Question

Find the mean and the mode of the following frequency distribution :

Hint:

The step-deviation method is highly recommended for grouped data with large numbers or decimal class marks.
Choosing the assumed mean \( A \) at the center of the distribution makes the values of \( u_i \) small symmetric integers, drastically reducing potential math errors.

Answer 1

Step 1: Understanding the Question:
The topic of this question is Statistics (Measures of Central Tendency).
We are given a grouped frequency distribution table and need to find both its Arithmetic Mean and Mode.

Step 2: Key Formula or Approach:
- For Mean (Step-deviation method):
\[ \bar{x} = A + h \times \frac{\sum f_i u_i}{\sum f_i} \]
Where \( A \) is the assumed mean, \( h \) is the class size, \( u_i = \frac{x_i - A}{h} \), and \( x_i \) is the class mark.
- For Mode:
\[ \text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \]
Where \( l \) is the lower limit of the modal class, \( f_1 \) is the modal class frequency, \( f_0 \) is the preceding frequency, and \( f_2 \) is the succeeding frequency.

Step 3: Detailed Explanation:
1. Calculate the Mean:
Let the class height \( h = 15 \). Let the assumed mean \( A = 52.5 \) (midpoint of class 45-60).
Let us set up the computation table:
- Class 0-15: \( x_1 = 7.5 \), \( f_1 = 9 \), \( u_1 = -3 \), \( f_1 u_1 = -27 \)
- Class 15-30: \( x_2 = 22.5 \), \( f_2 = 15 \), \( u_2 = -2 \), \( f_2 u_2 = -30 \)
- Class 30-45: \( x_3 = 37.5 \), \( f_3 = 35 \), \( u_3 = -1 \), \( f_3 u_3 = -35 \)
- Class 45-60: \( x_4 = 52.5 \), \( f_4 = 20 \), \( u_4 = 0 \), \( f_4 u_4 = 0 \)
- Class 60-75: \( x_5 = 67.5 \), \( f_5 = 11 \), \( u_5 = 1 \), \( f_5 u_5 = 11 \)
- Class 75-90: \( x_6 = 82.5 \), \( f_6 = 13 \), \( u_6 = 2 \), \( f_6 u_6 = 26 \)
- Class 90-105: \( x_7 = 97.5 \), \( f_7 = 17 \), \( u_7 = 3 \), \( f_7 u_7 = 51 \)
Sum of frequencies \( \sum f_i \):
\[ \sum f_i = 9 + 15 + 35 + 20 + 11 + 13 + 17 = 120 \]
Sum of products \( \sum f_i u_i \):
\[ \sum f_i u_i = -27 - 30 - 35 + 0 + 11 + 26 + 51 = -92 + 88 = -4 \]
Using the step-deviation formula:
\[ \bar{x} = 52.5 + 15 \times \left(\frac{-4}{120}\right) \]
\[ \bar{x} = 52.5 - 15 \times \frac{1}{30} = 52.5 - 0.5 = 52.0 \]
2. Calculate the Mode:
The maximum frequency is 35, which corresponds to the class interval 30-45.
Therefore, the modal class is 30-45.
Identify parameters:
- Lower limit \( l = 30 \)
- Frequency of modal class \( f_1 = 35 \)
- Frequency of preceding class \( f_0 = 15 \)
- Frequency of succeeding class \( f_2 = 20 \)
- Class width \( h = 15 \)
Substitute these into the mode formula:
\[ \text{Mode} = 30 + \left(\frac{35 - 15}{2(35) - 15 - 20}\right) \times 15 \]
\[ \text{Mode} = 30 + \left(\frac{20}{70 - 35}\right) \times 15 \]
\[ \text{Mode} = 30 + \left(\frac{20}{35}\right) \times 15 \]
\[ \text{Mode} = 30 + \frac{4}{7} \times 15 = 30 + \frac{60}{7} \approx 30 + 8.57 = 38.57 \]

Step 4: Final Answer:
The Arithmetic Mean of the distribution is 52.0 and the Mode is 38.57.