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To calculate mean of a grouped data, Rahul used assumed mean method. He used d = (x - A), where A is assumed mean. Then $\bar{x}$ is equal to
To calculate mean of a grouped data, Rahul used assumed mean method. He used d = (x - A), where A is assumed mean. Then $\bar{x}$ is equal to
a) $A+ \bar{d}$
b) $A+ h \bar{d}$
c) $h(A+ \bar{d})$
d) $A - h \bar{d}$
This Question has 1 answers.
In the assumed mean method, the mean $ \displaystyle \bar{x} $ is given by:
$ \displaystyle \bar{x} = A + \frac{\sum fd}{\sum f} $
where:
- $ \displaystyle A $ is the assumed mean,
- $ \displaystyle d = x - A $ (deviation of class mark $ \displaystyle x $ from assumed mean),
- $ \displaystyle f $ is the frequency of the class,
- $ \displaystyle \sum fd $ is the summation of the product of frequency and deviation,
- $ \displaystyle \sum f $ is the total frequency.
Since $ \displaystyle \bar{d} = \frac{\sum fd}{\sum f} $, we can rewrite the formula as:
$ \displaystyle \bar{x} = A + \bar{d} $
If the class interval has a common width $ \displaystyle h $, the deviation is taken as $ \displaystyle d = \frac{x - A}{h} $. In such cases, the mean formula becomes:
$ \displaystyle \bar{x} = A + h \bar{d} $
Thus, the correct answer is:
**(b) $ \displaystyle A + h \bar{d} $**
$ \displaystyle \bar{x} = A + \frac{\sum fd}{\sum f} $
where:
- $ \displaystyle A $ is the assumed mean,
- $ \displaystyle d = x - A $ (deviation of class mark $ \displaystyle x $ from assumed mean),
- $ \displaystyle f $ is the frequency of the class,
- $ \displaystyle \sum fd $ is the summation of the product of frequency and deviation,
- $ \displaystyle \sum f $ is the total frequency.
Since $ \displaystyle \bar{d} = \frac{\sum fd}{\sum f} $, we can rewrite the formula as:
$ \displaystyle \bar{x} = A + \bar{d} $
If the class interval has a common width $ \displaystyle h $, the deviation is taken as $ \displaystyle d = \frac{x - A}{h} $. In such cases, the mean formula becomes:
$ \displaystyle \bar{x} = A + h \bar{d} $
Thus, the correct answer is:
**(b) $ \displaystyle A + h \bar{d} $**
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