Answer :
Sure, let's walk through each of the steps required for the given age distribution data:
### 1. Construct Steam and Leaf Plot
For a stem-and-leaf plot, the "stem" represents the leading digit(s) and the "leaf" represents the trailing digit. The age data are:
[tex]\[ 23, 24, 18, 14, 20, 36, 24, 26, 23, 21, 16, 15, 19, 20, 22, 14, 13, 10, 19, 27, 29, 22, 38, 28, 34, 32, 23, 19, 21, 31, 16, 28, 19, 12, 27, 15, 21, 25, 16 \][/tex]
We sort the data and construct the stem-and-leaf plot as follows:
- 1: 0 2 3 4 4 5 5 6 6 6 8 9 9 9
- 2: 0 0 1 1 1 2 2 3 3 3 4 4 5 6 7 7 8 8 9
- 3: 1 2 4 6 8
This plot gives a visual representation of the dataset.
### 2. Mean for Ungrouped and Grouped Data
Ungrouped Data:
The mean for ungrouped data is calculated by dividing the sum of all age values by the total number of data points.
[tex]\[ \text{Mean}_{\text{ungrouped}} = \frac{\text{Sum of all ages}}{\text{Total number of entries}} \][/tex]
[tex]\[ \text{Mean}_{\text{ungrouped}} = 22.05128205128205 \][/tex]
Grouped Data:
For grouped data, we use the class midpoints and frequencies.
[tex]\[ \text{Mean}_{\text{grouped}} = \frac{\sum(\text{Midpoint} \times \text{Frequency})}{\sum(\text{Frequency})} \][/tex]
[tex]\[ \text{Mean}_{\text{grouped}} = 20.94736842105263 \][/tex]
### 3. Median for Ungrouped and Grouped Data
Ungrouped Data:
For the ungrouped data, the median is the middle value of the sorted dataset. If the number of observations is odd, the median is the middle number. If even, it is the average of the two middle numbers.
Sorted data:
[tex]\[ \{10, 12, 13, 14, 14, 15, 15, 16, 16, 16, 18, 19, 19, 19, 19, 20, 20, 21, 21, 21, 22, 22, 23, 23, 23, 24, 24, 25, 26, 27, 27, 28, 28, 29, 31, 32, 34, 36, 38\} \][/tex]
Since there are 39 numbers, the median is the 20th value,
[tex]\[ \text{Median}_{\text{ungrouped}} = 21 \][/tex]
Grouped Data:
For grouped data, we find the median class. The cumulative frequency that reaches or exceeds N/2 identifies the median class.
Calculations showed that the cumulative frequency equals or exceeds N/2 at the "15-19" group. Median class is 15-19.
Median for grouped data:
[tex]\[ \text{Median}_{\text{grouped}} = L + \left( \frac{\frac{N}{2} - CF}{f} \right) \times C \][/tex]
Where:
- [tex]\(L\)[/tex] is the lower boundary of the median class.
- [tex]\(N\)[/tex] is the total number of frequencies.
- [tex]\(CF\)[/tex] is the cumulative frequency before the median class.
- [tex]\(f\)[/tex] is the frequency of the median class.
- [tex]\(C\)[/tex] is the class interval.
[tex]\[ \text{Median}_{\text{grouped}} = 21.0 \][/tex]
### 4. Mode and Mode Type for Ungrouped Data
The mode is the value that appears most frequently:
For our dataset, the mode is:
[tex]\[ \text{Mode}_{\text{ungrouped}} = 19 \][/tex]
Our dataset has only one mode, so it is unimodal.
### 5. Quartiles for Ungrouped Data
Quartiles divide the data into four equal parts.
- [tex]\(Q_1\)[/tex] (25th percentile)
- [tex]\(Q_2\)[/tex] (50th percentile, which is the median)
- [tex]\(Q_3\)[/tex] (75th percentile)
For our dataset the values are:
[tex]\[ Q_1 = 17.0 \][/tex]
[tex]\[ Q_2 = 21.0 \][/tex]
[tex]\[ Q_3 = 26.5 \][/tex]
### 6. Mean Deviation for Ungrouped Data
The mean deviation is calculated as:
[tex]\[ \text{Mean Deviation} = \frac{\sum |x - \text{Mean}|}{n} \][/tex]
For the ungrouped data:
[tex]\[ \text{Mean Deviation}_{\text{ungrouped}} = 5.2886259040105195 \][/tex]
### 7. Variance for Ungrouped Data
Variance is the average of the squared differences from the Mean.
[tex]\[ \text{Variance}_{\text{ungrouped}} = \frac{\sum (x - \text{Mean})^2}{n-1} \][/tex]
For the ungrouped data:
[tex]\[ \text{Variance}_{\text{ungrouped}} = 44.89203778677463 \][/tex]
These steps detail how to calculate each required measure using the provided data.
