Question

Write the median class of the data.

Answer 1

Step 1: Understanding the given data: 

We are given a frequency distribution table with the following details:
The class intervals represent the length of leaves in mm, the frequency indicates the number of leaves in each class, and the cumulative frequency (CF) represents the total number of leaves up to that class.

The table is as follows:

\(\text{Length (in mm)}\)	\(\text{Number of Leaves}\)	\(\text{Cumulative Frequency (CF)}\)
70 – 80	3	3
80 – 90	5	8
90 – 100	9	17
100 – 110	12	29
110 – 120	5	34
120 – 130	4	38
130 – 140	2	40

Step 2: Determining the median class:

The total number of leaves is given as 40. The median class is the class where the median value lies.
To find the median class, we first calculate the position of the median using the formula:
\[ \text{Median Position} = \frac{N}{2} \] where \( N \) is the total number of leaves. Here, \( N = 40 \), so the median position is:
\[ \frac{40}{2} = 20 \] This means the median value lies at the 20th position in the cumulative frequency table.

 

Step 3: Identifying the median class:

From the cumulative frequency column in the table, we can observe the following:
- The cumulative frequency for the class 70–80 is 3 (less than 20).
- The cumulative frequency for the class 80–90 is 8 (less than 20).
- The cumulative frequency for the class 90–100 is 17 (less than 20).
- The cumulative frequency for the class 100–110 is 29 (greater than 20).
Therefore, the 20th leaf lies in the class interval **100–110**.

 

Step 4: Conclusion:

Thus, the median class is 100–110.

Answer 2

Step 1: Understanding the given data:

We are asked to find the number of leaves with a length equal to or more than 10 cm (100 mm).
The relevant classes are those where the length is greater than or equal to 100 mm.

Step 2: Identifying the relevant classes:

From the frequency distribution table, the classes with leaves of length 100 mm or more are:

• 100–110: 12 leaves
• 110–120: 5 leaves
• 120–130: 4 leaves
• 130–140: 2 leaves

Step 3: Calculating the total number of leaves:

To find the total number of leaves with a length equal to or more than 100 mm, we add the frequencies of the relevant classes:
\[ 12 + 5 + 4 + 2 = 23 \quad \text{leaves of length} \geq 10 \, \text{cm} \, (\text{100 mm}) \]

Step 4: Conclusion:

Thus, there are 23 leaves of length equal to or more than 10 cm (100 mm).

Answer 3

(a) To find the median, use the formula for grouped data:

\[ \text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h \]

Where:

• \( L = 100 \) (lower boundary of the median class)
• \( N = 40 \) (total number of leaves)
• \( F = 17 \) (cumulative frequency before median class)
• \( f = 12 \) (frequency of the median class)
• \( h = 10 \) (class width)

\[ \text{Median} = 100 + \left( \frac{20 - 17}{12} \right) \times 10 = 100 + 2.5 = 102.5 \, \text{mm} \]

Thus, the Median is \( 102.5 \, \text{mm} \).

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(b) The modal class is the class with the highest frequency, which is 100-110 with 12 leaves. To find the mode, use the formula:

\[ \text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \]

Where:

• \( L = 100 \) (lower boundary of the modal class)
• \( f_1 = 12 \) (frequency of the modal class)
• \( f_0 = 9 \) (frequency of the class before the modal class)
• \( f_2 = 5 \) (frequency of the class after the modal class)
• \( h = 10 \) (class width)

\[ \text{Mode} = 100 + \left( \frac{12 - 9}{2 \times 12 - 9 - 5} \right) \times 10 = 100 + \left( \frac{3}{10} \right) \times 10 = 100 + 3 = 103 \, \text{mm} \]

Thus, the Mode is \( 103 \, \text{mm} \).