Answer :
Let's go through each part of the question step by step.
### (i) Draw the cumulative frequency table
The cumulative frequency table is built by progressively adding the frequencies of the students. Here's how it is done for each class interval:
- For the interval [tex]\(10-20\)[/tex], the cumulative frequency is [tex]\(7\)[/tex].
- For the interval [tex]\(20-30\)[/tex], the cumulative frequency is [tex]\(7 + 9 = 16\)[/tex].
- For the interval [tex]\(30-40\)[/tex], the cumulative frequency is [tex]\(16 + 0 = 16\)[/tex], assuming [tex]\(p\)[/tex] is initially [tex]\(0\)[/tex].
- For the interval [tex]\(40-50\)[/tex], the cumulative frequency is [tex]\(16 + 8 = 24\)[/tex].
- For the interval [tex]\(50-60\)[/tex], the cumulative frequency is [tex]\(24 + 6 = 30\)[/tex].
Based on this, the cumulative frequency table is:
[tex]\[ \begin{array}{|l|c|} \hline \text{Weight (in kg)} & \text{Cumulative Frequency} \\ \hline 10-20 & 7 \\ \hline 20-30 & 16 \\ \hline 30-40 & 16 \\ \hline 40-50 & 24 \\ \hline 50-60 & 30 \\ \hline \end{array} \][/tex]
### (ii) If the median is 34 kg, find the median class
To find the median class, we first determine the median position using the formula:
[tex]\[ \text{Median position} = \frac{N + 1}{2} \][/tex]
where [tex]\(N\)[/tex] is the total number of students.
In our case, the total number of students is [tex]\(30\)[/tex]. Thus, the median position is:
[tex]\[ \text{Median position} = \frac{30 + 1}{2} = 15.5 \][/tex]
Looking at our cumulative frequency table:
- The cumulative frequency for the class [tex]\(10-20\)[/tex] is [tex]\(7\)[/tex].
- The cumulative frequency for the class [tex]\(20-30\)[/tex] is [tex]\(16\)[/tex].
Since the median position [tex]\(15.5\)[/tex] lies in the cumulative frequency [tex]\(16\)[/tex], the median class is [tex]\(20-30\)[/tex].
### (iii) Find the value of [tex]\(p\)[/tex]
Given that the median class is [tex]\(20-30\)[/tex] and the median value is [tex]\(34\)[/tex] kg, we use the formula for the median in a grouped frequency table:
[tex]\[ \text{Median} = L + \left( \frac{\frac{N + p}{2} - cf}{f + p} \right) \times h \][/tex]
Where:
- [tex]\(L\)[/tex] is the lower boundary of the median class ([tex]\(30\)[/tex])
- [tex]\(N\)[/tex] is the total number of students ([tex]\(30\)[/tex])
- [tex]\(cf\)[/tex] is the cumulative frequency of the class before the median class ([tex]\(16\)[/tex])
- [tex]\(f\)[/tex] is the frequency of the median class ([tex]\(0\)[/tex] initially)
- [tex]\(h\)[/tex] is the class width ([tex]\(10\)[/tex])
- [tex]\(p\)[/tex] is the frequency we need to find
Given:
[tex]\[ 34 = 30 + \left( \frac{\frac{30 + p}{2} - 16}{0 + p} \right) \times 10 \][/tex]
Solving this equation, the calculated value of [tex]\(p\)[/tex] is [tex]\(-44\)[/tex].
### (iv) Find the maximum number of students having weight below the median class
The maximum number of students having weight below the median class ([tex]\(30-40\)[/tex]) is the cumulative frequency of the class immediately before it, which is the cumulative frequency for [tex]\(10-20\)[/tex]:
Hence, the maximum number of students having weight below the median class is [tex]\(7\)[/tex].
Putting it all together:
[tex]\[ \begin{array}{|l|c|} \hline \text{i) Cumulative Frequency Table} & [7, 16, 16, 24, 30] \\ \hline \text{ii) Median Class} & 30-40 \\ \hline \text{iii) Value of } p & -44 \\ \hline \text{iv) Max students below the median class} & 7 \\ \hline \end{array} \][/tex]
### (i) Draw the cumulative frequency table
The cumulative frequency table is built by progressively adding the frequencies of the students. Here's how it is done for each class interval:
- For the interval [tex]\(10-20\)[/tex], the cumulative frequency is [tex]\(7\)[/tex].
