The mean of ungrouped data and the mean calculated when the same data is grouped are always the same. Do you agree with this statement? Give reason for your answer.
(B) The statement is not true.
When we calculate the mean of grouped data,we assume that the frequency of each class is centered at the mid-point of that class.
In reality,the individual observations within a class are rarely distributed exactly at the mid-point.
Therefore,the mean calculated from grouped data is an approximation,and it is rarely the same as the mean calculated from the original ungrouped data.
When we calculate the mean of grouped data,we assume that the frequency of each class is centered at the mid-point of that class.
In reality,the individual observations within a class are rarely distributed exactly at the mid-point.
Therefore,the mean calculated from grouped data is an approximation,and it is rarely the same as the mean calculated from the original ungrouped data.
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Similar Questions
In the following distribution:
Monthly income range (in $Rs.$) Number of families Income more than $10000$ $100$ Income more than $13000$ $85$ Income more than $16000$ $69$ Income more than $19000$ $50$ Income more than $22000$ $33$ Income more than $25000$ $15$
The number of families having an income range (in $Rs.$) of $16000-19000$ is:
| Monthly income range (in $Rs.$) | Number of families |
| Income more than $10000$ | $100$ |
| Income more than $13000$ | $85$ |
| Income more than $16000$ | $69$ |
| Income more than $19000$ | $50$ |
| Income more than $22000$ | $33$ |
| Income more than $25000$ | $15$ |
The number of families having an income range (in $Rs.$) of $16000-19000$ is:
Easy
View SolutionIn the formula $M = l + \frac{(\frac{n}{2} - cf)}{f} \times h$ for the median,$h = \ldots \ldots \ldots$
Easy
View SolutionIn calculating the mean of grouped data,grouped in classes of equal width,we may use the formula $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$,where $a$ is the assumed mean. $a$ must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.
Easy
View SolutionIn the formula $\bar{x} = A + \frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}$,$d_{i} =$ .........
Easy
View SolutionThe median of the following frequency distribution is $50$ and the total frequency is $60$. Find the missing frequencies $x$ and $y$.
Class $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ Frequency $5$ $x$ $20$ $y$ $2$
| Class | $0-20$ | $20-40$ | $40-60$ | $60-80$ | $80-100$ |
| Frequency | $5$ | $x$ | $20$ | $y$ | $2$ |
Medium
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