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In a frequency distribution, the mid value of the class is 10 and width of the class is 6. Find the lower limit of the class. |
Answer» Consider the lower limit of the class as a So the upper limit can be written as a + 6 We get (a + (a + 6))/ 2 = 10 By cross multiplication 2a + 6 = 20 On further calculation 2a = 14 By division a = 7 Class width = 6Mid value = 10 cm Half of Class width = = 3 Lower limit of a class = Mid value - half of class width = 10 - 3 = 7 ---------------------------------------------------------------------------------------------------- If you were asked to find the upper limit of the class , then , Upper limit = Mid value + half of class width = 10 + 3 = 13 |
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