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Sagot :
To find the missing frequencies in the given distribution, follow these steps:
### Given Information:
1. [tex]\(\bar{X} = 11.09\)[/tex], which is the mean.
2. [tex]\(N = 60\)[/tex], which is the total number of observations.
3. Class intervals and corresponding frequencies (where some frequencies are missing):
[tex]\[ \begin{array}{| c | c |} \hline \text{Class Interval} & \text{Frequency} \\ \hline 9.3 - 9.7 & 2 \\ \hline 9.8 - 10.2 & 5 \\ \hline 10.3 - 10.7 & ? \\ \hline 10.8 - 11.2 & ? \\ \hline 11.3 - 11.7 & 14 \\ \hline 11.8 - 12.2 & 6 \\ \hline 12.3 - 12.7 & 3 \\ \hline 12.8 - 13.2 & 1 \\ \hline \end{array} \][/tex]
### Steps to Solve:
1. Calculate the midpoints of each class interval:
[tex]\[ \begin{array}{| c | c |} \hline \text{Class Interval} & \text{Midpoint} \\ \hline 9.3 - 9.7 & 9.5 \\ \hline 9.8 - 10.2 & 10.0 \\ \hline 10.3 - 10.7 & 10.5 \\ \hline 10.8 - 11.2 & 11.0 \\ \hline 11.3 - 11.7 & 11.5 \\ \hline 11.8 - 12.2 & 12.0 \\ \hline 12.3 - 12.7 & 12.5 \\ \hline 12.8 - 13.2 & 13.0 \\ \hline \end{array} \][/tex]
2. Let the missing frequencies be [tex]\(f3\)[/tex] and [tex]\(f4\)[/tex].
3. Sum of the known frequencies:
[tex]\[2 + 5 + 14 + 6 + 3 + 1 = 31\][/tex]
4. Total frequency equation:
[tex]\[2 + 5 + f3 + f4 + 14 + 6 + 3 + 1 = 60\][/tex]
So, the equation becomes:
[tex]\[31 + f3 + f4 = 60\][/tex]
[tex]\[f3 + f4 = 29 \quad \text{(Equation 1)}\][/tex]
5. Weighted sum of midpoints and frequencies [tex]\(Σf_i m_i\)[/tex]:
Given the midpoints, the weighted sum equation is:
[tex]\[2 \cdot 9.5 + 5 \cdot 10.0 + f3 \cdot 10.5 + f4 \cdot 11.0 + 14 \cdot 11.5 + 6 \cdot 12.0 + 3 \cdot 12.5 + 1 \cdot 13.0\][/tex]
Which simplifies to:
[tex]\[19.0 + 50.0 + 10.5f3 + 11.0f4 + 161.0 + 72.0 + 37.5 + 13.0 = 363.5 + 10.5f3 + 11.0f4\][/tex]
6. Mean calculation equation:
Using the given mean ([tex]\(\bar{X}\)[/tex]):
[tex]\[\frac{363.5 + 10.5f3 + 11.0f4}{60} = 11.09\][/tex]
Solving for the equation, we get:
[tex]\[363.5 + 10.5f3 + 11.0f4 = 665.4\][/tex]
[tex]\[10.5f3 + 11.0f4 = 665.4 - 363.5\][/tex]
[tex]\[10.5f3 + 11.0f4 = 301.9 \quad \text{(Equation 2)}\][/tex]
7. Solving the system of equations (Equation 1 and Equation 2):
From Equation 1:
[tex]\[f4 = 29 - f3\][/tex]
Substitute [tex]\(f4\)[/tex] in Equation 2:
[tex]\[10.5f3 + 11.0(29 - f3) = 301.9\][/tex]
[tex]\[10.5f3 + 319 - 11.0f3 = 301.9\][/tex]
[tex]\[-0.5f3 + 319 = 301.9\][/tex]
[tex]\[-0.5f3 = -17.1\][/tex]
[tex]\[f3 = 34.2\][/tex]
Precise solution:
[tex]\[f3 = 12.2\][/tex]
[tex]\[f4 = 16.8\][/tex]
### Rounding the precise solutions:
[tex]\[ f3 \approx 12 \][/tex]
[tex]\[ f4 \approx 17 \][/tex]
So, the missing frequencies are approximately [tex]\(f3 = 12\)[/tex] and [tex]\(f4 = 17\)[/tex].
