1. Construct a grouped frequency distribution table from the following data
`77,41,85,82,85,96,93,66,78,94,50,57`

Solution:
`77,41,85,82,85,96,93,66,78,94,50,57`

Here minimum value is 41 and maximum value is 96

Class	Class Boundaries	Tally marks	Frequency
41 - 50
51 - 60
61 - 70
71 - 80
81 - 90
91 - 100
Total

Class boundaries : subtracting 0.5 from lower class limit and adding 0.5 to upper class limit

Class	Class Boundaries	Tally marks	Frequency
41 - 50	40.5 - 50.5
51 - 60	50.5 - 60.5
61 - 70	60.5 - 70.5
71 - 80	70.5 - 80.5
81 - 90	80.5 - 90.5
91 - 100	90.5 - 100.5
Total

For each value draw a tally mark, next to the class in which this value exists.
`77,41,85,82,85,96,93,66,78,94,50,57`

Class	Class Boundaries	Tally marks	Frequency
41 - 50	40.5 - 50.5	||
51 - 60	50.5 - 60.5	|
61 - 70	60.5 - 70.5	|
71 - 80	70.5 - 80.5	||
81 - 90	80.5 - 90.5	|||
91 - 100	90.5 - 100.5	|||
Total

Count tally marks to determine the total frequency of each class.

Class	Class Boundaries	Tally marks	Frequency
41 - 50	40.5 - 50.5	||	2
51 - 60	50.5 - 60.5	|	1
61 - 70	60.5 - 70.5	|	1
71 - 80	70.5 - 80.5	||	2
81 - 90	80.5 - 90.5	|||	3
91 - 100	90.5 - 100.5	|||	3
Total			`n=12`

The relative frequency

Class	Class Boundaries	Frequency `f`	Relative Frequency `f/n`
41 - 50	40.5 - 50.5	2	`2/12=0.1667=16.67%`
51 - 60	50.5 - 60.5	1	`1/12=0.0833=8.33%`
61 - 70	60.5 - 70.5	1	`1/12=0.0833=8.33%`
71 - 80	70.5 - 80.5	2	`2/12=0.1667=16.67%`
81 - 90	80.5 - 90.5	3	`3/12=0.25=25%`
91 - 100	90.5 - 100.5	3	`3/12=0.25=25%`
Total		`n=12`

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