In 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.

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(N/A) No,the statement is incorrect. While it is common practice to choose the assumed mean $a$ as one of the class marks (mid-points) to simplify calculations,it is not a mathematical requirement. The assumed mean $a$ can be any arbitrary value chosen from the data set or even outside it,provided it helps in simplifying the calculation of deviations $d_i = x_i - a$.

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Similar Questions

If $\bar{x} = Z - 3$ and $M = 22$,then the mean is ..........

The median class of the following frequency distribution is ...........
Class$0-10$$10-20$$20-30$$30-40$$40-50$$50-60$
Frequency$4$$5$$9$$10$$14$$8$

Total frequency $n = 50$.

The following table shows the cumulative frequency distribution of marks of $800$ students in an examination:
Marks Number of students
Below $10$ $10$
Below $20$ $50$
Below $30$ $130$
Below $40$ $270$
Below $50$ $440$
Below $60$ $570$
Below $70$ $670$
Below $80$ $740$
Below $90$ $780$
Below $100$ $800$

Construct a frequency distribution table for the data above.

For a given frequency distribution,$\Sigma f_{i} u_{i} = -13$,$n = \Sigma f_{i} = 100$,$A = 62.5$ and $c = 15$. Then,mean $\bar{x} = \dots$

In the formula $\bar{x} = A + \frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}} \times c$ for the mean,$u_{i} = \ldots \ldots \ldots$

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