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Sagot :
Let's construct the extended frequency distribution and then calculate the mean of the grouped data.
### Step 1: List the Data
The given data set is:
[tex]\[ 1.44, 1.13, 1.44, 1.19, 1.23, 1.46, 1.25, 1.30, 1.35, 1.27, 1.29, 1.40, 1.28, 1.08, 1.47, 1.38, 1.26, 1.22, 1.13, 1.07 \][/tex]
### Step 2: Determine the Intervals
We use class intervals of 0.09 starting from 1.00. The intervals are:
[tex]\[ 1.00-1.09, 1.09-1.18, 1.18-1.27, 1.27-1.36, 1.36-1.45, 1.45-1.54 \][/tex]
### Step 3: Calculate the Frequency Distribution
Count the number of data points falling into each interval:
- [tex]\(1.00-1.09\)[/tex]: 2 (1.08, 1.07)
- [tex]\(1.09-1.18\)[/tex]: 2 (1.13, 1.13)
- [tex]\(1.18-1.27\)[/tex]: 5 (1.19, 1.23, 1.25, 1.26, 1.22)
- [tex]\(1.27-1.36\)[/tex]: 5 (1.30, 1.35, 1.27, 1.29, 1.28)
- [tex]\(1.36-1.45\)[/tex]: 4 (1.40, 1.38, 1.44, 1.44)
- [tex]\(1.45-1.54\)[/tex]: 2 (1.46, 1.47)
### Frequency Distribution Table
[tex]\[ \begin{tabular}{|c|c|} \hline Class Intervals & Frequency, $F$ \\ \hline $1.00-1.09$ & 2 \\ \hline $1.09-1.18$ & 2 \\ \hline $1.18-1.27$ & 5 \\ \hline $1.27-1.36$ & 5 \\ \hline $1.36-1.45$ & 4 \\ \hline $1.45-1.54$ & 2 \\ \hline \end{tabular} \][/tex]
### Step 4: Calculate the Mean of the Grouped Data
First, find the midpoints of each class interval:
- Midpoint of [tex]\(1.00-1.09\)[/tex] = [tex]\(\frac{1.00 + 1.09}{2}=1.045\)[/tex]
- Midpoint of [tex]\(1.09-1.18\)[/tex] = [tex]\(\frac{1.09 + 1.18}{2}=1.135\)[/tex]
- Midpoint of [tex]\(1.18-1.27\)[/tex] = [tex]\(\frac{1.18 + 1.27}{2}=1.225\)[/tex]
- Midpoint of [tex]\(1.27-1.36\)[/tex] = [tex]\(\frac{1.27 + 1.36}{2}=1.315\)[/tex]
- Midpoint of [tex]\(1.36-1.45\)[/tex] = [tex]\(\frac{1.36 + 1.45}{2}=1.405\)[/tex]
- Midpoint of [tex]\(1.45-1.54\)[/tex] = [tex]\(\frac{1.45 + 1.54}{2}=1.495\)[/tex]
Next, multiply each midpoint by the corresponding frequency:
[tex]\[ \text{Sum of } ( \text{Midpoint} \times \text{Frequency}) = 1.045 \times 2 + 1.135 \times 2 + 1.225 \times 5 + 1.315 \times 5 + 1.405 \times 4 + 1.495 \times 2 = 2.09 + 2.27 + 6.125 + 6.575 + 5.62 + 2.99 = 25.67 \][/tex]
Sum of frequencies:
[tex]\[ \text{Total count} = 2 + 2 + 5 + 5 + 4 + 2 = 20 \][/tex]
Mean of the grouped data:
[tex]\[ \text{Mean} = \frac{\sum (\text{Midpoint} \times \text{Frequency})}{\text{Total count}} = \frac{25.67}{20} \approx 1.2835 \][/tex]
Thus, the extended frequency distribution table and mean are:
### Extended Frequency Distribution Table
[tex]\[ \begin{tabular}{|c|c|} \hline Class Intervals & Frequency, $F$ \\ \hline $1.00-1.09$ & 2 \\ \hline $1.09-1.18$ & 2 \\ \hline $1.18-1.27$ & 5 \\ \hline $1.27-1.36$ & 5 \\ \hline $1.36-1.45$ & 4 \\ \hline $1.45-1.54$ & 2 \\ \hline \end{tabular} \][/tex]
