Question
Rewrite the following frequency distribution by stating class length and Mid Value of each class :
Class 0-99 100-299 300-499 500-749 750-899 900-999
Frequency 10 12 14 16 8 10

Answer

  • Given frequency distribution is inclusive type. Here. the difference between the upper limit of each class and the lower limit of its immediate adjoining class is 1.
  • Subtracting = 0.5 from the lower limit of each class and adding it to the upper limit of each class. we yet lower boundary point and upper boundary point respectively for each class, which is shown in the table on page 28.
Class Class length = (Upper class boundary point – lower class boundary point ) Mid value = Frequency
-0.5-99.5 (99.5 – (-0.5) = 100 = 49.5 10
99.5 -299.5 (299.5 – 99.5) = 200 = 199.5 12
299.5 -499.5 (499.5 – 299.5) = 200 = 399.5 14
499.5 -749.5 (749.5 – 499.5 ) = 250 = 624.5 16
749.5 – 899.5 (899.5 – 749.5) = 150 = 824.5 8
899.5 – 999.5 (999. 5 – 899.5) = 100 = 949.5 10

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