Answer :
Sure! Let's address each part of the question step-by-step.
### 1.1 Median Height
The median of a dataset is the middle value when the data is arranged in ascending order. Since we have 20 measurements, an even number, the median will be the average of the 10th and 11th values.
Given heights in ascending order:
[tex]\[ 127, 128, 129, 130, 131, 133, 134, 134, 135, 136, 137, 138, 139, 140, 141, 142, 142, 143, 144, 145 \][/tex]
The 10th and 11th values are 136 and 137. Therefore, the median is:
[tex]\[ \text{Median} = \frac{136 + 137}{2} = \frac{273}{2} = 136.5 \][/tex]
### 1.2 Determine:
#### 1.2.1 Mean Height
The mean height is calculated by adding all the heights together and then dividing by the number of observations:
[tex]\[ \text{Mean} = \frac{\sum \text{heights}}{\text{number of heights}} \][/tex]
Summing the heights:
[tex]\[ 127 + 128 + 129 + 130 + 131 + 133 + 134 + 134 + 135 + 136 + 137 + 138 + 139 + 140 + 141 + 142 + 142 + 143 + 144 + 145 = 2744 \][/tex]
Number of heights = 20
Thus,
[tex]\[ \text{Mean} = \frac{2744}{20} = 137.2 \][/tex]
#### 1.2.2 The Range
The range is the difference between the maximum and minimum values in the dataset.
[tex]\[ \text{Range} = \text{Max height} - \text{Min height} \][/tex]
Max height = 145
Min height = 127
So,
[tex]\[ \text{Range} = 145 - 127 = 18 \][/tex]
#### 1.2.3 Interquartile Range (IQR)
The interquartile range is the difference between the third quartile (Q3) and the first quartile (Q1).
First, we find Q1 (25th percentile) and Q3 (75th percentile).
For 20 data points:
[tex]\[ Q1 \text{ position} = 0.25 \times (20 + 1) = 0.25 \times 21 = 5.25 \text{th value} \][/tex]
This means Q1 is slightly more than the 5th value (131), so estimate:
[tex]\[ Q1 \approx 131 + 0.25 \times (133 - 131) = 131 + 0.5 = 131.5 \][/tex]
For Q3:
[tex]\[ Q3 \text{ position} = 0.75 \times (20 + 1) = 0.75 \times 21 = 15.75 \text{th value} \][/tex]
This means Q3 is slightly less than the 16th value (142), so estimate:
[tex]\[ Q3 \approx 142 \][/tex]
Therefore,
[tex]\[ \text{IQR} = Q3 - Q1 = 142 - 131.5 = 10.5 \][/tex]
### 1.3 Draw a Box and Whisker Diagram
To draw a box and whisker diagram, we need the following five-number summary:
1. Minimum value (127)
2. Q1 (131.5)
3. Median (136.5)
4. Q3 (142)
5. Maximum value (145)
Steps to plot the box plot:
1. Draw a number line that covers the range of the data.
2. Draw a box from Q1 to Q3.
3. Inside the box, draw a line at the median.
4. Draw lines (whiskers) from the minimum value to Q1 and from Q3 to the maximum value.
Here's a simple textual representation:
```
Min Q1 Median Q3 Max
| | | | |
127 130 ---- 131.5 ---- 136.5 ---- 142 ---- 145
```
Of course, the actual box plot should be visually plotted on graph paper or using plotting software for accurate representation.
### 1.1 Median Height
The median of a dataset is the middle value when the data is arranged in ascending order. Since we have 20 measurements, an even number, the median will be the average of the 10th and 11th values.
Given heights in ascending order:
[tex]\[ 127, 128, 129, 130, 131, 133, 134, 134, 135, 136, 137, 138, 139, 140, 141, 142, 142, 143, 144, 145 \][/tex]
The 10th and 11th values are 136 and 137. Therefore, the median is:
[tex]\[ \text{Median} = \frac{136 + 137}{2} = \frac{273}{2} = 136.5 \][/tex]
### 1.2 Determine:
#### 1.2.1 Mean Height
The mean height is calculated by adding all the heights together and then dividing by the number of observations:
[tex]\[ \text{Mean} = \frac{\sum \text{heights}}{\text{number of heights}} \][/tex]
Summing the heights:
[tex]\[ 127 + 128 + 129 + 130 + 131 + 133 + 134 + 134 + 135 + 136 + 137 + 138 + 139 + 140 + 141 + 142 + 142 + 143 + 144 + 145 = 2744 \][/tex]
Number of heights = 20
Thus,
[tex]\[ \text{Mean} = \frac{2744}{20} = 137.2 \][/tex]
#### 1.2.2 The Range
The range is the difference between the maximum and minimum values in the dataset.
[tex]\[ \text{Range} = \text{Max height} - \text{Min height} \][/tex]
Max height = 145
Min height = 127
So,
[tex]\[ \text{Range} = 145 - 127 = 18 \][/tex]
#### 1.2.3 Interquartile Range (IQR)
The interquartile range is the difference between the third quartile (Q3) and the first quartile (Q1).
First, we find Q1 (25th percentile) and Q3 (75th percentile).
For 20 data points:
[tex]\[ Q1 \text{ position} = 0.25 \times (20 + 1) = 0.25 \times 21 = 5.25 \text{th value} \][/tex]
This means Q1 is slightly more than the 5th value (131), so estimate:
[tex]\[ Q1 \approx 131 + 0.25 \times (133 - 131) = 131 + 0.5 = 131.5 \][/tex]
For Q3:
[tex]\[ Q3 \text{ position} = 0.75 \times (20 + 1) = 0.75 \times 21 = 15.75 \text{th value} \][/tex]
This means Q3 is slightly less than the 16th value (142), so estimate:
[tex]\[ Q3 \approx 142 \][/tex]
Therefore,
[tex]\[ \text{IQR} = Q3 - Q1 = 142 - 131.5 = 10.5 \][/tex]
### 1.3 Draw a Box and Whisker Diagram
To draw a box and whisker diagram, we need the following five-number summary:
1. Minimum value (127)
2. Q1 (131.5)
3. Median (136.5)
4. Q3 (142)
5. Maximum value (145)
Steps to plot the box plot:
1. Draw a number line that covers the range of the data.
2. Draw a box from Q1 to Q3.
3. Inside the box, draw a line at the median.
4. Draw lines (whiskers) from the minimum value to Q1 and from Q3 to the maximum value.
Here's a simple textual representation:
```
Min Q1 Median Q3 Max
| | | | |
127 130 ---- 131.5 ---- 136.5 ---- 142 ---- 145
```
Of course, the actual box plot should be visually plotted on graph paper or using plotting software for accurate representation.
