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To solve the problem of completing the table with the missing frequencies [tex]\(a\)[/tex] and [tex]\(b\)[/tex] and the percentage [tex]\(c\)[/tex], we need to carefully analyze the information given and ensure everything adds up correctly. Here is our step-by-step solution:
1. Identify the given data:
- Frequency (f) for 0 caries: [tex]\( f_0 = 25 \)[/tex]
- Frequency (f) for 1 caries: [tex]\( f_1 = 20 \)[/tex]
- Frequency (f) for 3 caries: [tex]\( f_3 = 15 \)[/tex]
- Percentages (h) given are:
- [tex]\( h_0 = 0.25 \)[/tex]
- [tex]\( h_1 = 0.20 \)[/tex]
- [tex]\( h_3 = 0.15 \)[/tex]
- [tex]\( h_4 = 0.05 \)[/tex]
2. Find the missing percentage [tex]\(c\)[/tex] (h for 2 caries):
- Since the total percentage must add up to 1, we calculate:
[tex]\[ h_0 + h_1 + h_2 + h_3 + h_4 = 1 \][/tex]
Substituting the given percentages:
[tex]\[ 0.25 + 0.20 + c + 0.15 + 0.05 = 1 \][/tex]
Now, summing the known percentages:
[tex]\[ 0.25 + 0.20 + 0.15 + 0.05 = 0.65 \][/tex]
Therefore:
[tex]\[ c = 1 - 0.65 = 0.35 \][/tex]
3. Compute the missing frequencies:
- Total frequency is given as 100 (since the total should represent all individuals counted):
[tex]\[ f_{\text{total}} = 100 \][/tex]
- First, add the given frequencies to find out the remaining frequency that needs to be split between 2 caries and 4 caries:
[tex]\[ f_{\text{known}} = f_0 + f_1 + f_3 = 25 + 20 + 15 = 60 \][/tex]
- The remaining frequency to distribute is:
[tex]\[ f_{\text{remaining}} = f_{\text{total}} - f_{\text{known}} = 100 - 60 = 40 \][/tex]
- Calculate frequency [tex]\(a\)[/tex] (for 2 caries) using the percentage [tex]\(c\)[/tex]:
[tex]\[ a = h_2 \times f_{\text{total}} = 0.35 \times 100 = 35 \][/tex]
4. Determine frequency [tex]\(b\)[/tex] (for 4 caries):
- Since the remaining frequencies need to sum up to the leftover from the total:
[tex]\[ b = f_{\text{remaining}} - a = 40 - 35 = 5 \][/tex]
From above calculations, the table should be completed as follows:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{N° de caries} & f_1 & h_1 \\ \hline 0 & 25 & 0.25 \\ \hline 1 & 20 & 0.20 \\ \hline 2 & 35 & 0.35 \\ \hline 3 & 15 & 0.15 \\ \hline 4 & 5 & 0.05 \\ \hline \text{total} & 100 & 1 \\ \hline \end{array} \][/tex]
This clearly shows how the frequencies [tex]\(a\)[/tex] and [tex]\(b\)[/tex], and the percentage [tex]\(c\)[/tex] fit into the table, ensuring all the information adds up correctly.
1. Identify the given data:
- Frequency (f) for 0 caries: [tex]\( f_0 = 25 \)[/tex]
- Frequency (f) for 1 caries: [tex]\( f_1 = 20 \)[/tex]
- Frequency (f) for 3 caries: [tex]\( f_3 = 15 \)[/tex]
- Percentages (h) given are:
- [tex]\( h_0 = 0.25 \)[/tex]
- [tex]\( h_1 = 0.20 \)[/tex]
- [tex]\( h_3 = 0.15 \)[/tex]
- [tex]\( h_4 = 0.05 \)[/tex]
2. Find the missing percentage [tex]\(c\)[/tex] (h for 2 caries):
- Since the total percentage must add up to 1, we calculate:
[tex]\[ h_0 + h_1 + h_2 + h_3 + h_4 = 1 \][/tex]
Substituting the given percentages:
[tex]\[ 0.25 + 0.20 + c + 0.15 + 0.05 = 1 \][/tex]
Now, summing the known percentages:
[tex]\[ 0.25 + 0.20 + 0.15 + 0.05 = 0.65 \][/tex]
Therefore:
[tex]\[ c = 1 - 0.65 = 0.35 \][/tex]
3. Compute the missing frequencies:
- Total frequency is given as 100 (since the total should represent all individuals counted):
[tex]\[ f_{\text{total}} = 100 \][/tex]
- First, add the given frequencies to find out the remaining frequency that needs to be split between 2 caries and 4 caries:
[tex]\[ f_{\text{known}} = f_0 + f_1 + f_3 = 25 + 20 + 15 = 60 \][/tex]
- The remaining frequency to distribute is:
[tex]\[ f_{\text{remaining}} = f_{\text{total}} - f_{\text{known}} = 100 - 60 = 40 \][/tex]
- Calculate frequency [tex]\(a\)[/tex] (for 2 caries) using the percentage [tex]\(c\)[/tex]:
[tex]\[ a = h_2 \times f_{\text{total}} = 0.35 \times 100 = 35 \][/tex]
4. Determine frequency [tex]\(b\)[/tex] (for 4 caries):
- Since the remaining frequencies need to sum up to the leftover from the total:
[tex]\[ b = f_{\text{remaining}} - a = 40 - 35 = 5 \][/tex]
From above calculations, the table should be completed as follows:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{N° de caries} & f_1 & h_1 \\ \hline 0 & 25 & 0.25 \\ \hline 1 & 20 & 0.20 \\ \hline 2 & 35 & 0.35 \\ \hline 3 & 15 & 0.15 \\ \hline 4 & 5 & 0.05 \\ \hline \text{total} & 100 & 1 \\ \hline \end{array} \][/tex]
This clearly shows how the frequencies [tex]\(a\)[/tex] and [tex]\(b\)[/tex], and the percentage [tex]\(c\)[/tex] fit into the table, ensuring all the information adds up correctly.
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