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In the previous topic, we have learnt how to find the mean using the direct method. Consider if the data is very large and finding the products of the observations and then adding them becomes tedious and also it may result in errors. To avoid such complications, let us use the assumed mean method to find the mean of grouped frequency.
Steps:
1. Calculate the midpoint of the class interval and name it as \(x\).
2. From the data of \(x\), choose any value(preferably in the middle) as the assumed mean(\(A\)).
3. Determine the deviation \(d=x-A\) for each of the class.
4. Multiply the deviation and frequency of each class interval and name it as \(fd\).
5. Find the values \(\sum fd\) and \(\sum f\).
6. Calculate the mean by applying the formula \(\overline X = A + \frac{\sum fd}{\sum f}\)
Example:
Find the mean of the following frequency distribution:
| Class interval | \(10 - 20\) | \(20 - 30\) | \(30 - 40\) | \(40 - 50\) | \(50 - 60\) | \(60 - 70\) | \(70 - 80\) |
| Frequency | \(23\) | \(15\) | \(10\) | \(28\) | \(5\) | \(7\) | \(11\) |
Solution:
Let the assumed mean be \(A = 45\).
| Class interval | Frequency (\(f\)) | Midpoint (\(x\)) | Deviation \(d = x - A\) | \(fd\) |
| \(10 - 20\) | \(23\) | \(15\) | \(-30\) | \(-690\) |
| \(20 - 30\) | \(15\) | \(25\) | \(-20\) | \(-300\) |
| \(30 - 40\) | \(10\) | \(35\) | \(-10\) | \(-100\) |
| \(40 - 50\) | \(28\) | \(45\) | \(0\) | \(0\) |
| \(50 - 60\) | \(5\) | \(55\) | \(50\) | \(250\) |
| \(60 - 70\) | \(8\) | \(65\) | \(20\) | \(160\) |
| \(70 - 80\) | \(11\) | \(75\) | \(30\) | \(330\) |
| Total | \(\sum f = 100\) | \(\sum fd = -350\) |
We know that the mean of grouped data using the assumed mean method can be determined using the formula \(\overline X = A + \frac{\sum fd}{\sum f}\)
Substituting the known values in the above formula, we get:
\(\overline X = 45 + (\frac{-350}{100})\)
\(\overline X = 45 - 3.5\)
\(\overline X = 41.5\)
Therefore, the mean of the given data is \(41.5\).
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