- For the interval [tex]\(20-30\)[/tex], the cumulative frequency is [tex]\(7 + 9 = 16\)[/tex].
- For the interval [tex]\(30-40\)[/tex], the cumulative frequency is [tex]\(16 + 0 = 16\)[/tex], assuming [tex]\(p\)[/tex] is initially [tex]\(0\)[/tex].
- For the interval [tex]\(40-50\)[/tex], the cumulative frequency is [tex]\(16 + 8 = 24\)[/tex].
- For the interval [tex]\(50-60\)[/tex], the cumulative frequency is [tex]\(24 + 6 = 30\)[/tex].
Based on this, the cumulative frequency table is:
[tex]\[ \begin{array}{|l|c|} \hline \text{Weight (in kg)} & \text{Cumulative Frequency} \\ \hline 10-20 & 7 \\ \hline 20-30 & 16 \\ \hline 30-40 & 16 \\ \hline 40-50 & 24 \\ \hline 50-60 & 30 \\ \hline \end{array} \][/tex]
### (ii) If the median is 34 kg, find the median class
To find the median class, we first determine the median position using the formula:
[tex]\[ \text{Median position} = \frac{N + 1}{2} \][/tex]
where [tex]\(N\)[/tex] is the total number of students.
In our case, the total number of students is [tex]\(30\)[/tex]. Thus, the median position is:
[tex]\[ \text{Median position} = \frac{30 + 1}{2} = 15.5 \][/tex]
Looking at our cumulative frequency table:
- The cumulative frequency for the class [tex]\(10-20\)[/tex] is [tex]\(7\)[/tex].
- The cumulative frequency for the class [tex]\(20-30\)[/tex] is [tex]\(16\)[/tex].
Since the median position [tex]\(15.5\)[/tex] lies in the cumulative frequency [tex]\(16\)[/tex], the median class is [tex]\(20-30\)[/tex].
### (iii) Find the value of [tex]\(p\)[/tex]
Given that the median class is [tex]\(20-30\)[/tex] and the median value is [tex]\(34\)[/tex] kg, we use the formula for the median in a grouped frequency table:
[tex]\[ \text{Median} = L + \left( \frac{\frac{N + p}{2} - cf}{f + p} \right) \times h \][/tex]
Where:
- [tex]\(L\)[/tex] is the lower boundary of the median class ([tex]\(30\)[/tex])
- [tex]\(N\)[/tex] is the total number of students ([tex]\(30\)[/tex])
- [tex]\(cf\)[/tex] is the cumulative frequency of the class before the median class ([tex]\(16\)[/tex])
- [tex]\(f\)[/tex] is the frequency of the median class ([tex]\(0\)[/tex] initially)
- [tex]\(h\)[/tex] is the class width ([tex]\(10\)[/tex])
- [tex]\(p\)[/tex] is the frequency we need to find
Given:
[tex]\[ 34 = 30 + \left( \frac{\frac{30 + p}{2} - 16}{0 + p} \right) \times 10 \][/tex]
Solving this equation, the calculated value of [tex]\(p\)[/tex] is [tex]\(-44\)[/tex].
### (iv) Find the maximum number of students having weight below the median class
The maximum number of students having weight below the median class ([tex]\(30-40\)[/tex]) is the cumulative frequency of the class immediately before it, which is the cumulative frequency for [tex]\(10-20\)[/tex]:
Hence, the maximum number of students having weight below the median class is [tex]\(7\)[/tex].
Putting it all together:
[tex]\[ \begin{array}{|l|c|} \hline \text{i) Cumulative Frequency Table} & [7, 16, 16, 24, 30] \\ \hline \text{ii) Median Class} & 30-40 \\ \hline \text{iii) Value of } p & -44 \\ \hline \text{iv) Max students below the median class} & 7 \\ \hline \end{array} \][/tex]