### 1. Construct Steam and Leaf Plot
For a stem-and-leaf plot, the "stem" represents the leading digit(s) and the "leaf" represents the trailing digit. The age data are:
[tex]\[ 23, 24, 18, 14, 20, 36, 24, 26, 23, 21, 16, 15, 19, 20, 22, 14, 13, 10, 19, 27, 29, 22, 38, 28, 34, 32, 23, 19, 21, 31, 16, 28, 19, 12, 27, 15, 21, 25, 16 \][/tex]
We sort the data and construct the stem-and-leaf plot as follows:
- 1: 0 2 3 4 4 5 5 6 6 6 8 9 9 9
- 2: 0 0 1 1 1 2 2 3 3 3 4 4 5 6 7 7 8 8 9
- 3: 1 2 4 6 8
This plot gives a visual representation of the dataset.
### 2. Mean for Ungrouped and Grouped Data
Ungrouped Data:
The mean for ungrouped data is calculated by dividing the sum of all age values by the total number of data points.
[tex]\[ \text{Mean}_{\text{ungrouped}} = \frac{\text{Sum of all ages}}{\text{Total number of entries}} \][/tex]
[tex]\[ \text{Mean}_{\text{ungrouped}} = 22.05128205128205 \][/tex]
Grouped Data:
For grouped data, we use the class midpoints and frequencies.
[tex]\[ \text{Mean}_{\text{grouped}} = \frac{\sum(\text{Midpoint} \times \text{Frequency})}{\sum(\text{Frequency})} \][/tex]
[tex]\[ \text{Mean}_{\text{grouped}} = 20.94736842105263 \][/tex]
### 3. Median for Ungrouped and Grouped Data
Ungrouped Data:
For the ungrouped data, the median is the middle value of the sorted dataset. If the number of observations is odd, the median is the middle number. If even, it is the average of the two middle numbers.
Sorted data:
[tex]\[ \{10, 12, 13, 14, 14, 15, 15, 16, 16, 16, 18, 19, 19, 19, 19, 20, 20, 21, 21, 21, 22, 22, 23, 23, 23, 24, 24, 25, 26, 27, 27, 28, 28, 29, 31, 32, 34, 36, 38\} \][/tex]
Since there are 39 numbers, the median is the 20th value,
[tex]\[ \text{Median}_{\text{ungrouped}} = 21 \][/tex]
Grouped Data:
For grouped data, we find the median class. The cumulative frequency that reaches or exceeds N/2 identifies the median class.
Calculations showed that the cumulative frequency equals or exceeds N/2 at the "15-19" group. Median class is 15-19.
Median for grouped data:
[tex]\[ \text{Median}_{\text{grouped}} = L + \left( \frac{\frac{N}{2} - CF}{f} \right) \times C \][/tex]
Where:
- [tex]\(L\)[/tex] is the lower boundary of the median class.
- [tex]\(N\)[/tex] is the total number of frequencies.
- [tex]\(CF\)[/tex] is the cumulative frequency before the median class.
- [tex]\(f\)[/tex] is the frequency of the median class.
- [tex]\(C\)[/tex] is the class interval.
[tex]\[ \text{Median}_{\text{grouped}} = 21.0 \][/tex]
### 4. Mode and Mode Type for Ungrouped Data
The mode is the value that appears most frequently:
For our dataset, the mode is:
[tex]\[ \text{Mode}_{\text{ungrouped}} = 19 \][/tex]
Our dataset has only one mode, so it is unimodal.
### 5. Quartiles for Ungrouped Data
Quartiles divide the data into four equal parts.
- [tex]\(Q_1\)[/tex] (25th percentile)
- [tex]\(Q_2\)[/tex] (50th percentile, which is the median)
- [tex]\(Q_3\)[/tex] (75th percentile)
For our dataset the values are:
[tex]\[ Q_1 = 17.0 \][/tex]
[tex]\[ Q_2 = 21.0 \][/tex]
[tex]\[ Q_3 = 26.5 \][/tex]
### 6. Mean Deviation for Ungrouped Data
The mean deviation is calculated as:
[tex]\[ \text{Mean Deviation} = \frac{\sum |x - \text{Mean}|}{n} \][/tex]
For the ungrouped data:
[tex]\[ \text{Mean Deviation}_{\text{ungrouped}} = 5.2886259040105195 \][/tex]
### 7. Variance for Ungrouped Data
Variance is the average of the squared differences from the Mean.
[tex]\[ \text{Variance}_{\text{ungrouped}} = \frac{\sum (x - \text{Mean})^2}{n-1} \][/tex]
For the ungrouped data:
[tex]\[ \text{Variance}_{\text{ungrouped}} = 44.89203778677463 \][/tex]
These steps detail how to calculate each required measure using the provided data.
