Get the information you need with the help of IDNLearn.com's extensive Q&A platform. Discover trustworthy solutions to your questions quickly and accurately with help from our dedicated community of experts.
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]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for trusting IDNLearn.com with your questions. Visit us again for clear, concise, and accurate answers.
