Question

What is the lower limit of the median class ?

Answer 1

Step 1: Problem Definition:
Determine the lower limit of the median class from the provided frequency distribution. The median class encompasses the median value of the dataset.

Step 2: Cumulative Frequency Calculation:
Calculate the cumulative frequency for each class interval by summing the frequencies of preceding classes and the current class.
Frequency Distribution:\[\begin{array}{|c|c|}\hline\text{Number of students per Teacher} & \text{Number of Schools} \\\hline20 - 25 & 5 \\25 - 30 & 15 \\30 - 35 & 25 \\35 - 40 & 30 \\40 - 45 & 15 \\45 - 50 & 10 \\\hline\end{array}\]Cumulative Frequencies:

• Class \(20 - 25\): CF = \(5\)
• Class \(25 - 30\): CF = \(5 + 15 = 20\)
• Class \(30 - 35\): CF = \(20 + 25 = 45\)
• Class \(35 - 40\): CF = \(45 + 30 = 75\)
• Class \(40 - 45\): CF = \(75 + 15 = 90\)
• Class \(45 - 50\): CF = \(90 + 10 = 100\)

Step 3: Median Class Identification:
Total schools = 100. The median is located at the \( \frac{100}{2} = 50 \)-th data point.
The median class is the interval where the cumulative frequency first exceeds 50.
Observation: CF for \(30 - 35\) is 45, and CF for \(35 - 40\) is 75. Therefore, the median class is \(35 - 40\).

Step 4: Final Answer:
The lower limit of the median class is 35.

Answer 2

Step 1: Problem Definition:
The objective is to determine the upper limit of the modal class from the provided frequency distribution. The modal class is defined as the class interval with the highest frequency.

Step 2: Modal Class Identification:
The initial step involves identifying the class interval exhibiting the highest frequency. The frequency distribution is as follows:
\[ \begin{array}{|c|c|} \hline \text{Number of students per Teacher} & \text{Number of Schools} \\ \hline 20 - 25 & 5 \\ 25 - 30 & 15 \\ 30 - 35 & 25 \\ 35 - 40 & 30 \\ 40 - 45 & 15 \\ 45 - 50 & 10 \\ \hline \end{array} \] Based on the table, the class \(35 - 40\) has the highest frequency, with 30 schools.
Therefore, the modal class is \(35 - 40\).

Step 3: Upper Limit Determination:
The upper limit of the identified modal class (\(35 - 40\)) is the upper value of this interval, which is 40.

Step 4: Final Result:
The upper limit of the modal class is 40.

Answer 3

Step 1: Problem Definition: The objective is to determine the median and modal values from the provided frequency distribution table. This will be addressed in two distinct phases: calculating the median and subsequently, the mode.

Step 2: Median Calculation: To find the median, we must first identify the median class. This is the class interval where the cumulative frequency surpasses the midpoint of the total number of observations. The median is represented by the \( \frac{N}{2} \)-th data point, where \( N \) signifies the total count of observations.

Given that the total number of schools is 100, the median is the \( \frac{100}{2} = 50 \)-th data point.
The cumulative frequencies have been previously computed as follows:
\[\begin{array}{|c|c|c|}\hline\text{Number of students per Teacher} & \text{Number of Schools} & \text{Cumulative Frequency} \\\hline20 - 25 & 5 & 5 \\25 - 30 & 15 & 20 \\30 - 35 & 25 & 45 \\35 - 40 & 30 & 75 \\40 - 45 & 15 & 90 \\45 - 50 & 10 & 100 \\\hline\end{array}\]
The median class is identified as \( 35 - 40 \) because its cumulative frequency of 75 is the first to exceed 50.

The median is calculated using the formula:
\[\text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h\]
Where:
- \( L \) is the lower boundary of the median class.
- \( N \) is the total frequency (100).
- \( F \) is the cumulative frequency of the class preceding the median class.
- \( f \) is the frequency of the median class.
- \( h \) is the width of the class interval.

The values to be substituted are:
- \( L = 35 \)
- \( N = 100 \)
- \( F = 45 \)
- \( f = 30 \)
- \( h = 5 \)

The median calculation proceeds as follows:
\[\text{Median} = 35 + \left( \frac{\frac{100}{2} - 45}{30} \right) \times 5 = 35 + \left( \frac{50 - 45}{30} \right) \times 5\]
\[\text{Median} = 35 + \left( \frac{5}{30} \right) \times 5 = 35 + \frac{25}{30} = 35 + \frac{5}{6} \approx 35.8333\]
Therefore, the median is approximately \( 35.83 \).

Step 3: Mode Calculation: The mode is determined by the class with the highest frequency. In the given frequency distribution table, the class \( 35 - 40 \) exhibits the highest frequency, with 30 schools.

The formula for the mode is:
\[\text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\]
Where:
- \( L \) is the lower boundary of the modal class.
- \( f_1 \) is the frequency of the modal class.
- \( f_0 \) is the frequency of the class preceding the modal class.
- \( f_2 \) is the frequency of the class succeeding the modal class.
- \( h \) is the width of the class interval.

The values for substitution are:
- \( L = 35 \)
- \( f_1 = 30 \)
- \( f_0 = 25 \)
- \( f_2 = 15 \)
- \( h = 5 \)

The mode is calculated as follows:
\[\text{Mode} = 35 + \left( \frac{30 - 25}{2 \times 30 - 25 - 15} \right) \times 5 = 35 + \left( \frac{5}{60 - 25 - 15} \right) \times 5\]
\[\text{Mode} = 35 + \left( \frac{5}{20} \right) \times 5 = 35 + \frac{25}{20} = 35 + 1.25 = 36.25\]
Consequently, the mode is \( 36.25 \).

Step 4: Summary of Findings:
1. The calculated median for the dataset is approximately \( 35.83 \).
2. The calculated mode for the dataset is \( 36.25 \).