}

STEP 5

Calculate the standard deviation using the midpoints, frequencies, and the mean. The formula for the sample standard deviation is:
s=fi(xixˉ)2fi1s = \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 2434.98 and the standard deviation is 951.32 951.32 .

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