Question

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

• A. Mean = 40.83, Mode = 38.75
• B. Mean = 42.50, Mode = 37.50
• C. Mean = 40.83, Mode = 35.50
• D. Mean = 39.50, Mode = 38.75

Hint:

Always ensure that the computed values of mean, median, and mode lie within the range of the overall data set.
The mode must specifically fall within the modal class interval (35--40).
Since \(38.75\) lies within \(35-40\), this acts as a quick confirmation of your calculation accuracy!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a grouped frequency distribution table and need to compute two measures of central tendency: the Mean and the Mode of the distribution.

Step 2: Key Formula or Approach:
1.

Mean formula (Direct method):
\[ \text{Mean } (\overline{x}) = \frac{\sum f_i x_i}{\sum f_i} \]
where \(x_i\) is the classmark (midpoint) of each class interval, calculated as \(\frac{\text{Lower Limit} + \text{Upper Limit}}{2}\), and \(f_i\) is the corresponding frequency.
2.

Mode formula:
\[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \]
where \(l\) is the lower limit of the modal class, \(f_1\) is the frequency of the modal class, \(f_0\) is the frequency of the preceding class, \(f_2\) is the frequency of the succeeding class, and \(h\) is the class width.

Step 3: Detailed Explanation:
1.

Calculate the Mean:
First, construct a calculation table to find classmarks (\(x_i\)) and product values (\(f_i x_i\)):
- For Class 30--35: \(x_1 = \frac{30+35}{2} = 32.5\), \(f_1 = 3 \implies f_1 x_1 = 3 \times 32.5 = 97.5\)
- For Class 35--40: \(x_2 = \frac{35+40}{2} = 37.5\), \(f_2 = 9 \implies f_2 x_2 = 9 \times 37.5 = 337.5\)
- For Class 40--45: \(x_3 = \frac{40+45}{2} = 42.5\), \(f_3 = 7 \implies f_3 x_3 = 7 \times 42.5 = 297.5\)
- For Class 45--50: \(x_4 = \frac{45+50}{2} = 47.5\), \(f_4 = 3 \implies f_4 x_4 = 3 \times 47.5 = 142.5\)
- For Class 50--55: \(x_5 = \frac{50+55}{2} = 52.5\), \(f_5 = 2 \implies f_5 x_5 = 2 \times 52.5 = 105.0\)
Now, sum the columns:
\[ \sum f_i = 3 + 9 + 7 + 3 + 2 = 24 \]
\[ \sum f_i x_i = 97.5 + 337.5 + 297.5 + 142.5 + 105.0 = 980.0 \]
Calculate the Mean:
\[ \overline{x} = \frac{980}{24} = 40.83 \]

2.

Calculate the Mode:
- Identify the modal class (the class with the highest frequency):
The highest frequency is 9, which corresponds to the class interval 35--40.
Therefore, the modal class is 35--40.
- Identify the parameters for the formula:
Lower limit of modal class, \(l = 35\)
Frequency of modal class, \(f_1 = 9\)
Frequency of preceding class, \(f_0 = 3\)
Frequency of succeeding class, \(f_2 = 7\)
Class width, \(h = 5\)
- Substitute these values into the Mode formula:
\[ \text{Mode} = 35 + \left( \frac{9 - 3}{2(9) - 3 - 7} \right) \times 5 \]
\[ \text{Mode} = 35 + \left( \frac{6}{18 - 10} \right) \times 5 \]
\[ \text{Mode} = 35 + \left( \frac{6}{8} \right) \times 5 \]
\[ \text{Mode} = 35 + 0.75 \times 5 = 35 + 3.75 = 38.75 \]

Step 4: Final Answer:
The mean of the distribution is 40.83 and the mode is 38.75, which corresponds to option (A).