Question

The monthly wages of 30 workers in a factory are given below: 

832, 860, 820, 890, 898, 845, 869, 836, 835, 810, 890, 835, 830, 833, 855, 845, 804, 808, 812, 840, 885, 835, 836, 878, 840, 868, 890, 840, 806, 890.

Represent the data in the form of a frequency distribution with class interval 10.

Answer 1

Given:

Monthly wages of 30 workers: 832, 860, 820, 890, 898, 845, 869, 836, 835, 810, 890, 835, 830, 833, 855, 845, 804, 808, 812, 840, 885, 835, 836, 878, 840, 868, 890, 840, 806, 890.

Step-wise calculation:

1. Sort/inspect data (ascending):

804, 806, 808, 810, 812, 820, 830, 832, 833, 835, 835, 835, 836, 836, 840, 840, 840, 845, 845, 855, 860, 868, 869, 878, 885, 890, 890, 890, 890, 898.

2. Find bounds and range:

Minimum = 804

Maximum = 898

Range = 898 − 804 = 94

3. Class interval (given) = 10. Choose class limits covering the data (classes written as [lower, upper), i.e., upper limit not included):

800 – 810, 810 – 820, 820 – 830, 830 – 840, 840 – 850, 850 – 860, 860 – 870, 870 – 880, 880 – 890, 890 – 900.

4. Tally frequencies by class (count observations falling in each class, including lower boundary, excluding the upper):

800 – 810: 804, 806, 808 → 3

810 – 820: 810, 812 → 2

820 – 830: 820 → 1

830 – 840: 830, 832, 833, 835, 835, 835, 836, 836 → 8

840 – 850: 840, 840, 840, 845, 845 → 5

850 – 860: 855 → 1

860 – 870: 860, 868, 869 → 3

870 – 880: 878 → 1

880 – 890: 885 → 1

890 – 900: 890, 890, 890, 890, 898 → 5

5. Check: Total frequency = 3 + 2 + 1 + 8 + 5 + 1 + 3 + 1 + 1 + 5 = 30 (matches sample size).

Frequency distribution with class interval 10:

Class	Frequency
800 – 810	3
810 – 820	2
820 – 830	1
830 – 840	8
840 – 850	5
850 – 860	1
860 – 870	3
870 – 880	1
880 – 890	1
890 – 900	5