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Presentation of data — Statistics STD 11 Commerce — Question

Gujarat BoardEnglish MediumSTD 11 CommerceStatisticsPresentation of data3 Marks
Question
From the following data, obtain class boundary points from the class limits and write the frequency distribution :
Class 1 – 1.475 1.5 – 1.975 2 – 2.475 2.5 – 2.975 3 – 3.475 3.5 – 3.975 Total
Frequency 5 10 20 20 10 5 70
✓

Answer

  • In the given data. the difference between the upper limit of a class and the lower limit of its immediate next class is 0.025. e.g. upper limit of 1 - 1.475 is 1.475 and lower limit of 1.5 - 1.975 is 1.5.
difference = 1.5 - 1.475 = 0.025.
  • Subtracting = 0.0125 from the lower limit of the class and adding it to the upper limit, we get lower
  • boundary point and upper
  • boundary point respectively of the classes.
  • Thus, the frequency distribution according class boundary points is obtained as follows :
Class Boundary points Class Frequency
f
Lower boundary point Upper boundary point
1 – 1.475 1 – 0.0125 = 0.9875 1.475 + 0.0125 = 1.4875 0.9875 – 1.4875 5
1.5 – 1.975 1.5 -0.0125 = 1.4875 1.975 + 0.0125 = 1.9875 1.4875 – 1.9875 10
2-2.475 2-0.0125 = 1.9875 2.475 + 0.0125 = 2.4875 1.9875 – 2.4875 20
2.5 – 2.975 2.5-0.0125 = 2.4875 2.975 + 0.0125 = 2.9875 2.4875 – 2.9875 20
3 – 3.475 3-0.0125 = 2.9875 3.475 + 0.0125 = 3.4875 2.9875 – 3.4875 10
3.5 - 3.975 3.5-0.0125 = 3.4875 3.975 + 0.0125 = 3.9875 3.4875 – 3.9875 5
- - - Total n = 70

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