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.

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(N/A) To construct the frequency distribution table,we subtract the cumulative frequency of the previous class from the current class to find the frequency of each class interval.
Class interval Number of students (Frequency)
$0-10$ $10$
$10-20$ $50 - 10 = 40$
$20-30$ $130 - 50 = 80$
$30-40$ $270 - 130 = 140$
$40-50$ $440 - 270 = 170$
$50-60$ $570 - 440 = 130$
$60-70$ $670 - 570 = 100$
$70-80$ $740 - 670 = 70$
$80-90$ $780 - 740 = 40$
$90-100$ $800 - 780 = 20$

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Similar Questions

Consider the following frequency distribution of the heights of $60$ students of a class:
Height (in $cm$) Number of students
$150-155$ $15$
$155-160$ $13$
$160-165$ $10$
$165-170$ $8$
$170-175$ $9$
$175-180$ $5$

The sum of the lower limit of the modal class and upper limit of the median class is:

$Z - M = \ldots \ldots \ldots \times (M - \bar{x})$

For the following distribution:
Class $0-5$ $5-10$ $10-15$ $15-20$ $20-25$
Frequency $10$ $15$ $12$ $20$ $9$

the sum of the lower limits of the median class and the modal class is:

If $Z+M=34$ and $M+\bar{x}=40,$ then $M=\ldots \ldots \ldots . . .$

If $M = 72$ and $\bar{x} = 70$,then $Z = \ldots \ldots \ldots \ldots$

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