Question

Write the median class of the data.

Answer 1

Median Class

The cumulative frequency is calculated 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

The total number of leaves is 40, so the median will lie at the 20th position. From the cumulative frequency, the 20th leaf lies in the class 100–110, so the median class is 100–110.

Answer 2

Step 1: Understand the given task:
We are asked to determine how many leaves have a length equal to or more than 10 cm (100 mm), based on the information provided.

Step 2: Analyze the data:
The relevant classes given in the problem are:
- 100–110 mm: 12 leaves
- 110–120 mm: 5 leaves
- 120–130 mm: 4 leaves
- 130–140 mm: 2 leaves

These are the leaves that have a length of 100 mm or more.

Step 3: Count the leaves:
We simply need to add up the number of leaves in these classes:
- 12 leaves in the 100–110 mm range
- 5 leaves in the 110–120 mm range
- 4 leaves in the 120–130 mm range
- 2 leaves in the 130–140 mm range
Total number of leaves with length ≥ 10 cm (100 mm) is:
\[ 12 + 5 + 4 + 2 = 23 \]

Step 4: Conclusion:
The total number of leaves with a length equal to or more than 10 cm is \( \boxed{23} \).

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 \( 103 \, \text{mm} \).