Directions: Consider the following data and answer questions: S. No. Class limit Frequency 1. 101-200 4 2. 201-300 12 3. 301-400 24 4. 401-500 40 5. 501-600 16 6. 601-700 12 7. 701-800 10 8. 801-900 5 9. 901-1000 2
Which one of the following is the arithmetic mean value for the given data set
483
The question asks us to find the arithmetic mean value for the given data set, which is presented in the form of a frequency distribution table with class limits. When data is grouped into classes, we cannot find the exact mean because we don't know the individual data points within each class. Instead, we estimate the mean by assuming that the midpoint of each class represents all the data points within that class.
The formula for calculating the arithmetic mean (\(\bar{x}\)) for grouped data is:
\[ \bar{x} = \frac{\sum (f_i \times m_i)}{\sum f_i} \]
Where:
First, let's add the class midpoints and the product of frequency and midpoint (\(f_i \times m_i\)) to the given frequency distribution table.
The midpoint (\(m_i\)) of a class is calculated as: \(\frac{\text{Lower Limit} + \text{Upper Limit}}{2}\).
| S. No. | Class Limit | Frequency (\(f_i\)) | Class Midpoint (\(m_i\)) | \(f_i \times m_i\) |
|---|---|---|---|---|
| 1 | 101-200 | 4 | \(\frac{101+200}{2} = 150.5\) | \(4 \times 150.5 = 602.0\) |
| 2 | 201-300 | 12 | \(\frac{201+300}{2} = 250.5\) | \(12 \times 250.5 = 3006.0\) |
| 3 | 301-400 | 24 | \(\frac{301+400}{2} = 350.5\) | \(24 \times 350.5 = 8412.0\) |
| 4 | 401-500 | 40 | \(\frac{401+500}{2} = 450.5\) | \(40 \times 450.5 = 18020.0\) |
| 5 | 501-600 | 16 | \(\frac{501+600}{2} = 550.5\) | \(16 \times 550.5 = 8808.0\) |
| 6 | 601-700 | 12 | \(\frac{601+700}{2} = 650.5\) | \(12 \times 650.5 = 7806.0\) |
| 7 | 701-800 | 10 | \(\frac{701+800}{2} = 750.5\) | \(10 \times 750.5 = 7505.0\) |
| 8 | 801-900 | 5 | \(\frac{801+900}{2} = 850.5\) | \(5 \times 850.5 = 4252.5\) |
| 9 | 901-1000 | 2 | \(\frac{901+1000}{2} = 950.5\) | \(2 \times 950.5 = 1901.0\) |
Next, we calculate the sum of frequencies (\(\sum f_i\)) and the sum of \(f_i \times m_i\) (\(\sum (f_i \times m_i)\)).
Sum of frequencies (\(\sum f_i\)):
\( \sum f_i = 4 + 12 + 24 + 40 + 16 + 12 + 10 + 5 + 2 = 125 \)
Sum of \(f_i \times m_i\) (\(\sum (f_i \times m_i)\)):
\( \sum (f_i \times m_i) = 602.0 + 3006.0 + 8412.0 + 18020.0 + 8808.0 + 7806.0 + 7505.0 + 4252.5 + 1901.0 = 60312.5 \)
Now, we can calculate the arithmetic mean using the formula:
\[ \bar{x} = \frac{\sum (f_i \times m_i)}{\sum f_i} = \frac{60312.5}{125} \]
Performing the division:
\[ \bar{x} = 482.5 \]
The calculated arithmetic mean is 482.5. Comparing this value to the given options, 483 is the closest value.
Based on the calculations for the grouped data set, the arithmetic mean is approximately 482.5. The closest value among the options provided is 483.
| Concept | Description | Formula/Method |
|---|---|---|
| Arithmetic Mean (\(\bar{x}\)) | A measure of central tendency; the average value. | Sum of values divided by count (for ungrouped data). |
| Grouped Data | Data organized into classes or intervals with frequencies. | |
| Class Limit | The upper and lower boundaries of a data class. | |
| Frequency (\(f_i\)) | The number of data points falling into a specific class. | |
| Class Midpoint (\(m_i\)) | The middle value of a class, used to represent all values in the class for mean calculation. | \(\frac{\text{Lower Limit} + \text{Upper Limit}}{2}\) |
| Arithmetic Mean for Grouped Data | Estimated mean for data in frequency distribution. | \(\frac{\sum (f_i \times m_i)}{\sum f_i}\) |
The arithmetic mean is one of several measures of central tendency, which are values that describe the center of a data set. Other common measures include the median and the mode.
Each measure of central tendency provides a different perspective on the typical value of a data set and their suitability depends on the nature of the data and the purpose of the analysis. The arithmetic mean is sensitive to extreme values, while the median is not.
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