Question

Find mean and mode of the following data :

Hint:

The mode must always lie within the range of the modal class (40-50).
Our calculated mode is 47.5, which is inside the interval, giving us immediate confidence in our result!

Answer 1

Correct Answer: 47.5

Step 1: Understanding the Question:
This is a grouped data statistics question.
We are given a grouped frequency table and need to calculate both the Mean and the Mode of the distribution.

Step 2: Key Formula or Approach:
1. Mean (\(\bar{x}\)): Calculated using the direct method: \[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \] where \(x_i\) is the class mark of each interval: \[ x_i = \frac{\text{Upper Limit} + \text{Lower Limit}}{2} \] 2. Mode: Calculated using the 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.
- \(h\) is the class size.

Step 3: Detailed Explanation:
Let us create the calculation table for the Mean:

1. Calculate the Mean: \[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2970}{60} = 49.5 \] 2. Calculate the Mode: Identify the modal class, which is the class with the highest frequency (\(13\)).
The modal class is 40-50.
Write down the parameters: - Lower limit of modal class, \(L = 40\)
- Frequency of modal class, \(f_1 = 13\)
- Frequency of preceding class, \(f_0 = 10\)
- Frequency of succeeding class, \(f_2 = 12\)
- Class size, \(h = 10\)
Substitute these parameters into the mode formula: \[ \text{Mode} = 40 + \left( \frac{13 - 10}{2(13) - 10 - 12} \right) \times 10 \] \[ \text{Mode} = 40 + \left( \frac{3}{26 - 22} \right) \times 10 \] \[ \text{Mode} = 40 + \left( \frac{3}{4} \right) \times 10 \] \[ \text{Mode} = 40 + 7.5 = 47.5 \]

Step 4: Final Answer:
The calculated Mean of the data is 49.5, and the Mode is 47.5.