Question

Write the median class of the data.

Answer 1

Step 1: Data Analysis:

The provided data is a frequency distribution of leaf lengths in millimeters (mm). The table details the class intervals for leaf length, the number of leaves within each interval (frequency), and the cumulative frequency (CF) indicating the total number of leaves up to and including each class.

\(\text{Length (in mm)}\)	\(\text{Number of Leaves}\)	\(\text{Cumulative Frequency (CF)}\)
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: Median Class Calculation:

The total number of leaves (N) is 40. The median class is identified by determining where the median value falls within the cumulative frequency distribution. The position of the median is calculated as:

\[ \text{Median Position} = \frac{N}{2} \]

Substituting N = 40:

\[ \frac{40}{2} = 20 \]

The median is located at the 20th position.

Step 3: Median Class Identification:

By examining the cumulative frequency column:

• CF for 70–80 is 3 (<20)
• CF for 80–90 is 8 (<20)
• CF for 90–100 is 17 (<20)
• CF for 100–110 is 29 (>20)

The 20th leaf falls within the class interval where the cumulative frequency first exceeds 20, which is 100–110.

Step 4: Final Determination:

The median class is determined to be 100–110.

Answer 2

Step 1: Data Analysis Objective:

Determine the count of leaves measuring 10 cm (100 mm) or greater in length.
Focus on data where leaf length is $\geq$ 100 mm.

Step 2: Identification of Relevant Data Categories:

Based on the provided frequency distribution, the leaf length categories meeting the criteria are:

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

Step 3: Calculation of Total Leaf Count:

Sum the frequencies of the identified categories to obtain the total number of leaves with a length of 100 mm or more:
\[12 + 5 + 4 + 2 = 23 \quad \text{leaves of length} \geq 10 \, \text{cm} \, (\text{100 mm})\]

Step 4: Final Result:

Consequently, there are 23 leaves with a length of 10 cm (100 mm) or greater.

Answer 3

(a) The median is calculated using 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 preceding the median class)
• \( f = 12 \) (frequency of the median class)
• \( h = 10 \) (class interval width)

The calculation is:

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

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

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(b) The modal class is identified as the class with the highest frequency. This is the 100-110 class, with 12 leaves. The mode is calculated using 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 preceding the modal class)
• \( f_2 = 5 \) (frequency of the class succeeding the modal class)
• \( h = 10 \) (class interval width)

The calculation is:

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

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