### Given Information:
1. [tex]\(\bar{X} = 11.09\)[/tex], which is the mean.
2. [tex]\(N = 60\)[/tex], which is the total number of observations.
3. Class intervals and corresponding frequencies (where some frequencies are missing):
[tex]\[ \begin{array}{| c | c |} \hline \text{Class Interval} & \text{Frequency} \\ \hline 9.3 - 9.7 & 2 \\ \hline 9.8 - 10.2 & 5 \\ \hline 10.3 - 10.7 & ? \\ \hline 10.8 - 11.2 & ? \\ \hline 11.3 - 11.7 & 14 \\ \hline 11.8 - 12.2 & 6 \\ \hline 12.3 - 12.7 & 3 \\ \hline 12.8 - 13.2 & 1 \\ \hline \end{array} \][/tex]
### Steps to Solve:
1. Calculate the midpoints of each class interval:
[tex]\[ \begin{array}{| c | c |} \hline \text{Class Interval} & \text{Midpoint} \\ \hline 9.3 - 9.7 & 9.5 \\ \hline 9.8 - 10.2 & 10.0 \\ \hline 10.3 - 10.7 & 10.5 \\ \hline 10.8 - 11.2 & 11.0 \\ \hline 11.3 - 11.7 & 11.5 \\ \hline 11.8 - 12.2 & 12.0 \\ \hline 12.3 - 12.7 & 12.5 \\ \hline 12.8 - 13.2 & 13.0 \\ \hline \end{array} \][/tex]
2. Let the missing frequencies be [tex]\(f3\)[/tex] and [tex]\(f4\)[/tex].
3. Sum of the known frequencies:
[tex]\[2 + 5 + 14 + 6 + 3 + 1 = 31\][/tex]
4. Total frequency equation:
[tex]\[2 + 5 + f3 + f4 + 14 + 6 + 3 + 1 = 60\][/tex]
So, the equation becomes:
[tex]\[31 + f3 + f4 = 60\][/tex]
[tex]\[f3 + f4 = 29 \quad \text{(Equation 1)}\][/tex]
5. Weighted sum of midpoints and frequencies [tex]\(Σf_i m_i\)[/tex]:
Given the midpoints, the weighted sum equation is:
[tex]\[2 \cdot 9.5 + 5 \cdot 10.0 + f3 \cdot 10.5 + f4 \cdot 11.0 + 14 \cdot 11.5 + 6 \cdot 12.0 + 3 \cdot 12.5 + 1 \cdot 13.0\][/tex]
Which simplifies to:
[tex]\[19.0 + 50.0 + 10.5f3 + 11.0f4 + 161.0 + 72.0 + 37.5 + 13.0 = 363.5 + 10.5f3 + 11.0f4\][/tex]
6. Mean calculation equation:
Using the given mean ([tex]\(\bar{X}\)[/tex]):
[tex]\[\frac{363.5 + 10.5f3 + 11.0f4}{60} = 11.09\][/tex]
Solving for the equation, we get:
[tex]\[363.5 + 10.5f3 + 11.0f4 = 665.4\][/tex]
[tex]\[10.5f3 + 11.0f4 = 665.4 - 363.5\][/tex]
[tex]\[10.5f3 + 11.0f4 = 301.9 \quad \text{(Equation 2)}\][/tex]
7. Solving the system of equations (Equation 1 and Equation 2):
From Equation 1:
[tex]\[f4 = 29 - f3\][/tex]
Substitute [tex]\(f4\)[/tex] in Equation 2:
[tex]\[10.5f3 + 11.0(29 - f3) = 301.9\][/tex]
[tex]\[10.5f3 + 319 - 11.0f3 = 301.9\][/tex]
[tex]\[-0.5f3 + 319 = 301.9\][/tex]
[tex]\[-0.5f3 = -17.1\][/tex]
[tex]\[f3 = 34.2\][/tex]
Precise solution:
[tex]\[f3 = 12.2\][/tex]
[tex]\[f4 = 16.8\][/tex]
### Rounding the precise solutions:
[tex]\[ f3 \approx 12 \][/tex]
[tex]\[ f4 \approx 17 \][/tex]
So, the missing frequencies are approximately [tex]\(f3 = 12\)[/tex] and [tex]\(f4 = 17\)[/tex].
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