Certainly! Let's fill in the missing relative frequencies and the total values step-by-step!
Given the data, we have:
- Absolute frequencies for the various values.
- The total count is 25.
We need to fill in the relative frequencies and compute the totals.
Here's the table with the calculated relative frequencies:
\begin{array}{|c|c|c|}
\hline \text{Numero de banios} & \text{Frecuencia absoluta} & \text{Frecuencia relativa} \\
\hline 1 & 2 & \frac{2}{25}=0.08 \\
\hline 2 & 3 & \frac{3}{25}=0.12 \\
\hline 21/2 & 1 & \frac{1}{25}=0.04 \\
\hline 3 & 10 & \frac{10}{25}=0.40 \\
\hline 31/2 & 1 & \frac{1}{25}=0.04 \\
\hline 4 & 3 & \frac{3}{25}=0.12 \\
\hline 41/2 & 1 & \frac{1}{25}=0.04 \\
\hline 5 & 3 & \frac{3}{25}=0.12 \\
\hline 6 & 0 & \frac{0}{25}=0.00 \\
\hline \text{TOTAL} & 25 & 1.00 \\
\hline
\end{array}
Here's the detailed step-by-step process we followed:
1. Calculate relative frequencies:
- For 1 bathroom: [tex]\(\frac{2}{25} = 0.08\)[/tex]
- For 2 bathrooms: [tex]\(\frac{3}{25} = 0.12\)[/tex]
- For [tex]\(\frac{21}{2}\)[/tex] bathrooms: [tex]\(\frac{1}{25} = 0.04\)[/tex]
- For 3 bathrooms: [tex]\(\frac{10}{25} = 0.40\)[/tex]
- For [tex]\(\frac{31}{2}\)[/tex] bathrooms: [tex]\(\frac{1}{25} = 0.04\)[/tex]
- For 4 bathrooms: [tex]\(\frac{3}{25} = 0.12\)[/tex]
- For [tex]\(\frac{41}{2}\)[/tex] bathrooms: [tex]\(\frac{1}{25} = 0.04\)[/tex]
- For 5 bathrooms: [tex]\(\frac{3}{25} = 0.12\)[/tex]
- For 6 bathrooms: [tex]\(\frac{0}{25} = 0.00\)[/tex]
2. Calculate the total absolute frequency:
- The sum of all absolute frequencies is given as 25.
3. Ensure the total of the relative frequencies sums to 1.00:
- [tex]\(0.08 + 0.12 + 0.04 + 0.40 + 0.04 + 0.12 + 0.04 + 0.12 + 0.00 = 1.00\)[/tex]
The final table represents the frequency distribution accurately, with both absolute and relative frequencies correctly calculated and the totals summing to the expected values.