### Mean of the Grouped Data
[tex]\[ \text{Mean} \approx 1.2835 \][/tex]
### Step 1: List the Data
The given data set is:
[tex]\[ 1.44, 1.13, 1.44, 1.19, 1.23, 1.46, 1.25, 1.30, 1.35, 1.27, 1.29, 1.40, 1.28, 1.08, 1.47, 1.38, 1.26, 1.22, 1.13, 1.07 \][/tex]
### Step 2: Determine the Intervals
We use class intervals of 0.09 starting from 1.00. The intervals are:
[tex]\[ 1.00-1.09, 1.09-1.18, 1.18-1.27, 1.27-1.36, 1.36-1.45, 1.45-1.54 \][/tex]
### Step 3: Calculate the Frequency Distribution
Count the number of data points falling into each interval:
- [tex]\(1.00-1.09\)[/tex]: 2 (1.08, 1.07)
- [tex]\(1.09-1.18\)[/tex]: 2 (1.13, 1.13)
- [tex]\(1.18-1.27\)[/tex]: 5 (1.19, 1.23, 1.25, 1.26, 1.22)
- [tex]\(1.27-1.36\)[/tex]: 5 (1.30, 1.35, 1.27, 1.29, 1.28)
- [tex]\(1.36-1.45\)[/tex]: 4 (1.40, 1.38, 1.44, 1.44)
- [tex]\(1.45-1.54\)[/tex]: 2 (1.46, 1.47)
### Frequency Distribution Table
[tex]\[ \begin{tabular}{|c|c|} \hline Class Intervals & Frequency, $F$ \\ \hline $1.00-1.09$ & 2 \\ \hline $1.09-1.18$ & 2 \\ \hline $1.18-1.27$ & 5 \\ \hline $1.27-1.36$ & 5 \\ \hline $1.36-1.45$ & 4 \\ \hline $1.45-1.54$ & 2 \\ \hline \end{tabular} \][/tex]
### Step 4: Calculate the Mean of the Grouped Data
First, find the midpoints of each class interval:
- Midpoint of [tex]\(1.00-1.09\)[/tex] = [tex]\(\frac{1.00 + 1.09}{2}=1.045\)[/tex]
- Midpoint of [tex]\(1.09-1.18\)[/tex] = [tex]\(\frac{1.09 + 1.18}{2}=1.135\)[/tex]
- Midpoint of [tex]\(1.18-1.27\)[/tex] = [tex]\(\frac{1.18 + 1.27}{2}=1.225\)[/tex]
- Midpoint of [tex]\(1.27-1.36\)[/tex] = [tex]\(\frac{1.27 + 1.36}{2}=1.315\)[/tex]
- Midpoint of [tex]\(1.36-1.45\)[/tex] = [tex]\(\frac{1.36 + 1.45}{2}=1.405\)[/tex]
- Midpoint of [tex]\(1.45-1.54\)[/tex] = [tex]\(\frac{1.45 + 1.54}{2}=1.495\)[/tex]
Next, multiply each midpoint by the corresponding frequency:
[tex]\[ \text{Sum of } ( \text{Midpoint} \times \text{Frequency}) = 1.045 \times 2 + 1.135 \times 2 + 1.225 \times 5 + 1.315 \times 5 + 1.405 \times 4 + 1.495 \times 2 = 2.09 + 2.27 + 6.125 + 6.575 + 5.62 + 2.99 = 25.67 \][/tex]
Sum of frequencies:
[tex]\[ \text{Total count} = 2 + 2 + 5 + 5 + 4 + 2 = 20 \][/tex]
Mean of the grouped data:
[tex]\[ \text{Mean} = \frac{\sum (\text{Midpoint} \times \text{Frequency})}{\text{Total count}} = \frac{25.67}{20} \approx 1.2835 \][/tex]
Thus, the extended frequency distribution table and mean are:
### Extended Frequency Distribution Table
[tex]\[ \begin{tabular}{|c|c|} \hline Class Intervals & Frequency, $F$ \\ \hline $1.00-1.09$ & 2 \\ \hline $1.09-1.18$ & 2 \\ \hline $1.18-1.27$ & 5 \\ \hline $1.27-1.36$ & 5 \\ \hline $1.36-1.45$ & 4 \\ \hline $1.45-1.54$ & 2 \\ \hline \end{tabular} \][/tex]
### Mean of the Grouped Data
[tex]\[ \text{Mean} \approx 1.2835 \][/tex]
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