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Example 2: Class marks of a distribution are 37, 42, 47, 52, and 57. Determine the class intervals.

Solution: We know that the class size is the difference between the upper and the lower limits of
a class or the difference between the class marks of the successive class intervals.
Here, we have class marks 37, 42, 47, 52, and 57.
So, the class size = 42 – 37 = 5

Now, we can get the upper limit and the lower limit by adding
and subtracting half of the class size to the class marks, Quick Check
respectively. Find the range of the given
data: 42, 38, 56, 47, 63, 41, 54,
Lower limit = Class mark – Class size = 37 − 5 = 34 5 48, 60, 39, 57, 50, 44, 62, 49, 45,
.
2 2 55, 52, 43, 61.
and Upper limit = Class mark + Class size = 37 + 5 = 39 5 What will be the class mark
.
for the third class interval with
2 2 class size 10 if you group the
Thus, the first class interval = 34.5-39.5 above data?

 5  5
- 42 +  = 39.5-44.5
Similarly, the second class interval = 42 −   2 

 

2
Further, the third class interval = 44.5-49.5, the fourth class interval = 49.5 - 54.5, and the fifth class
interval = 54.5-59.5.
Example 3: The daily wages (in `) of 40 workers in a factory are as follows:
830, 835, 980, 810, 835, 836, 869, 845, 898, 890, 820, 860, 932, 833, 855, 845, 904, 958, 812, 840, 985,
835, 935, 836, 978, 840, 868, 890, 806, 840, 915, 895, 935, 879, 987, 912, 924, 864, 975, 888.
Using tally marks make a grouped frequency distribution table with class intervals 800-825, 825-
850 and so on.
( a) Find the class mark of the class that has the maximum frequency.

( b) How many workers have their daily wages in the highest class interval?
( c) In which class do the least number of workers have their daily wages?

( d) How many workers have their daily wages of `900 or more?

Solution: Class interval Tally marks Frequency

(Wages in `) (Number of workers)
800-825 |||| 4
825-850 |||| |||| || 12

850-875 |||| 5
875-900 |||| | 6

900-925 |||| 4
925-950 ||| 3
950-975 | 1

975-1000 |||| 5
Total 40

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