Answer :
To answer the given question, we'll perform the statistical analyses step-by-step.
i) Represent in Discrete Series Table
The first step is to create a frequency distribution table for the given weights.
1. List unique weights in ascending order.
2. Count the frequency of each weight.
The weights in the list are:
[tex]\[25, 27, 410, 29, 40, 27, 35, 40, 29, 38, 35, 25, 29, 35, 40, 35, 35, 25, 38, 39, 38, 29, 215, 29, 27, 29, 40, 27, 29, 35, 27, 29, 38, 38, 48, 29, 35, 38, 25, 35, 35, 40, 29, 27, 35, 25, 40, 27, 25, 27\][/tex]
| Weight (kg) | Frequency |
|:-----------:|:---------:|
| 25 | 6 |
| 27 | 7 |
| 29 | 10 |
| 35 | 10 |
| 38 | 6 |
| 39 | 1 |
| 40 | 6 |
| 48 | 1 |
| 215 | 1 |
| 410 | 1 |
ii) Mean
The mean weight can be calculated using the formula:
[tex]\[ \text{Mean} = \frac{\sum x_i}{n} \][/tex]
where \(x_i\) are the individual weights and \(n\) is the total number of weights.
[tex]\[ \text{Mean} = \frac{25 + 27 + 410 + 29 + 40 + 27 + 35 + 40 + \ldots + 25 + 27}{50} = \frac{1636}{50} = 32.72 \][/tex]
iii) Median
The median is the middle value of the data when it is ordered in ascending order. With 50 values, the median is the average of the 25th and 26th values.
Ordered weights:
[tex]\[25, 25, 25, 25, 25, 25, 25, 25, 27, 27, 27, 27, 27, 27, 27, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 35, 35, 35, 35, 35, 35, 35, 35, 35, 35, 38, 38, 38, 38, 38, 38, 39, 40, 40, 40, 40, 40, 40, 48, 215, 410\][/tex]
The 25th and 26th values are both 29.
[tex]\[ \text{Median} = 29 \][/tex]
iv) Mode
The mode is the value that appears most frequently. By inspecting the frequency table, we see that the weights 29 and 35 both have the highest frequency of 10.
[tex]\[ \text{Mode} = 29 \text{ and } 35 \][/tex]
v) Upper Quartile (75th percentile)
The upper quartile is the 75th percentile. For \(n = 50\), the ( \( \frac{3n+1}{4} \) )th value is the 37.75th value, so we take the average of the 37th and 38th values.
The ordered list:
[tex]\[ \ldots, 35, 35, 35, 35, 35, 35, 35, 35, 38, 38, 38 \ldots\][/tex]
The 37th and 38th values are both 38.
[tex]\[ \text{Upper Quartile} = 38 \][/tex]
vi) Lower Quartile (25th percentile)
The lower quartile is the 25th percentile. For \(n = 50\), the ( \(\frac{n+1}{4}\) )th value is the 12.75th value, so we take the average of the 12th and 13th values.
The ordered list:
[tex]\[ \ldots, 27, 27, 27, 27, 27, 27, 27, 29, 29 \ldots\][/tex]
The 12th and 13th values are both 27.
[tex]\[ \text{Lower Quartile} = 27 \][/tex]
To summarize:
1. Discrete Series Table:
[tex]\[ \begin{array}{|c|c|} \hline \text{Weight (kg)} & \text{Frequency} \\ \hline 25 & 6 \\ 27 & 7 \\ 29 & 10 \\ 35 & 10 \\ 38 & 6 \\ 39 & 1 \\ 40 & 6 \\ 48 & 1 \\ 215 & 1 \\ 410 & 1 \\ \hline \end{array} \][/tex]
2. Mean: \(32.72\)
3. Median: \(29\)
4. Mode: \(29 \text{ and } 35\)
5. Upper Quartile: \(38\)
6. Lower Quartile: [tex]\(27\)[/tex]
i) Represent in Discrete Series Table
The first step is to create a frequency distribution table for the given weights.
1. List unique weights in ascending order.
2. Count the frequency of each weight.
The weights in the list are:
[tex]\[25, 27, 410, 29, 40, 27, 35, 40, 29, 38, 35, 25, 29, 35, 40, 35, 35, 25, 38, 39, 38, 29, 215, 29, 27, 29, 40, 27, 29, 35, 27, 29, 38, 38, 48, 29, 35, 38, 25, 35, 35, 40, 29, 27, 35, 25, 40, 27, 25, 27\][/tex]
| Weight (kg) | Frequency |
|:-----------:|:---------:|
| 25 | 6 |
| 27 | 7 |
| 29 | 10 |
| 35 | 10 |
| 38 | 6 |
| 39 | 1 |
| 40 | 6 |
| 48 | 1 |
| 215 | 1 |
| 410 | 1 |
ii) Mean
The mean weight can be calculated using the formula:
[tex]\[ \text{Mean} = \frac{\sum x_i}{n} \][/tex]
where \(x_i\) are the individual weights and \(n\) is the total number of weights.
[tex]\[ \text{Mean} = \frac{25 + 27 + 410 + 29 + 40 + 27 + 35 + 40 + \ldots + 25 + 27}{50} = \frac{1636}{50} = 32.72 \][/tex]
iii) Median
The median is the middle value of the data when it is ordered in ascending order. With 50 values, the median is the average of the 25th and 26th values.
Ordered weights:
[tex]\[25, 25, 25, 25, 25, 25, 25, 25, 27, 27, 27, 27, 27, 27, 27, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 35, 35, 35, 35, 35, 35, 35, 35, 35, 35, 38, 38, 38, 38, 38, 38, 39, 40, 40, 40, 40, 40, 40, 48, 215, 410\][/tex]
The 25th and 26th values are both 29.
[tex]\[ \text{Median} = 29 \][/tex]
iv) Mode
The mode is the value that appears most frequently. By inspecting the frequency table, we see that the weights 29 and 35 both have the highest frequency of 10.
[tex]\[ \text{Mode} = 29 \text{ and } 35 \][/tex]
v) Upper Quartile (75th percentile)
The upper quartile is the 75th percentile. For \(n = 50\), the ( \( \frac{3n+1}{4} \) )th value is the 37.75th value, so we take the average of the 37th and 38th values.
The ordered list:
[tex]\[ \ldots, 35, 35, 35, 35, 35, 35, 35, 35, 38, 38, 38 \ldots\][/tex]
The 37th and 38th values are both 38.
[tex]\[ \text{Upper Quartile} = 38 \][/tex]
vi) Lower Quartile (25th percentile)
The lower quartile is the 25th percentile. For \(n = 50\), the ( \(\frac{n+1}{4}\) )th value is the 12.75th value, so we take the average of the 12th and 13th values.
The ordered list:
[tex]\[ \ldots, 27, 27, 27, 27, 27, 27, 27, 29, 29 \ldots\][/tex]
The 12th and 13th values are both 27.
[tex]\[ \text{Lower Quartile} = 27 \][/tex]
To summarize:
1. Discrete Series Table:
[tex]\[ \begin{array}{|c|c|} \hline \text{Weight (kg)} & \text{Frequency} \\ \hline 25 & 6 \\ 27 & 7 \\ 29 & 10 \\ 35 & 10 \\ 38 & 6 \\ 39 & 1 \\ 40 & 6 \\ 48 & 1 \\ 215 & 1 \\ 410 & 1 \\ \hline \end{array} \][/tex]
2. Mean: \(32.72\)
3. Median: \(29\)
4. Mode: \(29 \text{ and } 35\)
5. Upper Quartile: \(38\)
6. Lower Quartile: [tex]\(27\)[/tex]
